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Given: Sector BAC with r=8

Radius of inscribed circle O is 2

Find the area of the sector BAC

Answer:

a) Q = 122 units/order

b) Number of orders = 2.05 orders/year

c) Average inventory = 61 units

d) Ordering costs = 125 $/order

Step-by-step explanation:

The economic quantity order (EOQ) formula allow us to minimize the ordering cost, in function of the demand, ordering cost and holding cost.

The EOQ formula is:

[tex]EOQ=\sqrt{\frac{2DS}{H} }[/tex]

where:

D: demand in units/year

S: Order costs, per order

H: holding or carrying cost, per unit a year

a) In this case:

D: 250 u/year

S: 30 $/order

H: 1 $/year-unit

[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*30}{1} }=\sqrt{15000}=122.47\approx122[/tex]

b) If we have a demand of 250 units/year and we place orders of 122 units, the amount of orders/year is:

[tex]\#orders=\frac{D}{EOQ}=\frac{250\,units/year}{122\,units/order}=2.05\, \frac{orders}{year}[/tex]

c) We assume that there is no safety stock, so everytime the stock hits 0 units, a new order enter the inventory.

In this case, the average inventory can be estimated as the average between the inventory when a new order enters the inventory (122 u.) and the inventory right before a order enters (0 u.)

[tex]\#av.inventory=\frac{122+0}{2}=61[/tex]

The average inventory is 61 units.

d) If 250 units is the optimal quantity for an order, it means it is equal to the EOQ. We can calculate the new ordering costs as:

[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*S}{1} }=250\\\\2*250*S=250^2\\\\S=250/2=125\,\$/order[/tex]

A) To minimize cost, the number of units that should be ordered each time an order is placed is; 122 units

B) The number of orders per year needed with the optimal policy is; 2 orders per year.

C)  The average inventory if costs are minimized is; 61 units

D) For the order policy of Q = 250 to be optimal, the ordering cost would have to be; $125 per order

The formula for economic quantity order (EOQ) is given as;

EOQ = √(2DS/H)

Where;

We are given;

Annual demand; D = 250 units/year

Holding Cost; H = $1 per unit per year

Set up costs; S = $30 per order

EOQ = √(2 × 250 × 30/1)

EOQ = 122.47

But EOQ has to be a whole number and so we approximate to the nearest whole number to get;

EOQ = 122

n = D/EOQ

Plugging in the relevant values gives;

n = 250/122

n = 2.049

But number of orders has to be a whole number. Thus, we approximate to the nearest whole number to get;

n = 2 orders per year

Before a new order enters the inventory is 0 if we assume that there is no safe stock. Thus;

average inventory = (122 + 0)/2 = 61 units

EOQ = √(2DS/H)

⇒ 250 = √(2 × 250 × S/1)

Square both sides to get;

250² = 500S

S = 250²/500

S = $125 per order

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Answer:

Step-by-step explanation:

1) in a rhombus, the diagonals bisect each other at the midpoint, forming 4 right angles.

MK = NK + NM

Since MK = 24 and NK = NM, then,

NK = 24/2 = 12

JL = NJ + NL

Since JL = 20 and NJ = NMlL, then,

NL = 20/2 = 10

Looking at triangle LMN, it is a right angle triangle. Applying Pythagoras theorem,

ML^2 = 10^2 + 12^2 = 100 + 144 = 244

ML = √244 = 15.62

Since all the sides of the rhombus are equal, then

MJ = ML = 15.62

Angle KNL = angle JML = 90 degrees. This is so because the diagonals are perpendicular)

Angle KJL = 90 - angle MJL

Angle KJL = 90 - 50 = 40 degrees.

