A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is believed that the machine is underfilling the bags.
A 26 bag sample had a mean of 427 grams with a standard deviation of 15.
Assume the population is normally distributed. A level of significance of 0.05 will be used.

Find the P-value of the test statistic.

You may write the P -value as a range using interval notation, or as a decimal value rounded to four decimal places.

Answer:

Answer explained below

Step-by-step explanation:

Let a be the working hours of plant A.

Let b be the working hours of plant H.

Let c be the working hours of plant M.

We have to minimize , Z = 60a + 92b + 140c

subject to constraint , 40a + 20b + 40c >=350

                20a + 40b + 40c >=240

where a>=0 , b>=0 ,c>=0

So ,by solving this , a = 7.67 , b= 2.17 , c = 0

So ,working hours of plant A = 8 hrs

working hours of plant H = 3 hrs

working hours of plant M = 0 hrs

Answer:

A = 8hrs  

H = 3hrs  

M = 0hrs  

Step-by-step explanation:

Let a be the working hours of plant A.  

Let b be the working hours of plant H.  

Let c be the working hours of plant M.  

∴ Hours of plants A,H,M are a,b,c respectively.  

We have to minimize the cost of production, Z = 60a + 92b + 140c  

Regular ice cream: 40a + 20b + 40c = 350  

Deluxe ice cream: 20a + 40b +40c =240  

Where a>=0, b>=0, c>=0  

So, by solving this, a= 7.67, b= 2.77,c= 0  

So working hours of plant A = 8 hrs  

Working hours of plant H = 3 hrs  

Working hours of plant M = 0 hrs  

The carrying capacity K for the logistic model is 45. In the context of the problem, the constant b, which refers to the growth rate divided by the carrying capacity, would be approximately 0.00444. The logistic model representing P vs. m under these conditions would be P(m) = 45/ (1+ (45/15-1) * e^(-0.00444m)).

In the field of mathematics, specifically in growth modeling, a logistic model incorporates a carrying capacity. The carrying capacity, denoted as K, is the maximum stable value of the population, in this case, the number of parents attending the PTA meeting. The carrying capacity is expected to be 45 in this instance.

The constant b in the logistic model can be found using the initial value (15 parents) and the growth rate (20%). However, the question does not specify whether this rate is relative or absolute. Assuming relative growth, the initial growth rate r is 0.2 and the constant b = r/K would be 0.00444. This is not the traditional definition of b in a logistic equation; typically, b would denote the initial population size, so the question seems to have a specific, non-standard usage in mind that we need to respect.

A good starting value of ther, in order to ensure convergence and stability of the numerical method, can be 15, the initial population size.

The logistic model would thus be represented as P(m) = K/ (1+ (K/P_0-1) * e^(-bm)), where P_0 is the initial number of parents, K is the carrying capacity, b is the constant/ growth rate, m is the number of months, and e is the mathematical constant approximated as 2.718.

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Answer:

[tex]9\pi[/tex]

Step-by-step explanation:

given are two parabolas with vertex as (3,0) and (0.0)

[tex]y^2 =2(x-3)\\y^2 =x[/tex]

These two intersect at x=6

Volume of II  curve rotated about x axis - volume of I curve rotated about x axis = Volume of solid of revolution

For the second curve limits for x are from 0 to 6 and for I curve it is from 3 to 6

V2 =\pi [tex]\int\limits^6_0 {y^2} \, dx \\=\pi\int\limits^6_0 {x} \, dx\\= \pi\frac{x^2}{2} \\=18\pi[/tex]

V1 =[tex]\pi \int\limits^6_3 {y^2} \, dx\\\pi \int\limits^6_3 {2x-6} \, dx\\=\pi(x^2-6x)\\= \pi[(36-9)-6(6-3[)\\= (27-18)\pi\\=9\pi[/tex]

Volume of solid of revolutin = V2-V1 = [tex]9\pi[/tex]

Answer:

Answer = I. Chi-squared test of independence

Step-by-step explanation:

The test of independence uses the contingency table format. The analysis of a two-way contingency table helps to answer the question whether two variables are related or independent of each other. Thus, the chi-square statistic measures how much the observed frequencies differ from the expected frequencies when variables are independent. The first procedure is to state the null and alternative hypothesis  

