A study was conducted on the amount of time drivers wait for a stoplight to change at a particular intersection. The amount of time spent by 300 drivers was recorded and the resulting data were used to create boxplot.

a. What is approximately the median amount of time spent at this traffic light?
b. The top 25% of drivers waited at least how long?
c. The mean amount of time spent at this traffic light was bigger or smaller than the median? Explain.

Answer:

Step-by-step explanation:

) Do you expect the distribution of number of friends made online to be symmetric, skewed to the right, or skewed to the left?

Skewed to the left.

Symmetric

Skewed to the right.

(b) Two measures of center for this distribution are 1 friend and 5.3 friends. Which is most likely to be the mean and which is most likely to be the median?

Mean=

Median=

a)

as proportion of people with 0 friends is 43% whcih is on left side and maximum ; and % decrease with increasing number of friends

skewed to the right

b)

as it is skewed to the right ; therefore mean is greater than median

mean=5.3

median=1

[since for skwewed to the right distribution :mean is always greater than median, therefore higher value should be mean which is 5.3 and lower value is median which is 1]

Part(a): Skewed to the right

Part(b) The required values are,

mean=5.3

median=1

a)

As a proportion of people with 0 friends are 43% which is on the left side and maximum; and % decrease with an increasing number of friends

skewed to the right

b)

As it is skewed to the right; therefore mean is greater than the median

mean=5.3

median=1

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Answer:

Step-by-step explanation:

F(x) results in a parabola with vertex (0,0) wich mean there is only one x-int at that point. g(x) has been shifted the grapgh of f(x) to the right by to units and down by three unites. Now our vertex lies in the point (2,-3) and since the graph was move dow i=of the x-axis we now have two different x-intercepts.

Answer:

The answer is "[tex]\sqrt{12}[/tex] is not a perfect square".

Step-by-step explanation:

12 is not a perfect square because it is the natural number, and no other natural number would square the number 12, that's why it is not a perfect square.

If we calculate the square root of [tex]\sqrt{12}[/tex]. so, it is will give [tex]2\sqrt{3}[/tex] that is not a perfect square root which can be described as follows:

[tex]\Rightarrow \sqrt{12}= \sqrt{2\times 2\times 3}[/tex]

            [tex]= \sqrt{2^2\times 3}\\\\= 2\sqrt{3}\\\\[/tex]

[tex]\bold{\sqrt{12}}[/tex] is not a perfect square root.

Answer:

Here's a picture

Step-by-step explanation:

Answer:

a) Probability the fatality resulted from a fall = 0.1422

b) Probability the fatality resulted from a transportation incident = 0.3958

c(i) The cause of fatality that is least likely to occur is fatality due to fires and explosions with the lowest number of fatalities, 113.

c(ii) Probability that the fatality resulted from fires and explosions = 0.0249

Step-by-step explanation:

Data on U.S, Work-Related fatalities by cause follow (The World Almanac, 2012)

Cause of Fatality | Number of Fatalities Transportation Incidents | 1795

Assaults and violent acts | 837

Contacts with objects and equipment | 741

Falls | 645

Exposure to harmful substances | 404

Fires and explosions | 113

Total number of fatalities = 1795+837+741+645+404+113 = 4,535

Assume that a fatality will be randomly chosen from this population.

a. What is the probability the fatality resulted from a fall?

The probabilty of an event is given mathematically as the number of elements in that event divided by the Total number of elements in the sample space.

P(E) = n(E) ÷ n(S)

Probability the fatality resulted from a fall = (645/4535) = 0.14223 = 0.1422

b. What is the probability the fatality resulted from a transportation incident?

Probability the fatality resulted from a transportation incident = (1795/4535) = 0.39581 = 0.3958

c(i) What cause of fatality is least likely to occur?

The cause of fatality that is least likely to occur is the cause of fatality with the lowest frequency or number of fatalities. And this is fatality due to fires and explosions with the lowest fatalities of 113

c(ii). What is the probability the fatality resulted from this cause?

This is the probability that the fatality resulted from fires and explosions.

Probability that the fatality resulted from fires and explosions = (113/4535) = 0.02492 = 0.0249

Hope this Helps!!!

Before Andrei adds the marbles to the glass tank, the glass tank was empty. This means that the volume of the empty tank is when n = 0 and the volume is 32 liters.

Given that:

[tex]W = 32 - 0.05n[/tex]

A linear function is represented as:

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

Where

[tex]b \to[/tex] y intercept

Literally, the y intercept is the initial value of the function.

In this function, the y intercept means the initial volume of the glass tank before filling it with marbles.

Compare [tex]y = b + mx[/tex] and [tex]W = 32 - 0.05n[/tex]

[tex]b = 32[/tex]

This means that the volume of the glass tank is 32 liters.