Since the opposite angles of the rhombus are equal,

Angle MLK = angle angle MJK = 50 + 40 = 90 degrees

2) since all the sides of a rhombus are equal,

5x + 16 = 9x - 32

9x - 5x = 32 + 16

4x = 48

x = 48/4 = 12

Since PQ = NR = 5x + 16, then

PQ = 5×12 + 16 = 60 + 16

PQ = 76

3) Triangle XYZ is an isosceles triangle. This means that its base angles, XZY and ZXY are equal. The sum of angles in a triangle is 180 degrees. Therefore,

Angle XZY + angle ZXY = 180 - 136

Angle XZY + angle ZXY = 44

Angle XZY = 44/2 = 22

Therefore,

10x - 8 = 22

10x = 22 + 8 = 30

x = 30/10 = 3

Answer:

0.001%

Step-by-step explanation:

Given that a recent study conducted by the state government attempts to determine whether the voting public supports a further increase in cigarette taxes to help the public schools

n = 1500

favour x = 622

Sample proportion p = [tex]622/1500 = 0.4147[/tex]

[tex]H_0: p = 0.66\\H_a: p >0.66[/tex]

(right tailed test)

Assume H0 to be true.

Std error = [tex]\sqrt{\frac{0.66*0.34}{1500} } \\=0.01223[/tex]

p difference = -0.2453

Test statistic Z = [tex]\frac{-0.2453}{0.01223} \\=-20.05[/tex]

p value <0.00001

Hence fon ull hypothesis not to be rejected significant level should be greater than 0.00001 or 0.001%

Answer:

D. Discrete

Step-by-step explanation:

The number of reported car accidents is a countable variable, that can only be natural values.

You can have 0, 1, 2, 3, 4, ..., 100, ..., 1000 reported car accidents in a given month.

You cannnot have 4.5 reported car accidents in a given month, for example.

So the correct answer is:

D. Discrete

The type of variable that represents the number of auto accidents reported in a given month is a discrete variable, as it can only take certain values (0, 1, 2, etc.), and never a fraction or decimal.

The type of variable that best describes the number of auto accidents reported in a given month is Discrete.

In statistics, variables are classified into different types based on their properties. A discrete variable is one that can only take certain values. This is the case for the number of auto accidents in a month, which could be 0, 1, 2, 3, etc., but never a fraction or decimal value like 2.5.

In contrast, a continuous variable could take any value within a certain range, including fractions and decimals. Meanwhile, interval and ratio are types of data measurements that don't apply in this context.

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Substituting [tex]z[/tex] from the cone's equation,

[tex]z=\dfrac13\sqrt{x^2+y^2}[/tex]

into the equation of the sphere,

[tex]x^2+y^2+z^2=4[/tex]

gives the intersection of the two surfaces,

[tex]x^2+y^2+\left(\dfrac13\sqrt{x^2+y^2}\right)^2=4\implies x^2+y^2=\dfrac{18}5[/tex]

which is a circle of radius [tex]\sqrt{\frac{18}5}[/tex] centered at [tex]\left(0,0,\frac13\sqrt{\frac{18}5}\right)[/tex].

We parameterize this part of the sphere outside the cone (call it [tex]S[/tex]) by

[tex]\vec s(u,v)=\langle2\cos u\sin v,2\sin u\sin v,2\cos v\rangle[/tex]

with [tex]0\le u\le2\pi[/tex] and [tex]\cos^{-1}\frac1{\sqrt{10}}\le v\le\pi[/tex].

Take the normal vector to [tex]S[/tex] to be

[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle4\cos u\sin^2v,4\sin u\sin^2v,4\cos v\sin v\rangle[/tex]

Then the area of [tex]S[/tex] is

[tex]\displaystyle\iint_S\mathrm dA=\int_0^{2\pi}\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|\,\mathrm dv\,\mathrm du[/tex]

[tex]=\displaystyle2\pi\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi4\sin v\,\mathrm dv=\boxed{\frac{40+4\sqrt{10}}5\pi}[/tex]

Answer: The point estimate of the population mean is 25.

The margin of error for the confidence interval is 3 .

Step-by-step explanation:

The confidence interval for population mean is given by :-

[tex](\overline{x}-E , \overline{x}+E)[/tex] , here [tex]\overline{x}[/tex] is the point estimate of the population mean and E is the margin of error .

As per given , we have

Lower bound of CI = [tex]\overline{x}-E =22[/tex]   (1)

Upper bound of CI =  [tex]\overline{x}+E =28[/tex]  (2)

Add (1) and (2) , we get

[tex]2\overline{x}=50\\\Rightarrow\ \overline{x}=25[/tex]

Subtract (1) from (2) , we get

[tex]2E=6\\\Rightarrow\ E=3[/tex]

Hence, the point estimate of the population mean is 25.