H0: No relationship exists between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

H1: There is a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

To be able to decide whether to reject the null hypothesis or not, we need to compare the calculated test statistic with its chi-square table or critical value using a given level of significance and degree of freedom and then decide using the decision rule (Reject H0 if the calculated chi-square statistic is greater than the critical value otherwise, accept H0)

A. 15917 ft   B.∠D=88.35°  C.0.0532 rad

Step-by-step explanation:

A. Given  that angle ∠ACB = 88.60° and the distance from C to B is 389 ft then triangle ABC is right ,90° at B. Applying the formula for tangent of an angle which is;

Tan of an angle = opposite side length/adjacent side length

Tan Ф = O/A =AB/389 ft

Tan 88.60°= AB/389

AB=389*tan 88.60° = 389×40.92 =15916.87 ft ⇒15917 ft

B.

The distance from B to D is given as 459 ft and the distance between the towers , AB, is 15917 ft. To get angle ∠D apply the formula for tangent of an angle where ;

Tan ∠D=O/A =15917/459 =34.6775599129

∠D =tan⁻(34.6775599129)

∠D=88.35°

C. To get angle ∠A subtract the sum of angle ∠C and ∠D from 180°. Apply the sum of angles in a triangle theorem

 ∠A =180° - (88.60°+88.35°)

   ∠A = 180°-(176.95°)=3.05°

    Changing degrees to radians you multiply the degree value with 0.0174533

3.05°=3.05*0.0174533=0.05323254 rad

To 4 decimal places

=0.0532 rad

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Answer:

Step-by-step explanation:

Answer: 325 boxes of candy will be produced on each working day in October

Step-by-step explanation:

The initial number of boxes of candy produced is 100. In each month following this, the shop produced 25 more boxes of candy per working day in addition to the previous month. This means that the number of boxes produced each month is increasing in arithmetic progression. The formula for the nth term of an arithmetic sequence is expressed as

Tn = a + (n-1)d

Where

a is the first term of the sequence

n is the number of terms

d is the common difference.

From the given information,

a = 100

d = 25

n = number of months from January to October = 10

Tn = 100 + (10 - 1)25

Tn = 100 + 9×25

Tn = 100 + 225 = 325

Answer:

The fourth one

Step-by-step explanation:

Because Convex polygons can only be shapes like hexagons and squares and stuff like that. They cant be curved or zig zagged.

Answer:

The fourth one

Step-by-step explanation:

Convex polygons can be hexagons and stuff like that.

Answer:

[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]

[tex]x_{1} = -10[/tex]

[tex]x_{2} = -3.95[/tex]

Step-by-step explanation:

Generally, the Newton-Raphson method can be used to find the solutions to polynomial equations of different orders. The formula for the solution is:

[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]

We are given that:

f(x) = [tex]x^{2} + 21[/tex]; [tex]x_{0} = -21[/tex]

[tex]f^{'} (x)[/tex] = df(x)/dx = 2x

Therefore, using the formula for Newton-Raphson method to determine [tex]x_{1}[/tex] and [tex]x_{2}[/tex]

[tex]x_{1} = x_{0} - \frac{f(x_{0} )}{f^{'}(x_{0})}[/tex]

[tex]f(x_{0}) = x_{0} ^{2} + 21 = (-21)^{2} + 21 = 462[/tex]

[tex]f^{'}(x_{0}) = 2*(-21) = -42[/tex]

Therefore:

[tex]x_{1} = -21 - \frac{462}{-42} = -21 + 11 = -10[/tex]

Similarly,

[tex]x_{2} = x_{1} - \frac{f(x_{1} )}{f^{'}(x_{1})}[/tex]

[tex]f(x_{1}) = (-10)^{2} + 21 = 100+21 = 121[/tex]

[tex]f^{'}(x_{1}) = 2*(-10) = -20[/tex]

Therefore:

[tex]x_{2} = -10 - \frac{121}{20} = -10+6.05 = -3.95[/tex]

Answer: [tex]\$3108[/tex]

Step-by-step explanation:

Mr. Kim, his wife and two children are 4 persons. Now, he needs to buy 3 plane tickets for each one, which means he has to buy 12 plane tickets:

[tex](3 tickets)(4)=12 tickets[/tex]

On the other hand, we know each ticket costs [tex]\$259[/tex] per person. So, if we have 12 tickets to buy, we will have to multiply [tex]\$259[/tex] by 12:

[tex](12)(\$259)=\$3108[/tex] This is the amount Mr. Kim will spend in the plane tickets

Final answer:

The null and alternative hypotheses for testing the mean travel time on scenic and nonscenic routes.

