Using the logistic model f(x)=1501+9e−2x, evaluate the function at f(4). Round your answer to the nearest tenth.

Answer: It's attached.

Step-by-step explanation:

The table is attached.

The ratio is:

[tex]ratio=\frac{24}{8}\\\\ratio=3[/tex]

Knowing tha ratio, you can complete the table.

The steps are:

1. Multiply the number of panes given in the table by the ratio find above, in order to find the number of windows.

3. Divide the number of windows given in the table by the ratio find above, in order to find the number of panes.

Given [tex]Panes=3[/tex]:

[tex]Windows=3*3=9[/tex]

 Given [tex]Windows=3[/tex]:

[tex]Panes=\frac{3}{3}=1[/tex]

Given [tex]Windows=5[/tex]:

[tex]Panes=\frac{5}{3}[/tex]

Given [tex]Panes=18[/tex]:

[tex]Windows=18*3=54[/tex]

Answer:windows x6

Step-by-step explanation:

Answer:

effective memory access = 658 ns

Step-by-step explanation:

GIven data:

Effective memory access time is given as

[tex] = [H_1*T_1]+[(1-H_1)*H_2*T_2]+[(1-H_1)(1-H_2)*H_m*T_m][/tex]

from the data given above we have

[tex]H_1 = 0.1[/tex]

[tex]H_2 = 0.3[/tex]

[tex]T_1 = 10 ns[/tex]

[tex]T_2 = 100 ns[/tex]

hit rate, [tex]H_m = 1 ns[/tex]

access time [tex]= T_m = 1000 ns[/tex]

Plugging all information in above formula to get the effective memory access

[tex]= 0.1\times 10 + 0.9\times 100+ 0.9 \times 0.7\times 1 \times 1000[/tex]

= 1+27+ 630

=658 ns

Answer:

Percentage score will be 83.33 %

Step-by-step explanation:

We have given Antonette gets 70 % on 10 problem test

Let consider here here total problem = total marks

So marks get 10 10 problem test = 10×0.7 = 7

Marks get in 20 problem test = 20×0.8 = 16

And marks get in 30 problem test = 30×0.9 = 27

Now total marks get get by Antonette = 7 +16 + 27 = 50

And total marks = 60

So percentage score of Antonette [tex]=\frac{50}{60}\times 100=83.33[/tex] %

Answer:

a.

With n = 25, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549[/tex]

With n = 50, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594[/tex]

b. [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943[/tex]

c.

The absolute error in the trapezoid rule is 0.08047

The absolute error in the Simpson's rule is 0.00008

Step-by-step explanation:

a. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using n = 25 with the trapezoid rule you must:

The trapezoidal rule states that

[tex]\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)[/tex]

where [tex]\Delta{x}=\frac{b-a}{n}[/tex]

We have that a = 0, b = 1, n = 25.

Therefore,

[tex]\Delta{x}=\frac{1-0}{25}=\frac{1}{25}[/tex]

We need to divide the interval [0,1] into n = 25 sub-intervals of length [tex]\Delta{x}=\frac{1}{25}[/tex], with the following endpoints:

[tex]a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b[/tex]

Now, we just evaluate the function at these endpoints:

[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]

[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]

[tex]2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579[/tex]

...

[tex]2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701[/tex]

[tex]f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]

Applying the trapezoid rule formula we get

[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549[/tex]

We have that a = 0, b = 1, n = 50.

Therefore,

[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]

We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:

[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]

Now, we just evaluate the function at these endpoints:

[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]

[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875[/tex]

[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]

...

[tex]2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705[/tex]

[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]

Applying the trapezoid rule formula we get

[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594[/tex]

b. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using 2n with the Simpson's rule you must:

The Simpson's rule states that

[tex]\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)[/tex]

where [tex]\Delta{x}=\frac{b-a}{n}[/tex]

We have that a = 0, b = 1, n = 50

Therefore,

[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]

We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:

[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]

Now, we just evaluate the function at these endpoints:

[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]

[tex]4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175[/tex]

[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]

...