The margin of error for the confidence interval is 3 .

Answer:

1. The point estimate for population mean is 25.

2)

[tex]\text{Margin of error} = \pm 3[/tex]

Step-by-step explanation:

We are given the following information in the question:

Confidence interval: (22,28)

Confidence interval is calculated as:

[tex]\text{Sample mean }\pm \text{ Margin of error}[/tex]

Thus, we can write the equations:

[tex]\bar{x} - \text{Margin of error} = 22\\\bar{x} + \text{Margin of error} = 28[/tex]

1) The point estimate of the population mean

To calculate the point estimate of the population mean we solve the two equations, to find the sample mean

Adding the two equations we get:

[tex]2\bar{x} = 22+ 28 = 50\\\bar{x} = 25[/tex]

Thus, the point estimate for population mean is 25.

2) The margin of error for the confidence interval

Putting the values from the equation, we get:

[tex]\text{Margin of error} = 28 - 25 = 3[/tex]

Thus, the margin of error f the given confidence interval is

[tex]\text{Margin of error} = \pm 3[/tex]

Answer:

The standard deviation of the distribution is 492.847

Step-by-step explanation:

Consider the provided information.

A city has 1,091,953 residents. A recent census showed that 364,656 of these residents regularly use the city's public transportation system.

Therefore the value of n is 1,091,953

The probability of success is: [tex]\frac{364,656 }{1,091,953}=0.33395[/tex]

Calculate the standard deviation using the formula: [tex]\sigma=\sqrt{np(1-p)}[/tex]

Substitute the respective values.

[tex]\sigma=\sqrt{1091953(0.334)(1-0.334)}[/tex]

[tex]\sigma=\sqrt{242898.3931}[/tex]

[tex]\sigma=492.847[/tex]

Hence, the standard deviation of the distribution is 492.847

Answer:

z-test statistic is 3.38

Step-by-step explanation:

Null and alternative hypotheses are:

[tex]H_{0}[/tex]: the proportion of papers that lacked statistical assistance  sent back without review is the same as the proportion of papers with statistical help sent back without review

[tex]H_{a}[/tex]: the proportion of papers that lacked statistical assistance sent back without review is different than the proportion of papers with statistical help sent back without review

Test statistic can be found using the equation:

[tex]z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

Then [tex]z=\frac{0.71-0.57}{\sqrt{{0.61*0.39*(\frac{1}{190} +\frac{1}{514}) }}}[/tex] ≈ 3.38

Answer:

All of them are polynomial functions

Step-by-step explanation:

Remember that a polynomial function of x is a function whose value f(x) is always equal to [tex]f(x)=a_0+a_1x+a_2x^2+\cdots a_nx^n[/tex] for a fixed n≥0 (the degree of f) and fixed coefficients [tex]a_i\in\mathbb{R}[/tex]

For example, [tex]f(x)=x^2+3x[/tex] is a polynomial function, but [tex]g(x)=2^x+x[/tex] is not because [tex]2^n[/tex] is not a nonnegative power of x. Another example of a non-polynomial function is [tex]g(x)=x^{-1}=\frac{1}{x}[/tex].

f(x)=4⋅11x is polynomial with degree 1 and [tex]a_0=0,a_1=4\cdot 11[/tex]. For the same reasons, f(x)=3⋅18x and f(x)=10⋅17x are polynomial functions.

f(x)=−4x³−4x²+5x+1 is a polynomial function of degree 3 with [tex]a_0=1,a_1=5, a_2=a_3=-4[/tex]. and f(x)=−2x−1 is a polynomial function of degree 1 and coefficients [tex]a_0=-1,a_1=-2[/tex].

In mathematics, a polynomial function involves operations of addition, subtraction, multiplication, and non-negative, whole-number exponents of variables. Here, the polynomial functions from the given list are: f(x)=−4x3−4x2+5x+1 and f(x)=−2x−1.