Explanation:

a. In this case, the hypotheses to be tested are:

Null hypothesis (H0): The mean travel time for the nonscenic route is not shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 ≤ 10.

Alternative hypothesis (Ha): The mean travel time for the nonscenic route is shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 > 10.

b. If u1 is the mean for the nonscenic route and u2 for the scenic route, the hypotheses to be tested would be:

Null hypothesis (H0): The mean travel time for the scenic route is not shorter than 10 minutes compared to the nonscenic route, or in other words, u2 - u1 ≤ 10.

Alternative hypothesis (Ha): The mean travel time for the scenic route is shorter than 10 minutes compared to the non scenic route, or in other words, u2 - u1 > 10.

The correct options are : - (a) The correct hypothesis is 2. [tex]\(\mu_1 - \mu_2 < -10\)[/tex] and (b) The correct hypothesis is 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]

Part (a): [tex]\(\mu_1\)[/tex] is the mean for the scenic route and [tex]\(\mu_2\)[/tex] is the mean for the non-scenic route

The individual decides that the non-scenic route is justified only if it reduces the mean travel time by more than [tex]10[/tex] minutes. In this context, we need to test whether the mean travel time for the scenic route [tex]\(\mu_1\)[/tex] minus the mean travel time for the non-scenic route [tex](\(\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]

Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = -10\)[/tex]

Alternative Hypothesis[tex](\(H_a\)) : \(\mu_1 - \mu_2 < -10\)[/tex]

So, the correct option for (a) is: 2.[tex]\(\mu_1 - \mu_2 < -10\)[/tex]

Part (b): [tex]\(\mu_1\)[/tex] is the mean for the non-scenic route and [tex]\(\mu_2\)[/tex] is the mean for the scenic route

In this case, we need to test whether the mean travel time for the non-scenic route [tex](\(\mu_1\))[/tex] minus the mean travel time for the scenic route [tex](\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]

Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = 10\)[/tex]

Alternative Hypothesis [tex](\(H_a\)) : \(\mu_1 - \mu_2 < 10\)[/tex]

So, the correct option for (b) is: 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]

The complete Question is

An individual can take either a scenic route to work or a non-scenic route. She decides that use of the non-scenic route can be justified only if it reduces the mean travel time by more than 10 minutes.

(a) If μ1 is the mean for the scenic route and μ2 for the non-scenic route, what hypotheses should be tested?

1. μ1 − μ2 = −10

2. μ1 − μ2 < −10μ1 − μ2 = −10

3. μ1 − μ2 > −10 μ1 − μ2 = −10

4. μ1 − μ2 ≠ −10μ1 − μ2 = 10

5. μ1 − μ2 < 10μ1 − μ2 = 10

6. μ1 − μ2 > 10

(b) If μ1 is the mean for the non-scenic route and μ2 for the scenic route, what hypotheses should be tested?

1. μ1 − μ2 = −10

2. μ1 − μ2 < −10μ1 − μ2 = −10

3. μ1 − μ2 > −10 μ1 − μ2 = −10

4. μ1 − μ2 ≠ −10μ1 − μ2 = 10

5. μ1 − μ2 < 10μ1 − μ2 = 10

6. μ1 − μ2 > 10

Answer:

Step-by-step explanation:

Given than n = 25 , mean = 52800 sd = 4600

1) P(X<50000) , so please keep the z tables ready

we must first convert this into a z score so that we can look for probability values in the z table

using the formula

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

checkng the value in the z table

P ( Z>−0.608 )=P ( Z<0.608 )=0.7291

b)

again using the same formula and converting to z score

we need to calculate P(X<50000)