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Answer:

0.05

Step-by-step explanation:

Answer:

1, 3, 5, 6

Step-by-step explanation:

Your solution has to be less than the number they are giving you for example if you have -3 one solution could be -16

The numbers that are solutions to the inequality are as follows: 20%, 17%, 29.5%, 1.5%.

The solution of an inequality is the set of all possible values that could serve as the result of the expression. So, for the given problem, the set of values that would correspond to the likelihood of snow is 20%, 17%, 29.5%, and 1.5%.

In other words, these percentages are less than 30% and can be rightly represented by the variable s.

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#SPJ2

Answer:

-The value of [tex]y[/tex]:

[tex]y = 1[/tex]

Step-by-step explanation:

-Find the value of [tex]y[/tex]:

[tex]11- 9y = -6y + 8[/tex]

-Add both sides by [tex]6y[/tex] and combine [tex]9y[/tex] and [tex]6y[/tex] together:

[tex]11- 9y + 6y = -6y + 6y+ 8[/tex]

[tex]11 - 3y = 8[/tex]

-Subtract both sides by [tex]11[/tex]:

[tex]11 - 11 - 3y = 8 -11[/tex]

[tex]-3y = -3[/tex]

-Divide both sides by [tex]-3[/tex]:

[tex]\frac{-3y}{-3} = \frac{-3}{-3}[/tex]

[tex]y = 1[/tex]

So the final answer is [tex]y = 1[/tex] .

Answer:

Option 2 is correct

Step-by-step explanation:

One number is 2 times another number plus 3. Their sum is 21.

"One number is 2 times another number plus 3" translated to

x = smaller number = another number

It is also given that: Their sum is 21.

Combine like terms:

3x+3 = 21

Answer:

I do questions like these everyday so I have too much experience. Let me explain step by step for you.

Brainliest?

First lets set 2 variables x and y

Lets make 2 equations.

x=3+2*y

Thats because it says 'x' is 3 more (+) than 2 times(*) 'y'

Now lets set second, we know both of them add up to 27.

x+y = 27

Since we know what x is equal to (look above equation)

We can replace it.

x is replaced with 3+2*y

3+2y+y = 27

3+4y = 27

Simplify 27-3 = 24

24/4 = 6

Now lets plug in for x

3+2*6 = 15

15 - x

6 - y

:))

Answer:

a) P(X < 30) = 0.0392.

b) P(28 < X < 32) = 0.2760

c) P(X > 35) = 0.1190

d) P(X > 31) = 0.8810

e) At least 35.7965 mpg

Step-by-step explanation:

Problems of normally distributed samples are 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 question, we have that:

[tex]\mu = 33, \sigma = 1.7[/tex]

a. P(X<30)

This is the pvalue of Z when X = 30. So

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

[tex]Z = \frac{30 - 33}{1.7}[/tex]

[tex]Z = -1.76[/tex]

[tex]Z = -1.76[/tex] has a pvalue of 0.0392.

Then

P(X < 30) = 0.0392.

b) P(28 < X < 32)

This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So

X = 32

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

[tex]Z = \frac{32 - 33}{1.7}[/tex]

[tex]Z = -0.59[/tex]

[tex]Z = -0.59[/tex] has a pvalue of 0.2776.

X = 28

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

[tex]Z = \frac{28 - 33}{1.7}[/tex]

[tex]Z = -2.94[/tex]

[tex]Z = -2.94[/tex] has a pvalue of 0.0016.

0.2776 - 0.0016 = 0.2760.

So

P(28 < X < 32) = 0.2760

c) P(X>35)

This is 1 subtracted by the pvalue of Z when X = 35. So

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

[tex]Z = \frac{35 - 33}{1.7}[/tex]

[tex]Z = 1.18[/tex]

[tex]Z = 1.18[/tex] has a pvalue of 0.8810.

1 - 0.8810 = 0.1190

So

P(X > 35) = 0.1190

d. P(X>31)

This is 1 subtracted by the pvalue of Z when X = 31. So

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

[tex]Z = \frac{31 - 33}{1.7}[/tex]

[tex]Z = -1.18[/tex]

[tex]Z = -1.18[/tex] has a pvalue of 0.1190.

1 - 0.1190 = 0.8810

So

P(X > 31) = 0.8810

e. the mileage rating that the upper 5% of cars achieve.