In the realm of mathematics, a polynomial function is an expression consisting of variables and coefficients, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents of variables. When asked to determine which of the given functions are polynomial functions, we can apply this definition. Your options were: f(x)=4⋅11x, f(x)=3⋅18x, f(x)=10⋅17x, f(x)=−4x3−4x2+5x+1, and f(x)=−2x−1.

By our definition, the polynomial functions from the list are:  f(x)=−4x3−4x2+5x+1 and f(x)=−2x−1. The other three functions are not polynomial functions as they do not comply with the characteristics of polynomial functions.

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Answer:

$6

Explanation:

The standard deviation gives an idea of the dispertion of values in those 12 month. A hight value of standard deviation means that the prices changed in big increments ( for example one month is $20, other month is $60 and other $85). If that  would be the case, is important to investors to know it and the deviation should be reported.

So, given that it isnt reported, we can say that the price vary but only a few dollars, resulting in a small standard deviation such as $6.

Final answer:

A standard deviation of $6 for a blue chip stock is more likely than $26 or $60, as blue chip stocks are typically stable with less price volatility.

Explanation:

Regarding the standard deviation of the blue chip stock price, it is unlikely that the standard deviation would be very high considering the nature of blue chip stocks. Blue chip stocks are known for their stability and are typically less volatile than other types of stocks. Therefore, a standard deviation of $60 would suggest a very high level of volatility, which is uncharacteristic of blue chip stocks. A standard deviation of $6 is more plausible because it indicates minor fluctuations around the average price of $72, which is more in line with the expected behavior of blue chip stocks. A standard deviation of $26 is possible but less likely than $6, given that such a value still represents a fairly high level of volatility for blue chip stocks, which are commonly considered as safe, long-term investments with relatively steady prices.

Answer:

B. Rational functions

Step-by-step explanation:

The partial fraction decomposition is used for functions there are described by fractions, and for which the substitution method is not possible. These are rational functions, in which both the numerator and the denominator are polynomials.

So the correct answer is:

B. Rational functions

Partial fraction decomposition is used to simplify complex rational functions to make them easier to integrate. Rational function is the correct answer.

The correct answer is B. Rational functions. Partial fraction decomposition is a mathematical technique used essentially to simplify complex rational functions. A rational function is a function that can be defined as the ratio of two polynomials. By breaking down a complex rational function into simpler fractions (which is what partial fraction decomposition does), integration becomes more manageable. For example, it's easier to integrate simple fractions like 1/x or 2/x^2, which would be the result of a partial fraction decomposition, than complex, intertwined expressions.

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Answer: At-least 88.89%

Step-by-step explanation:

As per given , we have

Population mean : [tex]\mu=\$3.73[/tex]

Standard deviation :  [tex]\sigma=\$0.10[/tex]

Now , $3.43= $⁢3.73- 3(0.10) = [tex]\mu-3\sigma[/tex]

$⁢4.03 = $⁢3.73+3(0.10) = [tex]\mu+3\sigma[/tex]

i.e. $3.43 is 3 standard deviations below mean and $⁢4.03 is 3 standard deviations above mean .

To find :  the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03.

i.e. to find minimum percentage of stores that sell a gallon of milk  lies within 3 standard deviations from mean.

According to Chebyshev, At-least [tex](1-\dfrac{1}{k^2})[/tex] of the values  lies with in [tex]k\sigma[/tex] from mean.

For k= 3

At-least [tex](1-\dfrac{1}{3^2})[/tex] of the values lies within [tex]3\sigma[/tex] from mean.

[tex]1-\dfrac{1}{3^2}=1-\dfrac{1}{9}=\dfrac{8}{9}[/tex]

In percent = [tex]\dfrac{8}{9}\times100\%\approx88.89\%[/tex]

Hence, the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03 =  At-least  88.89%

Final answer:

Using Chebyshev's Theorem, the minimum percentage of stores selling a gallon of milk between $3.43 and $4.03 is at least 88.9% when the mean price is $3.73 and the standard deviation is $0.10.