[tex]4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541[/tex]

[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]

Applying the Simpson's rule formula we get

[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943[/tex]

c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by [tex]|A-B|[/tex]

The absolute error in the trapezoid rule is

The calculated value is

[tex]\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225[/tex]

and our estimate is 67.1519320308594

Thus, the absolute error is given by

[tex]|67.0714655821225-67.1519320308594|=0.08047[/tex]

The absolute error in the Simpson's rule is

[tex]|67.0714655821225-67.0715427161943|=0.00008[/tex]

Answer:

Confidence interval for the population mean lifetime of the Amazon Kindle DX is (33.30 months to 42.70 months)

Step-by-step explanation:

Given;

Mean lifespan x = 38 months

Standard deviation r = 12 months

Number of kindle selected n = 25

Confidence range = 95%

Z*(95%) = 1.96

Confidence interval = x+/-Z*(r/√n)

= 38 +/- 1.96(12/√25)

= 38 +/- 4.70

Confidence interval = (33.30 months to 42.70 months)

Answer:

C. Fail to Reject H0.

Step-by-step explanation:

If the P-value is 0.06, that means that the result enters in the acceptance region. It is a value that is expected to happen if both proportions are equal (null hypothesis) at this significance level.

The conclusion when the P-value is bigger than the significance level is that the effect is not significant and it failed to reject the null hypothesis.

Final answer:

The null and alternative hypotheses for the claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents are H0: p1-p2 = 0 (p1 = p2) and Ha: p1-p2 > 0 (p1 > p2).

The p-value of 0.06 is larger than the significance level of 0.05, so the appropriate conclusion is C. Fail to Reject H0.

Explanation:

The null and alternative hypotheses for the claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents are:

H0: p1-p2 = 0 (p1 = p2)

Ha: p1-p2 > 0 (p1 > p2)

The p-value of 0.06 is larger than the significance level of 0.05. Therefore, we fail to reject the null hypothesis. Hence, an appropriate conclusion is:

C. Fail to Reject H0.

Answer:

[tex]t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893[/tex]  

[tex]p_v =P(t_{50}>0.893)=0.1881[/tex]  

If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.  

Step-by-step explanation:

Data given and notation

[tex]\bar X=0.505[/tex] represent the sample mean  

[tex]s=0.12[/tex] represent the standard deviation for the sample

[tex]n=51[/tex] sample size  

[tex]\mu_o =0.49[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the average nitrogen level dos not exced 0.49 units, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 0.49[/tex]  

Alternative hypothesis:[tex]\mu > 0.49[/tex]  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893[/tex]  

Now we need to find the degrees of freedom for the t distirbution given by:

[tex]df=n-1=51-1=50[/tex]

What do you conclude?  

Compute the p-value  

Since is a right tailed test the p value would be:  

[tex]p_v =P(t_{50}>0.893)=0.1881[/tex]  

If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.  

The expression 9p − 8q for p = 4 and q = −8 is equal to 100

Given that we have to evaluate expression for the given values of the variables

Expression is:

⇒ 9p − 8q

Given that p = 4 and q = -8

Let us substitute the given values of p = 4 and q = -8 in given expression and evaluate it

⇒ 9p − 8q = 9(4) - 8(-8)

⇒ 9p - 8q = 9(4) + 8(8)

Upon multiplying the terms we get,

⇒ 9p - 8q = 36 + 64 = 100

Thus the expression is equal to 100

Answer:

The chemist needs 50 mL of 51% solution and 70 mL of 87% solution.

Step-by-step explanation:

If x is the volume of 51% solution, and y is the volume of 87% solution, then:

x + y = 120

0.51x + 0.87y = 0.72(120)

Solve the system of equations.

0.51x + 0.87(120 − x) = 0.72(120)

0.51x + 104.4 − 0.87x = 86.4

0.36x = 18

x = 50

y = 70

The chemist needs 50 mL of 51% solution and 70 mL of 87% solution.

Answer:

Consider the following explanation

Step-by-step explanation:

Completely Randomized Design :-

A completely randomized design is probably the simplest experimental design, in terms of data analysis and convenience. With this design, subjects are randomly assigned to treatments.

This completely randomized design relies on randomization to control for the effects of extraneous variables. The experimenter assumes that, on average, extraneous factors will affect treatment conditions equally; so any significant differences between conditions can fairly be attributed to the independent variable.

Double Blind Design :-

In an experiment, if subjects in the control group know that they are receiving a placebo, the placebo effect will be reduced or eliminated; and the placebo will not serve its intended control purpose.