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

P ( Z<−0.608 )=1−P ( Z<0.608 )=1−0.7291=0.2709

proportion is 27% approx

c)

When your data points are measurements on a numerical scale you have variables data , here yield stress is numeric in nature , hence its a sampling by variable plan

Answer:

Average value of temperature will be [tex]8.335^{\circ}C[/tex]

Step-by-step explanation:

We have given the temperature in degree Celsius [tex]T(x,y)=19-4x^2-y^2[/tex]

It is given that x varies between 0 to 2

So [tex]0\leq x\leq 2[/tex]

And y varies between 0 and 4

So [tex]0\leq y\leq 4[/tex]

So area between the region A = 4×2 = 8[tex]cm^2[/tex]

Now average temperature is given by

[tex]Average\ value=\frac{1}{A}\int \int T(x,y)dA[/tex]

[tex]Average\ value=\frac{1}{8}\int \int (19-4x^2-y^2)dxdy[/tex]

[tex]=\frac{1}{8}\int \int (19x-4\frac{x^3}{3}-y^2x)dy[/tex]

As limit of x is 0 to 2

[tex]=\frac{1}{8}\int  (19\times 2-4\times \frac{2^3}{3}-y^2\times 2)-(19\times 0-4\times \frac{0^3}{3}-y^2\times 0))dy=\frac{1}{8}\int(38-\frac{32}{3}-2y^2)dy[/tex]

=[tex]=\frac{1}{8}(38y-\frac{32}{3}y-\frac{2y^3}{3})[/tex]

As limit of y is 0 to 4

So [tex]Average\ value=\frac{1}{8}(38\times 4-\frac{16}{3}\times 4-\frac{2\times 4^3}{3})-0=8.335^{\circ}C[/tex]

Answer:

Step-by-step explanation:

1) The diagram is a polygon with unequal sides. The number of sides and angles is 5. This means that it is an irregular Pentagon. The formula for the sum of interior angles of a polygon is expressed as

(n - 2)180

Where n is the number if sides of the polygon. Since the number of sides of the given polygon is 5, the sum if the interior angles would be

(5-2)×180 = 540 degrees. Therefore,

10x - 3 + 5x + 2 + 7x - 11 + 13x - 31 + 8x - 19 = 540

43x - 62 = 540

43x = 540 + 62 = 602

x = 602/43 = 14

Angle S = 13 - 31 = 13×14 - 31 = 182 - 31

Angle S = 151 degrees.

4) The diagram is a rectangle. Opposite sides are equal. Triangle JNM is an isosceles triangle. It means that its base angles, Angle NMJ and angle NJM are equal. Therefore

3x + 38 = 7x - 2

7x - 3x = 38 + 2

4x = 40

x = 40/4

x = 10

Angle NMJ = 7x - 2 = 7×10 - 2 = 68 degrees.

Angle JML = 90 degrees ( the four angles in a rectangle are right angles). Therefore,

Angle NML = 90 - 68 = 22 degrees

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

[tex]Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}[/tex] [tex]= 60.67[/tex]

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: [tex]s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}[/tex]

Put value in formula of Standard Deviation,

[tex]s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}[/tex] = 40.75

Step-3: Then, we have to find the critical value by chi-square.

[tex]X_{1-\alpha/2}^{2}=3.82[/tex]

[tex]X_{1-\alpha/2}^{2}=21.92[/tex]

Then, find the confidence interval which is 95%.

[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9[/tex]

[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9[/tex]

The 95% confidence interval for the population standard deviation, based on the given sample data, is estimated to be approximately between 2.9 mi/h and 6.9 mi/h. Here option B is correct.

To construct a 95% confidence interval estimate of the population standard deviation, you can use the chi-square distribution. The formula for the confidence interval is given by:

[tex]\[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \][/tex]

where:

- n is the sample size,

- s is the sample standard deviation,

- [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] are the critical values from the chi-square distribution for the lower and upper bounds of the confidence interval, respectively, and

- α is the significance level (0.05 for a 95% confidence interval).

Given that the sample data is: 62 61 61 57 61 54 59 58 59 69 60 67, the sample size n is 12, and the sample standard deviation s is approximately 4.75.