Explanation:

Using Chebyshev's Theorem, we can determine the minimum percentage of stores selling a gallon of milk within a certain range of prices given the mean and standard deviation. The theorem states that for any number k, where k > 1, at least (1 - 1/k²) of the data values will fall within k standard deviations of the mean. In this case, the range of prices is from $3.43 to $4.03, which is $0.30 away from the mean of $3.73 on either side.

To find k, we divide the distance from the mean by the standard deviation: k = 0.30 / 0.10 = 3. Thus, at least (1 - 1/3²) or (1 - 1/9) of the stores sell a gallon of milk within this range. Calculating this, we get at least (1 - 1/9) = 8/9 or approximately 88.9% of stores.

Therefore, according to Chebyshev's Theorem, the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03 is at least 88.9%.

Answer:

Experimental smoothing forecast for the next period will be 58.9

Step-by-step explanation:

We have given actual demand is [tex]A_{t-1}=61[/tex]

Initial forecast [tex]f_{t-1}=61[/tex] and [tex]\alpha =0.3[/tex]

We have to find the experimental smoothing forecast [tex]f_t[/tex]

Experimental forecast is given by

[tex]f_t=(1-\alpha )f_{t-1}+\alpha A_{t-1}=(1-0.3)\times 58+0.3\times 61=58.9[/tex]

So experimental smoothing forecast for the next period will be 58.9

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

[tex]n_1 [/tex] represent the sample after the preparation

[tex]n_2 [/tex] represent the sample without preparation  

[tex]\bar x_1 =678[/tex] represent the mean sample after preparation

[tex]\bar x_2 =1837[/tex] represent the mean sample without preparation

[tex]s_1 =197[/tex] represent the sample deviation after preparation

[tex]s_2 =328[/tex] represent the sample deviation without preparation

[tex]\alpha=0.1[/tex] represent the significance level

Confidence =90% or 0.90

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_s}+\frac{s^2_2}{n_s})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex]

The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2[/tex]

Since the Confidence is 0.90 or 90%, 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: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

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

After replace in the formula for the confidence interval we got this:

[tex]3.0 < \mu_1 -\mu_2 <13.0 [/tex]

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Answer:

(7, 8)

Step-by-step explanation:

If Sidney places each piece 3 units east and 2 units north of the last one, we can figure out which coordinate comes first by subtracting the position of the second piece (S) by the position of the first piece (F):

[tex]S - F = (4,6) - (1,4) = (3,2)[/tex]

We can conclude that coordinates are in the (East, North) format.

Therefore, the location of the third piece (T) is:

[tex]T = S(E,N) +(3,2)= (4,6)+(3,2)\\T = (7,8)[/tex]

Answer:

Paired  t test

Step-by-step explanation:

Given that you collect a random sample of 28 adult golfers and record two scores for each: one taken before the subject receives professional coaching and one taken after.

Here subject of interest is to study whether the professional coaching really improves the scores.

For this two groups are taken from golfers and they were given chances to play and scores were recorded before and after coaching.  The same subject with two different scores recorded and the difference calculated.  Hence here the appropriate test is paired t test

Statistic should be t because population std devition is not known.

Hence paired t test for comparison of mean differences before and after should be done.

Answer:

The answer is b): Yes, and it is highly probable.

Step-by-step Explanation:

Yes, it is highly probable that the mean BAC of all drivers involved in fatal accidents and found to have positive BAC values, is less than the legal intoxication level because blood alcohol concentration (BAC) level is not the most/only factor that determines fatal accidents. Other types of human errors are the main causes of fatal accidents; they include: side distractions, avoiding the use of helmets and seat belts, over-speeding, beating traffic red lights, overtaking in an inappropriate manner, using the wrong lane, etc. Most times, these errors are not caused by BAC levels.

Answer:

Option a is right

Step-by-step explanation:

Given that a random sample of 250 men yielded 175 who said they'd ridden a motorcycle at some time in their lives, while a similar sample of 215 women yielded only 43 that had done so.

For proportions since binomial and sample size large we can use  z critical values.