Blinding is the practice of not telling subjects whether they are receiving a placebo. In this way, subjects in the control and treatment groups experience the placebo effect equally. Often, knowledge of which groups receive placebos is also kept from analysts who evaluate the experiment. This practice is called double blinding. It prevents the analysts from "spilling the beans" to subjects through subtle cues; and it assures that their evaluation is not tainted by awareness of actual treatment conditions.

Matched Pairs Design :-

A matched pairs design is a special case of a randomized block design. It can be used when the experiment has only two treatment conditions; and subjects can be grouped into pairs, based on some blocking variable. Then, within each pair, subjects are randomly assigned to different treatments.

Randomized Block Design :-

With a randomized block design, the experimenter divides subjects into subgroups called blocks, such that the variability within blocks is less than the variability between blocks. Then, subjects within each block are randomly assigned to treatment conditions. Compared to a completely randomized design, this design reduces variability within treatment conditions and potential confounding, producing a better estimate of treatment effects.

The experimental design that is given in the problem, is an example of a Randomized Block Design. Here, the different type of soils can be considered different blocks. In these blocks, the variability among within the blocks is minimum and between the blocks, it is maximum. Also, the treatments ( fertilizers ) are assigned randomly to the plts in each block (different soil type ).

Answer:

17%

Step-by-step explanation:

If 83% is removed than

total is always 100%

that's why

so the new sound level will sound

100%-83%=17% remains.

hence 17 % is correct answer .

Answer:

3.216%

Step-by-step explanation:

This bond sells at a higher price or value, which means that its coupon is bogus of market interest rate. Therefore, the minimum yield rate that accounts for the possibility of the bond being called is calculated at the earliest possible call date. Let say exactly 15 years from the date of purchase, because that would be the most disadvantageous date for the bondholder for the call to occur.

The minimum semiannual yield:

j= i²/2

i² = 2j

which therefore satisfies the expression below for the worst possible case scenario yield:

1722.25 = 0.04*1100*[tex]a]_30[/tex]+[tex]\frac{1100}{(1+j)^30}[/tex]

Also, with the use of a financial calculator (making sure that the calculator is not in BGN mode)

1722.25 PV, -44 PMT, -1100 FV, 30 N, CPT 1/Y.

j can be found to be 1.608245%. The corresponding nominal annual rate compounded semiannually is (X) = i² = 2j =3.216%

Answer:

Ray Flagg will pay $5,272 at the time of his 30th installment.

Step-by-step explanation:

Ray took $12,000 load for 60 months. As he paid no amount as down payment so his monthly payment will be $200:

[tex]=12000/60\\=200[/tex]

Instead of $200 per month, he used to pay $232 per month. So, before his 30th installment, he paid 29 installments each of $232 which is $6,728:

[tex]=232*29\\=6728[/tex]

As the business does better, he wishes to payback remaining amount at once so he will pay $5,272 as:

[tex]12000-6728\\=5272[/tex]

To calculate the remaining balance of a fixed installment loan after a certain number of payments, use the formula provided in the detailed answer.

Ray Flagg took out a 60-month fixed installment loan of $12,000 to open a new pet store. He paid no money down and began making monthly payments of  $232. Ray's business does better than expected and instead of making his 30th payment, Ray wishes to repay his loan in full.

To calculate the remaining balance after 29 months, we can use the formula for the remaining balance of a fixed installment loan:

Remaining Balance = Balance × (1 + Monthly Interest Rate)Number of Payments Made - (Monthly Payment × ((1 + Monthly Interest Rate)Number of Payments Made - 1) / Monthly Interest Rate)

Using the given values, the monthly interest rate can be calculated by dividing the annual interest rate by 12 and converting it to a decimal.

Finally, substitute the values into the formula to find the remaining balance after 29 months.

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Answer:

the answer is D

Step-by-step explanation:

Think of the Pythagorean Theorem which states that a^2 + b^2 = c^2. The Pythagorean Identities used in trigonometry are the angle version which can be used to simplify expressions.  

HOPE THIS HELPED ;3

The formula of half angle identities is sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.

An angle is the formed when two straight lines meet at one point, it is denoted by θ.

The given terms are,

sin²x and cos²x

To integrate sin²x, the formula is used,

sin²x = (1-cos(x/2))/2

To integrate cos²x, the formula is used,

cos²x = (1+cos(x/2))/2

The formula is, sin²x = (1-cos(x/2))/2 and  cos²x = (1+cos(x/2))/2.