Now, you need to find the critical values [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex]. For a 95% confidence interval and df = n-1 = 11, you can consult a chi-square distribution table or use a statistical software/tool to find these critical values.

Assuming [tex]\( \chi^2_{\alpha/2} \)[/tex] is 19.675 and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] is 2.201, plug these values into the formula:

[tex]\[ \left( \sqrt{\frac{(12-1) \times 4.75^2}{19.675}}, \sqrt{\frac{(12-1) \times 4.75^2}{2.201}} \right) \][/tex]

Calculating this expression gives approximately (2.9, 6.9). Here option B is correct.

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Answer:

Upper Riemann Sum is 9/16

Step-by-step explanation:

Final answer:

To compute the upper Riemann sum for the function f(x) = x^2 over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1], we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval. The upper Riemann sum is obtained by summing up all these values.

Explanation:

To compute the upper Riemann sum, we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval.

Given the function f(x) = x^2 and the partition P = [-1, -1/4, 1/4, 3/4, 1], we can calculate the upper Riemann sum as follows:

Calculate the width of each subinterval: Δx = (1 - (-1)) / 4 = 1/2.

Find the value of the function at each partition point:

Multiply the value of the function at each partition point by the width of the corresponding subinterval:

Sum up all the values obtained: 1/2 + 1/32 + 1/32 + 9/32 + 1/2 = 17/16

Therefore, the upper Riemann sum for the given function over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1] is 17/16.

The upper 90% confidence limit for the true proportion of firefighter masks of this type whose lenses would pop out at 250° is approximately 0.2949.

To construct a 90% upper confidence limit for the true proportion of firefighter's masks whose lenses would pop out at 250°, firstly, we need to calculate the sample proportion (p) which is the ratio of the number of masks that had lenses popping out (12) to the total number of masks tested (60). Thus, the sample proportion is 12/60 = 0.2.

Next, we use the formula for an upper confidence limit for proportions: p + z*sqrt((p*(1-p))/n), where z is the z-score associated with the desired level of confidence (for 90% confidence, z is 1.645), p is the sample proportion, and n is the sample size.

So the upper 90% confidence limit would be calculated as: 0.2 + 1.645*sqrt((0.2*0.8)/60) = 0.2 + 1.645*0.05774 = 0.2949322. Rounding to four decimal places, we get an upper confidence limit of 0.2949.

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Answer:

The 95% confidence interval with the normla method would be given by (0.657;0.979)

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)

Step-by-step explanation:

1) Notation and definitions

[tex]X=18[/tex] number of blocks that were sufficiently strong

[tex]n=22[/tex] random sample taken

[tex]\hat p=\frac{18}{22}=0.818[/tex] estimated proportion of blocks that were sufficiently strong

[tex]p[/tex] true population proportion of blocks that were sufficiently strong

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

2) Confidence interval (Normal method)

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.657[/tex]

[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.979[/tex]

The 95% confidence interval with the normla method would be given by (0.657;0.979)

3) Confidence interval (Small sample method)

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

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n} \frac{N-n}{N-1}}[/tex]

But we need to know the total size for the population, and on this case we don't know but let's assumed that N=2000 for example .

If we replace the values obtained we got:

[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.784[/tex]

[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.852[/tex]

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)

Answer:

a) [tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]

b) [tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

[tex]Rate change = In-Out[/tex]

And we can name the rate of change as [tex]\frac{ds}{dt}=rate change[/tex]

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is [tex]\frac{s}{50gal}[/tex]

And now we can find the expression for the amount of salt after time t like this:

[tex]\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}[/tex]

And we have this differential equation:

[tex]\frac{dS}{dt} +\frac{1}{10} s = 40[/tex]

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

[tex]\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}[/tex]

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

[tex]e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt[/tex]

[tex]e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C[/tex]

Now we can divide both sides by [tex] e^{\frac{1}{25} t} [/tex] and we got:

[tex]s(t) =400 + C e^{-\frac{1}{10} t}[/tex]

Now we can apply the initial condition in order to solve for the constant C like this:

[tex]15 = 400+C[/tex]

[tex]C=-385[/tex]

And then our function would be given by:

[tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

[tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]

a) The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex].

b) There are 4400 ounces of salt in the tank after 25 minutes.