Sample             I            II

N                    250        215     465

X                     175           43    218

p                       0.7          0.2    0.4688

p difference = 0.5

Std error of difference = [tex]\sqrt{p(1-p)(\frac{1}{n_1}+  \frac{1}{n_2} }\\=\sqrt{0.4688*0.5312)(\frac{1}{250} + \frac{1}{215} )}\\=0.0409[/tex]

Margin of error for 99% = 2.58*std error =  0.105

Confidence interval 99% = (0.5±0.105)            

Option a is right.

Answer: 0.6

Step-by-step explanation:

Formula to get the standardized value :

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]  (1)

, where x= random variable , [tex]\mu[/tex] = Population mean and [tex]\sigma[/tex] = population standard deviation.

As per given , we have

[tex]\mu=25[/tex]

[tex]\sigma=5[/tex]

To find standardized value for the score of x=28

[tex]z=\dfrac{28-25}{5}=0.6[/tex]   (substitute values in (1))

Hence, the standardized value for the score of 28 is 0.6 .

Answer:

[tex] y= 2x +1[/tex]

By direct comparison we can see that m =2 (slope) and b =1 (intercept)

Step-by-step explanation:

Assuming the following function:

[tex] y-2x =1[/tex]

We want to find the following general equation for a linear model:

[tex]y = mx +b [/tex]

On this case we just need to add 2x on both sides of the original equation and we got:

[tex] y= 2x +1[/tex]

By direct comparison we can see that m =2 and b =1

[tex] mx + b = 2x +1[/tex]

[tex]mx=2x , m =2[/tex]

[tex] b =1[/tex]

the value of m on this case represnt the slope and b the intercept.

The slope is defined by the following formula:

[tex] m =\frac{\Delta y}{\Delta x}[/tex]

And is the interpretation is the rate of change of y respect to x, can be positive or negative. Or the increase/decrease of y when x increase 1 unit.

And the value b=1 represent the y intercept, that means if x=0 then y =1

Answer:

a) [tex]y = 3.8 x +100[/tex]

b) [tex]Abs. change= |168.4-167.5|=0.9[/tex]

So the calculated value is 0.9 points above the actual value.

[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%

And the calculated value it's 0.537% higher than the actual value.

c) For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.

Step-by-step explanation:

Data given

1982 , CPI=100

1986, CPI = 191.2

Notation

Let CPI the dependent variable y. And the time th independent variable x.

For this case we want to adjust a linear model givn by the following expression:

[tex]y=mx+b[/tex]

Solution to the problem

Part a

For this case we can find the slope with the following formula:

[tex] m =\frac{CPI_{2006}-CPI_{1982}}{2006-1982}[/tex]

And if we replace we got:

[tex] m =\frac{191.2-100}{2006-1982}=3.8[/tex]

Let X represent the number of years after. Then for 1982 t = 0, and if we replace we can find b:

[tex] 100 = 3.8(0)+b[/tex]

And then [tex]b=100[/tex]

So then our linear model is given by:

[tex]y = 3.8 x +100[/tex]

Part b

For this case we need to find the years since 1982 and we got x = 2000-1982=18, and if we rpelace this into our linear model we got:

[tex]y = 3.8(18) +100=168.4[/tex]

And the actual value is 167.5 we can compare the result using absolute change or relative change like this:

[tex]Abs. change= |168.4-167.5|=0.9[/tex]

So the calculated value is 0.9 points above the actual value.

And we can find also the relative change like this:

[tex]Relative. Change =\frac{|Calculated -Real|}{Real}x100[/tex]

And if we replace we got:

[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%

And the calculated value it's 0.537% higher than the actual value.

Part c

For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.

Answer:

Equation: [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]

Intersection: [tex](y-2)^2+(z-5)^2=7[/tex]

Step-by-step explanation:        

We are asked to write an equation of the sphere with center center [tex](-3,2,5)[/tex] and radius 4.    

We know that equation of a sphere with radius 'r' and center at [tex](h,k,l)[/tex] is in form:

[tex](x-h)^2+(y-k)^2+(z-l)^2=r^2[/tex]

Since center of the given sphere is at point [tex](-3,2,5)[/tex], so we will substitute [tex]h=-3[/tex], [tex]k=2[/tex], [tex]l=5[/tex] and [tex]r=4[/tex] in above equation as:

[tex](x-(-3))^2+(y-2)^2+(z-5)^2=4^2[/tex]

[tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]

Therefore, our required equation would be [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex].