To learn more about Angle on:

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The equation to represent this situation is 8x = 32

Solution:

Given that ureg placed 32 chairs in the auditorium

There are 8 chairs in each row

To find: equation used to represent this situation

From given information,

1 row = 8 chairs

So let us find an expression to determine the number of rows to place 32 chairs

Let "x" be the number of rows required to place 32 chairs

Since 1 row contains 8 chairs, expression to determine the number of rows to place 32 chairs is given as:

[tex]8 \times \text{ number of row } = 32[/tex]

8x = 32

Thus the equation to represent this situation is 8x = 32

Answer:

The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.

Step-by-step explanation:

The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

This is the question of normal distribution:

First w calculate the value of Z corresponding to X = 60,000 miles

We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles

Now, for Z, we know that:

Z = (x-μ)/σ

Z = (60,000 - 50,000)/5,000

Z = 2

Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.

P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)

P(X > 60,000) = 1 - 0.9772

P(X > 60,000) = 0.0228

The expected value of the received voltage is 4. The variance of the received voltage is 64/3. The covariance of the transmitted and received voltage is 0.

(a) To find the expected value of the received voltage, we need to use the linearity property of the expectation and the fact that V and X are independent. The expected value of R is given by:

E[R] = E[V+X] = E[V] + E[X]

Since V and X are independent, we have E[X] = 4 and E[V] = 0 (by symmetry of the uniform distribution). Therefore, E[R] = 0 + 4 = 4.

(b) To find the variance of the received voltage, we can use the properties of variance. Variance is additive for independent random variables, so:

Var[R] = Var[V+X] = Var[V] + Var[X]

Since V and X are independent, we have Var[X] = 4^2 = 16 and Var[V] = (16^2)/12 = 64/12 = 16/3. Therefore, Var[R] = 16/3 + 16 = 64/3.

(c) The covariance of the transmitted and received voltage is given by:

Cov[V, R] = E[(V - E[V])(R - E[R])]

Since E[V] = 0 and E[R] = 4, this simplifies to:

Cov[V, R] = E[VR] - E[V]E[R]

Since V and R are independent, we have Cov[V, R] = E[V]E[R] - E[V]E[R] = 0.

(d) The optimal linear estimate VL(R) is given by:

VL(R) = E[V] + Cov[V, R]/Var[R] * (R - E[R])

Since Cov[V, R] = 0, the optimal linear estimate becomes:

VL(R) = E[V] + 0/Var[R] * (R - E[R]) = E[V] = 0.

(e) The minimum mean square error of the estimate is given by:

e* = Var[V - VL(R)]

Since VL(R) = E[V] = 0, this simplifies to:

e* = Var[V] = 16/3.

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Answer:

Step 2 involved distributive property and the value of y is equal to 8.

Step-by-step explanation:

In step 2, there has been an error in applying the Distributive Property correctly.

Distributive Property-  a(b+c)

                                       = a x b + a x c

Step 2: [tex]2y+16=4y[/tex]

Step 3:[tex]2y=16[/tex]

Step 4: [tex]y=8[/tex]

y=8

Answer:

2 should be distributed as 2y + 16; y + 8

Answer:

The statement which is true is:

B. This is a double blind study

Step-by-step explanation:

Answer:

C. Observational Study

Step-by-step explanation:

There arent any treatments being manipulated so it can't be an experiment. The researcher is aware of the experiment being performed so its not double-blind. The only option it could be is C, because the researcher is observing the people eating and isn't manipulating their eating habits in any way.

Answer:

If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 15

Standard Deviation, σ = 1

Sample size = 4

Total lifetime of 4 batteries = 40 hours

We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

Standard error due to sampling:

[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5[/tex]

We have to find the value of x such that the probability is 0.05

P(X > x)  = 0.05

[tex]P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95 [/tex]  

Calculation the value from standard normal z table, we have,  

[tex]\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83[/tex]  

Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

Answer: 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.

Step-by-step explanation:

The Range Rule of Thumb tells that the range is approximately four times the standard deviation.

95% of the data lies within 2 standard deviations from the mean .