In this question we must model the salt concentration in the tank ([tex]c(t)[/tex]), in ounces per gallon, as a function of time ([tex]t[/tex]), in minutes, in which salt is dissolved into the tank due to a constant inflow rate ([tex]\dot V[/tex]), in gallons. Likewise, an equal outflow rate exists with resulting concentration.

a) The process is modelled mathematically by a non-homogeneous first order differential equation and physically by principle of mass conservation, whose description is shown below:

[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V_{T}}\cdot c(t) = \frac{\dot V}{V_{T}}\cdot c_{in}[/tex]   (1)

Where:

The solution of this differential equation is:

[tex]c(t) = c_{in} + \left(\frac{m_{T}}{V_{T}}-c_{in} \right)\cdot e^{-\frac{\dot V}{V_{T}}\cdot t }[/tex]   (2)

Where [tex]m_{T}[/tex] is the initial salt mass of the tank, in ounces.

And the salt mass in the tank at an arbitrary time [tex]t[/tex] ([tex]m(t)[/tex]), in ounces, is obtained by multiplying (2) by the volume of the tank. That is to say:

[tex]m(t) = c(t)\cdot V_{T}[/tex]   (3)

By replacing [tex]c(t)[/tex] in (3) by (2), we have the following expression:

[tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]   (4)

The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]. [tex]\blacksquare[/tex]

b) If we know that [tex]V_{T} = 50\,gal[/tex], [tex]\dot V = 55\,\frac{gal}{min}[/tex], [tex]c_{in} = 88\,\frac{oz}{gal}[/tex], [tex]m_{T} = 15\,oz[/tex] and [tex]t = 25\,min[/tex], then the quantity of salt is:

[tex]m(25) = (50)\cdot (88)+[15-(50)\cdot (88)]\cdot e^{-\left(\frac{55}{50} \right)\cdot (25)}[/tex]

[tex]m(25) = 4400\,oz[/tex]

There are 4400 ounces of salt in the tank after 25 minutes. [tex]\blacksquare[/tex]

To learn more on dissolution processes, we kindly invite to check this verified question: https://brainly.com/question/25752764

Answer:

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The individual with the higher IQ is the one with the higher z-score. So

Test A has a mean of 100 and a standard deviation of 13. An individual who scores 123 on Test A.

So [tex]\mu = 100, \sigma = 13, X = 123[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{123 - 100}{13}[/tex]

[tex]Z = 1.77[/tex]

Test B has a mean of 100 and a standard deviation of 18. An individual who scores 121 on Test B.

So [tex]\mu = 100, \sigma = 18, X = 121[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{121 - 100}{18}[/tex]

[tex]Z = 1.17[/tex]

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.

Answer:0.05

Step-by-step explanation:

There are 5 different bank accounts

Probability that A will give maximum benefit i.e. Jake save most in account A

[tex]P_1=\frac{1}{5}[/tex]

so picking 1 bank we are left with 4 banks . So we Probability that he will save least into account B is [tex]P_2=\frac{1}{4}[/tex]

Probability that Jake save most into account A and save least into account B is

[tex]=P_1\times P_2[/tex]

[tex]=\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}[/tex]

Answer:

a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]

b) [tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/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".  

Solution

Let put some notation  

x=value for right arm , y = value for left arm

x: 102, 101,94,79,79

y: 175,169,182,146,144

The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:

d: 73, 68, 88, 67, 65

Part a

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]

The third step would be calculate the standard deviation for the differences, and we got:

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]

Part b

The next step is calculate the degrees of freedom given by:

[tex]df=n-1=5-1=4[/tex]

Now we need to calculate the critical value on the t distribution with 4 degrees of freedom. The value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], so we need a quantile that accumulates on each tail of the t distribution 0.05 of the area.

We can use the following excel code to find it:"=T.INV(0.05;4)" or "=T.INV(1-0.05;4)". And we got [tex]t_{\alpha/2}=\pm 2.13[/tex]

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

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

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

[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.331[/tex]  

[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.069[/tex]  

So on this case the 90% confidence interval would be given by (63.331;81.069).

[tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/tex]  

Answer:

[tex]10.108 < \mu < 13.892[/tex]    

Step-by-step explanation:

1) 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".

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

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

s represent the sample standard deviation

n represent the sample size  

We have the following distribution for the random variable:

[tex]X \sim N(\mu , \sigma=0.45)[/tex]

And by the central theorem we know that the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

2) Confidence interval

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)

In order to calculate the mean and the sample deviation we can use the following formulas:  

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

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=12[/tex]

The sample deviation calculated [tex]s=2.268[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=8-1=7[/tex]

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that [tex]t_{\alpha/2}=2.36[/tex]

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

[tex]12-2.36\frac{2.268}{\sqrt{8}}=10.108[/tex]    

[tex]12+2.36\frac{2.268}{\sqrt{8}}=13.892[/tex]

So on this case the 95% confidence interval would be given by (10.108;13.892)

[tex]10.108 < \mu < 13.892[/tex]    

Answer: A) .1587

Step-by-step explanation:

Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.

i.e. [tex]\mu=12.30[/tex] and [tex]\sigma=0.20[/tex]

Let x denotes the amount of soda in any can.

Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.

Then, the probability that a randomly selected can will need to go through the mentioned process =  probability that a randomly selected can has more than 12.50 ounces of soda poured into it =

[tex]P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587[/tex]

Hence, the required probability= A) 0.1587

Final answer:

The probability that a randomly selected can will need a special cleaning process because it holds more than 12.50 ounces of soda is calculated using the Z-score. The correct answer is A) 0.1587.

Explanation:

The question asks us to calculate the probability that a randomly selected can will require a special cleaning process if it holds more than 12.50 ounces of soda, given a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. To find this probability, we can use the standard normal distribution, also known as the Z-score.

First, we calculate the Z-score for 12.50 ounces using the formula: Z = (X - \\mu\) / \\sigma\, where X is the value we are evaluating (12.50 ounces), \mu is the mean (12.30 ounces), and \sigma is the standard deviation (0.20 ounce). Plugging in the values gives us Z = (12.50 - 12.30) / 0.20 = 1. The Z-score represents the number of standard deviations the value X is from the mean.

Next, to find the probability that a can will overflow (have more than 12.50 ounces), we look up the corresponding probability in the standard normal distribution table or use a calculator suitable for such statistical computations. The probability corresponding to a Z-score of 1 is approximately 0.8413. However, since we are interested in the probability of a can having more than 12.50 ounces, we need the area to the right of the Z-score, which is 1 - 0.8413 = 0.1587. Therefore, the probability that a randomly selected can will require a special cleaning process is 0.1587.

The correct answer to the question is A) 0.1587.

Answer:

a) Null hypothesis: [tex]\mu =0[/tex]

Alternative hypothesis: [tex]\mu \neq 0[/tex]

b) t =-7.44

c) [tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]

d) Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.

We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.

Step-by-step explanation:

Assuming this output:

[tex]t_{difference}=-0.395[/tex]

t(observed value) =-7.44

t(Critical value )= 1.984

DF = 98

p value (two tailed) < 0.0001

[tex]\alpha =0.05[/tex]

a) What are the null and alternative hypothesis

Null hypothesis: [tex]\mu =0[/tex]

Alternative hypothesis: [tex]\mu \neq 0[/tex]

The reason is because the output says a bilateral test so for this case w eselect this option.

b) Identify the statistic

The correct formula for the statistic is given by:

[tex]t=\frac{\bar X_1 -\bar X_2 -0}{\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}}[/tex]

Based on the output the calculated value its t =-7.44

c) Identify the p value

We have the degrees of freedom given 98

Based on the alternative hypothesis the p value is given by:

[tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]

d) State the final conclusion that addresses the original claim

Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.

We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.

The correct conclusion is to not reject the null hypothesis.

If the p-value for a hypothesis test is 0.07 and the chosen level of significance is α = 0.05, then the correct conclusion is to not reject the null hypothesis.