To find the intersection of our sphere with the yz-plane, we will substitute [tex]x=0[/tex] in our equation as:

[tex](0+3)^2+(y-2)^2+(z-5)^2=16[/tex]

[tex]9+(y-2)^2+(z-5)^2=16[/tex]

[tex]9-9+(y-2)^2+(z-5)^2=16-9[/tex]

[tex](y-2)^2+(z-5)^2=7[/tex]

Therefore, the intersection of the given sphere with the yz-plane would be [tex](y-2)^2+(z-5)^2=7[/tex].

Answer:both students are incorrect

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

Slope = (y2 - y1)/(x2 - x1)

y2 = 1

y1 = 4

x2 = 6

x1 = - 3

Slope = (1 - 4)/(6 - -3) = -3/9 = -1/3

To determine the intercept, we would substitute m = -1/3, x = 6 and y = 1 into y = mx + c. It becomes

1 = -1/3 × 6 + c = -2 + c

c = 1 + 2 = 3

The equation becomes

y = -x/3 + 3

If the equation was written in the slope intercept form which is expressed as

y - y1 = m(x - x1)

It becomes

y - 4 = -1/3(x - - 3)

y - 4 = -1/3(x + 3)

Both students are incorrect

Multiply his rate by number of hours:

30.40 x 30 = $912

Shane would make $912 if he works 30 hours at his normal work rate.

To calculate how much Shane makes if he works 30 hours at an hourly wage of $30.40, we can multiply his hourly wage by the number of hours he works:

$30.40/hour x 30 hours = $912

Therefore, Shane would make $912 if he works 30 hours at his normal work rate.

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Answer:

3.573 to 4.127

Step-by-step explanation:

Given

Sample size = 15

Mean = Sum of ratings/ sample size

Mean = 57.7/15

Mean = 3.85

Degree of freedom = sample size - 1

Degree of freedom = 15 - 1 = 14

df = 14

Then we calculate the standard deviation

(x - mean)² ||

(3.6 - 3.85)² || 0.0625

(2.9 - 3.85)² || 0.9025

(3.8 - 3.85)² || 0.0025

(4.5 - 3.85)² || 0.4225

(3.2 - 3.85)² || 0.4225

( 3.9 - 3.85)² || 0.0025

( 3.3 - 3.85)² || 0.3025

( 4.6 - 3.85)² || 0.5625

(4.1 - 3.85)² || 0.0625

(4.3 - 3.85)² || 0.2025

4.4 - 3.85)² || 0.3025

( 3.9 - 3.85)² || 0.0025

(3.2 - 3.85)² || 0.4225

( 4.2 - 3.85)² || 0.1225

( 3.8 - 3.85)² || 0.0025

Total || 3.7975

Variance = 3.7975/15 = 0.253167

Standard Deviation = √0.253167 = 0.50315703314174194

Standard Deviation = 0.5 ------- Approximated

The next step is to subtract the confidence level from 1, then divide by two.

i.e (1 - 0.95)/2 = 0.025

α = 0.025

Then we look up this answer to step in the t-distribution table.

For 14 degrees of freedom (df) and α = 0.025, my result is 2.145

The next step is to divide the sample standard deviation by the square root of the sample size.

0.5 / √15 = 0.129

Next is to multiply this result by step 2.145 (from the t table)

0.129 × 2.45 = 0.277

For the lower end of the range, subtract 0.277 from the sample mean.

3.85 – 0.277 = 3.573

Step 7: For the upper end of the range, add step 0.277 to the sample mean.

3.85 + 0.277 = 4.127

Answer: C. 461

Step-by-step explanation:

We know that the formula to find the sample size is given by :-

[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]

, where [tex]\sigma[/tex] = Population standard deviation from prior study.

E = margin of error.

z* = Critical value.