Maximum usual value = Mean +2 (Standard deviation )

Minimum usual value = Mean - 2 (Standard deviation)

Given : Mean =  29.9 degrees Celsius

Standard deviation = 2.31 degrees Celsius

Then, according to the Range Rule of Thumb , we have

Maximum usual value = 29.9 +2 (2.31) = 34.52 degrees Celsius

Minimum usual value = 29.9 - 2 (2.31) = 25.28 degrees Celsius

i.e. 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.

Using the empirical rule for normal distribution, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius, given the mean is 29.9°C and the standard deviation is 2.31°C.

The question asks for the range of temperatures within which 95% of the maximum monthly temperatures fall, given that the average maximum monthly temperature is 29.9 degrees Celsius with a standard deviation of 2.31 degrees, assuming a normal distribution. To find this, we can apply the empirical rule that states approximately 95% of the data in a normal distribution falls within two standard deviations of the mean. Therefore:

Thus, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius.

Answer:

Step-by-step explanation:

Given that many couples believe that it is getting too expensive to host an "average" wedding in the United States.

Population mean =24066

Sample mean =  23224

Sample size = 40

Sample std dev = 2903

Since sample std dev is known, we use t critical value.

df =39

Sample mean follows a normal distribution with mean = 23224, and std dev = [tex]\frac{s}{\sqrt{n} } \\=\frac{2903}{\sqrt{40} } \\=459.005[/tex]

t critical value = 2.023

Margin of error = 2.023*459.005

Confidence interval[tex](22295.43, 24152.57)[/tex]

Final answer:

The question involves calculating a 95% confidence interval for the average cost of weddings in the U.S., resulting in a range of $22,296 to $24,152, based on recent sample data.

Explanation:

The question pertains to the formation of a confidence interval for the average cost of weddings in the U.S. based on a sample. The provided data asserts that the average cost of a wedding in 2009 was $24,066, while a more recent sample reveals an average of $23,224 with a standard deviation of $2,903. Calculating a 95% confidence interval results in a range of $22,296 to $24,152. This implies that we can be 95% confident that the true average cost of weddings in the whole population falls within this interval.

Two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent.

In order to prove that two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent, we can use the side-side-side (SSS) congruence theorem. This theorem states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

In this case, we can show that the corresponding altitudes and angle bisectors through the right angles are congruent for both triangles. Since both triangles have congruent corresponding altitudes and congruent angle bisectors through the right angles, we can conclude that the triangles are congruent.

Therefore, the statement is proven.

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Answer:

Step-by-step explanation:

Answer:

Null hypothesis: [tex]p\leq 0.3[/tex]

Alternative hypothesis: [tex]p > 0.3[/tex]

Step-by-step explanation:

1) Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

2) Solution to the problem

On this case we want to test is [tex]p>0.3[/tex] since we want to check if the new model has improved the warranty rate, we can express it like this:

[tex]p-0.3<0[/tex] since are equivalent expressions.

And the alternative hypothesis should be the complement:

Null hypothesis: [tex]p\leq 0.3[/tex] or [tex]p=0.3[/tex]

So the correct system of hypothesis for this case would be:

Null hypothesis: [tex]p\leq 0.3[/tex]

Alternative hypothesis: [tex]p > 0.3[/tex]

The alternative way should be:

Null hypothesis: [tex]p = 0.3[/tex]

Alternative hypothesis: [tex]p > 0.3[/tex]

Answer:

a) [tex]\mu_1 -\mu_2[/tex] parameter of interest.

Where [tex]\mu_1[/tex] represent the mean response for adults

[tex]\mu_2[/tex] represent the mean response for teenegers

b) The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]

Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]

c) The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]

d) The 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Let group 1 be adults and group 2 be teenagers.

[tex]\bar X_1 =0.225[/tex] represent the sample mean 1

[tex]\bar X_2 =0.059[/tex] represent the sample mean 2

n1 represent the sample 1 size  

n2 represent the sample 2 size  

[tex]s_1 [/tex] sample standard deviation for sample 1

[tex]s_2 [/tex] sample standard deviation for sample 2

SE =0.091 represent the standard error for the estimate

(a) Give notation for the quantity that is being estimated.

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

(b) Give notation for the quantity that gives the best estimate.

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]

Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]

(c) Give the value for the quantity that gives the best estimate.

The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]

(d) Give a confidence interval for the quantity being estimated. Assuming 95% of confidence

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]

We can assume that since we know the standard error the deviations are known and we can use the z distribution instead of the t distribution for the confidence interval.