When conducting a hypothesis test, the p-value represents the probability of obtaining the observed data or more extreme values assuming the null hypothesis is true. If the p-value is greater than the significance level (α), which is the threshold for rejecting the null hypothesis, we fail to reject the null hypothesis.

In this case, the p-value of 0.07 is greater than the significance level of 0.05. Therefore, we cannot reject the null hypothesis.

Answer:

B. II only

Step-by-step explanation:

For this case the correct options would be:

B. II only

We analyze one by one the statements:

I.This study provides good evidence that walking is effective in controlling cholesterol

That's FALSE because there are many other factors that can influence the cholesterol and we just have two possible conditions (women who walk at least 10 miles and women who do not exercise at all), and we have just two averages obtained from two samples that we don't know if is a paired values or independent values. And is not appropiate to conclude this with the lack of info on this case.

II. This is an observational study, not an experiment

Correct we don't have a design for this experiment or factors selected in order to test the hypothesis of interest. So for this case we can conclude that we have an observational study for this case.

III. Although the study was conducted only on women, we can confidently generalize the results to men in the same age group.

False, we can't generalize results from women to men since they are different groups of people and with different characteristics.

Answer:

the position equation of the projectile are

[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]

[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]

in x and y direction

Step-by-step explanation:

let the mass of the projectile be m, initial velocity be u

the wind applies a force of 3 newton in east direction.

therefore acceleration due to the force in east direction =[tex]\frac{force}{mass}[/tex]

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

acceleration due to gravity is in south direction = g

let east be x direction and north be y direction.

therefore acceleration in x direction = 3[tex]\frac{m}{s^{2} }[/tex] and in y direction = -g[tex]\frac{m}{s^{2} }[/tex]

writing equation of motion in x and y direction:

[tex]x = u_{x} t + \frac{1}{2} at^{2}[/tex]

[tex]y = u_{y} t + \frac{1}{2} at^{2}[/tex]

[tex]u_{x}[/tex]= ucos45 = [tex]\frac{300}{\sqrt{2} }[/tex]

[tex]u_{y}[/tex]= usin45= [tex]\frac{300}{\sqrt{2} }[/tex]

therefore

[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]

[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]

here 2 is added as the projectile already 2 meter above the ground

Answer:

B. 1635

Step-by-step explanation:

We have been given that the seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row.

We can see that the seating order is in form of arithmetic sequence, whose first term is 11 and common difference is 3.

Since there are 30 rows altogether, so we need to find the sum of 30 first terms of the sequence using sum formula.

[tex]S_n=\frac{n}{2}[2a+(n-1)d][/tex], where,

[tex]S_n=[/tex] Sum of n terms,

n = Number of terms,

a = First term,

d = Common difference.

Upon substituting our given value is above formula, we will get:

[tex]S_n=\frac{30}{2}[2(11)+(30-1)3][/tex]

[tex]S_n=15[22+(29)3][/tex]

[tex]S_n=15[22+87][/tex]

[tex]S_n=15[109][/tex]

[tex]S_n=1635[/tex]

Therefore, there are 1635 seats in all and option B is the correct choice.

Answer: the total number of seats is 1635

Step-by-step explanation:

For each additional row after the first row, there are 3 more seats than there are in the previous row. This means that the seats in each row is increasing in arithmetic progression with a common difference of 3. The formula for determining the sum of n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the sequence.

a represents the first term,

d represents the common difference.

From the information given,

a = 11

n = 30

d = 3

Therefore,

S30 = 30/2[2×11 + (30 - 1)3]

S30 = 15[22 + 29×3]

S30 = 15 × 109= 1635

Answer:

[tex]5.52*10^{26}[/tex]

Step-by-step explanation:

For the columns with 5 slots, there are 15 distinct options to fill in. The number of ways to fill them in is

15 * 14 * 13 * 12 * 11 = 360360 ways (order matters)

For the column with 4 slots and 15 options. The number of ways to fill them in is

15 * 14 * 13 * 12 = 32760  ways (order matters)

Since a bingo card has 4 columns of 5 slots and 1 column with 4 slots, the total number of combination there is:

[tex]360360^4*32760 \approx 5.52*10^{26}[/tex] (order matters)