As per given , we have

[tex]\sigma=125[/tex]

E= 15

Significance level : [tex]\alpha=0.01[/tex]

Critical value (Two tailed)=[tex]z^*=z_{\alpha/2}=z_{0.005}=2.576[/tex]

Now , Required sample size = [tex]n=(\dfrac{(2.576)\cdot 125}{15})^2[/tex]

[tex]n=(21.4666666667)^2\\\\ n=460.817777779\approx461[/tex] [Round to next integer]

Hence, the required sample size to be taken is 461.

Correct answer = C. 461

Sampling is most appropriate

option c

In the given multiple choice situations, the use of sampling would most probably be appropriate in the situation where 'The need for precise information is less important'. Sampling is a procedure of statistical analysis where a predetermined number of observations are taken from a larger population. It is typically used when it is impracticable or expensive to study the entire population. It's an ideal method when having precise and exact information is not as crucial, however, the results will be sufficient to make accurate predictions or conclusions about a larger group.

On the other hand, sampling may not be substantially beneficial for smaller populations or where the likelihood of selecting a representative sample is low because the results might be skewed or non-representative of the larger group. In such conditions, examining the entire population or utilizing another form of data collection would be more advantageous.

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Answer:

Based on the model, the length of the wall is [tex]\frac{9}{8}[/tex] ft, the width of the wall is [tex]\frac{1}{2}[/tex] ft,  and the height of the wall is [tex]\frac{11}{8}[/tex] ft. The volume of the portion of security wall that  Tim has constructed so far is [tex]\frac{99}{128}[/tex]  cu ft.

Step-by-step explanation:

Given:

The figure constructed shows a rectangular prism made up of small wooden cubes of length [tex]\frac{1}{8}\ ft[/tex].

Width of the prism = [tex]\frac{1}{2}\ ft[/tex].

To find length , height and volume of the figure.

Solution:

From the figure we can conclude that :

Length side of the prism counts 9 cubes.

Thus, length of the prism will be given as :

⇒ [tex]\textrm{Length of each cube}\times \textrm{Number of cubes}[/tex]

⇒ [tex]\frac{1}{8}\ ft\times 9[/tex]

⇒ [tex]\frac{9}{8}\ ft[/tex]  (Answer)

Height side of the prism counts 11 cubes.

Thus, height of the prism will be given as :

⇒ [tex]\textrm{Length of each cube}\times \textrm{Number of cubes}[/tex]

⇒ [tex]\frac{1}{8}\ ft\times 11[/tex]

⇒ [tex]\frac{11}{8}\ ft[/tex] (Answer)

Volume of the prism can be given as :

⇒ [tex]Length\times width\times height[/tex]

⇒ [tex]\frac{9}{8}\ ft\times \frac{1}{2}\ ft\times \frac{11}{8}\ ft  [/tex]

⇒ [tex]\frac{99}{128}\ ft^3[/tex] (Answer)

Answer: a. 4

Step-by-step explanation:

The number of degrees of freedom represents the number of data-values in the evaluation of a test-statistic that are independent to vary.

The degree of freedom for chi-square test = n-1 , n= Sample size.

Given : A study was conducted to examine whether the proportion of females was the same for five groups (Groups A, B, C, D, and E).

i.e. n= 5

Then, the number of degrees of freedom the χ2 test statistic have when testing the hypothesis that the proportions in each group are all equal = 5-1=4

Hence, the correct answer is OPTION a. 4 .

Final answer:

The cutoff time to be in the fastest 10% of job applicants is approximately 118.0 seconds or less, calculated using a z-score of the 10th percentile in a normal distribution.

Explanation:

The student is asking about finding the cutoff time for the fastest 10% of job applicants performing a certain task, given that the time taken is normally distributed with a mean of 150 seconds and a standard deviation of 25 seconds. To solve this problem, we will use the z-score corresponding to the fastest 10% (the 10th percentile) in a normal distribution.

First, we look up the z-score for the 10th percentile in the z-table, which is approximately -1.28. Then we use the z-score formula:

z = (X - μ) / σ

Plugging in the known values:

-1.28 = (X - 150) / 25

Now, we solve for X:

X = -1.28 × 25 + 150

X = -32 + 150

X = 118 seconds

So, job applicants must complete the task in approximately 118.0 seconds or less to qualify for the advanced training.