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=0.091[/tex]

Given by the problem

Now we have everything in order to replace into formula (1):  

[tex]0.166-1.96(0.091)=-0.012[/tex]  

[tex]0.166+1.96(0.091)=0.344[/tex]  

So on this case the 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]  

Final answer:

The quantity being estimated in the study is the difference in mean response to unlearn fear associations between adults and teenagers, denoted by Δμ = μ1 - μ2, where μ1 and μ2 represent the mean responses for adults and teenagers, respectively. This study contributes to understanding how fear associations are formed and unlearned, with implications on evolutionary predisposition towards certain fears.

Explanation:

The quantity being estimated in the study between adolescents and adults regarding their ability to unlearn fear associations tied to a specific sound is captured by the notation Δμ = μ1 - μ2. Here, μ1 represents the mean response for adults, and μ2 represents the mean response for teenagers. In this context, a higher physiological measure indicates less fear, with adults showing a mean response of 0.225 and teenagers showing a mean response of 0.059. The standard error of the estimate provided is 0.091, which helps in understanding the variability or precision of our estimated difference between the two groups' mean responses.

This study hints at the broader theory of preparedness, suggesting that humans are evolutionarily predisposed to easily associate certain stimuli with fear. Notably, the differentiation in fear response unlearning between age groups aligns with observations in social and developmental psychology about the specificity of fear acquisition and the challenges in modify these responses once established, especially during the teenage years.

Answer:the horizontal change of the line is 3

Step-by-step explanation:

The horizontal change is the change in the value of x on the horizontal axis. It is expressed as

x2 - x1

Where

x2 represents the final value of x

x1 represents the initial value of x

An x intercept at (3,0) means that the line cut across the x axis at the point when x = 3 and y = 0

A y intercept at (0,2) means that the line cut across the y axis at the point when 0 = 3 and y = 2

Change in the horizontal axis would be x2 - x1 = 3 - 0 = 3

Answer:

[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]

Step-by-step explanation:

Data given

[tex]\bar X=20[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

[tex]s^2=0.02[/tex] represent the sample variance

[tex]s=0.141[/tex] represent the sample deviation

n=25 represent the sample size  

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 0.01, so the system of hypothesis are:

H0: [tex]\sigma \leq 0.1[/tex]

H1: [tex]\sigma >0.1[/tex]

In order to check the hypothesis we need to calculate the statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1=25-1=24 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 24 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,24)". And our critical value would be [tex]\chi^2 =36.415[/tex]

Since our calculated value is higher than the critical value we reject the null hypothesis at 5% of significance.

The variance hypothesis test for a cancer treatment drug with a sample mean of 20 mg and sample variance of 0.02 mg results in a chi-square test statistic of 48. This test statistic will be used to determine if the drug's variance exceeds the acceptable limit.

The question at hand is concerning a hypothesis test of the variance in dosage of a cancer treatment drug. The null hypothesis (H0) claims that the standard deviation of the drug content is not more than 0.1 mg, which corresponds to a variance of 0.01 mg² since variance = standard deviation². The alternative hypothesis (Ha) is that the variance is greater than 0.01 mg². Given the sample variance as 0.02 mg² and a sample size of 25, the test statistic for the chi-square test can be calculated using the formula:

Test statistic (chi-square) = (n - 1)*sample variance / hypothesized variance

Test statistic = (25 - 1) * 0.02 / 0.01 = 24 * 2 = 48

The calculated test statistic is 48. Since the sample variance is greater than the hypothesized variance, we have a test statistic that would fall in the rejection region based on the selected significance level in a Chi-square distribution, suggesting that the drug dosage may indeed have greater variability than the company's standard.

Answer:the number of dimes is 22

the number of quarters is 11

Step-by-step explanation:

A dime is worth 10 cents. Converting to dollars, it becomes 10/100 = $0.1

A quarter is worth 25 cents. Converting to dollars, it becomes 25/100 = $0.25

Let x represent the number of dimes.

Let y represent the number of quarters.

Daniele has 33 quarters and dimes in her piggy bank. This means that

x + y = 33

The piggy bank contains a total of $4.95. This means that

0.1x + 0.25y = 4.95 - - - - - - - - -1

Substituting x = 33 - y into equation 1, it becomes

0.1(33 - y) + 0.25y = 4.95

3.3 - 0.1y + 0.25y = 4.95

- 0.1y + 0.25y = 4.95 - 3.3

0.15y = 1.65

y = 1.65/0.15 = 11

Substituting y = 11 into x = 33 - y, it becomes

x= 33 - 11 = 22

Final answer:

To find the number of dimes and quarters in Daniele's piggy bank, a system of equations can be solved by substitution method.

Explanation:

System of Equations:

Let x be the number of dimes and y be the number of quarters.

We have the equations: 0.10x + 0.25y = 4.95 and x + y = 33.

Solving by substitution, we first solve x + y = 33 for x to get x = 33 - y.

Substitute x = 33 - y into 0.10x + 0.25y = 4.95 and solve to find the values of x and y.

The solution is x = 15, y = 18, so there are 15 dimes and 18 quarters.

Answer:

a) The 90% confidence interval would be given by (73.56;79.84)  

b) The 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].

c) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance.  So the claim of the teacher makes sense.

d) Since is a one-side upper test the p value would given by:  

[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]  

Step-by-step explanation:

Data given: 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We can calculate the sample mean and deviation with the following formulas:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]s=\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And we got the following results:

[tex]\bar X=76.7[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]s=8.112[/tex] represent the population standard deviation  

n=20 represent the sample size  

90% confidence interval  

Part a

The confidence interval for the mean is given by the following formula:  

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)  

The degrees of freedom are given by:

[tex]df=n-1=20-1=19[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,19)".And we see that [tex]t_{\alpha/2}=1.73[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]76.7-1.73\frac{8.112}{\sqrt{20}}=73.56[/tex]  

[tex]76.7+1.73\frac{8.112}{\sqrt{20}}=79.84[/tex]  

So on this case the 90% confidence interval would be given by (73.56;79.84)  

Part b

The confidence interval for the population variance is given by the following formula:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=30.143[/tex]

[tex]\chi^2_{1- \alpha/2}=10.117[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(19)(8.112^2)}{30.143} \leq \sigma^2 \leq \frac{(19)(8.112^2)}{10.117}[/tex]

[tex] 41.472 \leq \sigma^2 \leq 123.574[/tex]

So the 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].

Part c

Null hypothesis:[tex]\mu \leq 70[/tex]  

Alternative hypothesis:[tex]\mu > 70[/tex]  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{76.7-70}{\frac{8.112}{\sqrt{20}}}=3.69[/tex]  

Part d

P-value  

Since is a one-side upper test the p value would given by:  

[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance.  

Answer:

(a) 95% confidence interval for the population mean score on Long Island is (73, 80.40)

(b) 90% confidence interval for the population standard deviation of the scores on Long Island is (4.85, 10.97)

(c) The teacher's claim that the mean score on Long Island is higher than the mean for New York State is reasonable

(d) p-value is 0.01

Step-by-step explanation:

From the data values from Long Island,

Mean is 76.7 and standard deviation is 7.91

Confidence Interval (CI) = mean + or - (t × sd)/√n

(a) mean = 76.7, sd = 7.91, n = 20, degree of freedom = n-1 = 20-1 = 19, t-value corresponding to 19 degrees of freedom and 95% confidence level is 2.093

Lower bound = 76.7 - (2.093×7.91)/√20 = 76.7 - 3.70 = 73

Upper bound = 76.7 + (2.093×7.91)/√20 = 76.7 + 3.70 = 80.40

95% CI is (73, 80.40)

(b) CI = sd + or - (t×sd)/√n

sd = 7.91, t-value corresponding to 19 degrees of freedom and 90% confidence level is 1.729

Lower bound = 7.91 - (1.729×7.91)/√20 = 7.91 - 3.06 = 4.85

Upper bound = 7.91 + (1.729×7.91)/√20 = 7.91 + 3.06 = 10.97

90% CI is (4.85, 10.97)

(c) Null hypothesis: The mean score on Long Island is 70

Alternate hypothesis: The mean score on Long Island is greater than 70

Z = (sample mean - population mean)/(sd/√n) = (70 - 76.7)/(7.91/√20) = -6.7/1.77 = -3.79

Using 0.01 significance level, the critical value is 2.326

Since -3.79 is less than 2.326, reject the null hypothesis. The teacher's claim is reasonable

(d) p-value = 1 - cumulative area of test statistic = 1 - 0.9900 = 0.01