At least the 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then

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

[tex]1.645 = \frac{X - 33}{1.7}[/tex]

[tex]X - 33 = 1.645*1.7[/tex]

[tex]X = 35.7965[/tex]

At least 35.7965 mpg

The upper 5% of cars have a mileage rating of 35.805 mpg

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given;  mean of 33 mpg and a standard deviation of 1.7

a) For < 30:

z = (30 - 33)/1.7 = -1.76

P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392

b) For < 28:

z = (28 - 33)/1.7 = -2.94

P(x < 28) = P(z < -2.94) = 0.0016

c) For > 35:

z = (35 - 33)/1.7 = 1.18

P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119

d) For > 31:

z = (31 - 33)/1.7 = -1.18

P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810

e) The  upper 5% of cars achieve have a z score of 1.65, hence:

1.65 = (x - 33)/1.7

x = 35.805 mpg

The upper 5% of cars have a mileage rating of 35.805 mpg

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Answer:

63.45

Step-by-step explanation:

First, it should be noted that the question is incorrect because 64 can't be divided by 11 but assuming that the question is correct, the solution is as follows

Given

LCM = 368

HCF = 11

One number = 64

Required

The other number

Let both numbers be represented by m and n, such that

[tex]m = 64[/tex]

From laws of HCF and LCM

The product of both numbers = Product of HCF and LCM

i.e.

[tex]m * n = HCF * LCM[/tex]

By substituting 68 for m; 368 for LCM and 11 for HCF

[tex]m * n = HCF * LCM[/tex] becomes

[tex]64 * n = 368 * 11[/tex]

[tex]64n = 4048[/tex]

Divide both sides by 64

[tex]\frac{64n}{64} = \frac{4048}{64}[/tex]

[tex]n = \frac{4048}{64}[/tex]

[tex]n = 63.25[/tex]

Answer:

9

Step-by-step explanation:

[tex]2x+3y=36\\\\2x+3(6)=36\\\\2x+18=36\\\\2x=18\\\\x=9[/tex]

Hope this helps!

Answer:

X= 9

Step-by-step explanation:

2x+3y=36

2x+3(6)=36

2x+18=36

    -18  -18

2x=18

----------

   2

x=9

Answer:

Step-by-step explanation:

Given that:

68.87, 78.25, 70.44, 84.67, 79.79, 86.33, 100.24, 98.26

we calculate sample mean and standard deviation from given data

Sample Mean

[tex]\bar x = \frac{\sum (x)}{n} =\frac{666.85}{8} \\\\=83.35625[/tex]

Sample Variance

[tex]s^2= \frac{\sum (x- \bar x )^2}{n-1} \\\\=\frac{933.224787}{7} =133.317827[/tex]

sample standard deviation

[tex]s=\sqrt{s^2} \\=\sqrt{133.317827} \\ =11.546334[/tex]

95% CI for [tex]\mu[/tex] using t - dist

Sample mean = 83.35625

Sample standard deviation = 11.546334

Sample size = n = 8

Significance level = α = 1 - 0.95 = 0.05

Degrees of freedom for t - distribution

d-f = n - 1 = 7

Critical value

[tex]t_{\alpha 12, df}= t_{0.025, df=7}=2.365[/tex] ( from t - table , two tails, d.f =7)

Margin of Error

[tex]E = t_{\alpha 12, df}\times \frac{s_x}{\sqrt{n} } \\\\=2.365 \times \frac{11.546334}{\sqrt{8} } \\\\=2.365 \times 4.082246\\\\E=9.654512[/tex]

Limits of 95% Confidence Interval are given by:

Lower limit

[tex]\bar x - E = 83.35625-9.654512\\\\=73.701738\approx 73.702[/tex]

Upper Limit

[tex]= \bar x + E\\=83.35625+ 9.654512\\=93.010762 \approx 93.011[/tex]

95% Confidence interval is

[tex]\bar x \pm E = 83.35625 \pm 9.654512\\\\=(73.701738,93.010762)[/tex]

95% CI using t - dist (73.70 < μ < 93.01)

D. The lower bound is $ and the upper bound is One can be [95]% confident that the mean travel tax for all cities is between these values.

c.What would you recommend to a researcher who wants to increase the precision of the interval, but does not have access to additional data?

A. The researcher could decrease the level of confidence.

Answer:

Step-by-step explanation:

4x+5x=27

9x=27

x=27/9

x=3

4x3=12

5x3=15

The total number of cats were 12.

Based on the ratio of dogs to cats in the shelter, we know that out of 27 animals, there are 12 cats.

The ratio of cats to dogs is 4:5 which means that there are 5 dogs for every 4 cats.

This means that out of 9 animals, 4 would be cats and 5 would be dogs. If there was 27 animals therefore:

= 4 / 9 x 27

= 108 / 9

= 12 cats

In conclusion, there are 12 cats.

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Answer:

y=-3x-2

Step-by-step explanation:

There is enough information to make a point-slope form equation that which we can convert into slope-intercept form.

Point-slope form is: [tex]y-y_1=m(x-x_1)[/tex]

We are given the slope of -3 and the point of (1,-5).

[tex]y-y_1=m(x-x_1)\rightarrow y+5=-3(x-1)[/tex]

Convert into Slope-Intercept Form:

[tex]y+5=-3(x-1)\\y+5-5=-3(x-1)-5\\\boxed{y=-3x-2}[/tex]

Answer:

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

Step-by-step explanation:

So first calculate what fraction of the circumference the arc is.

[tex]\frac{20}{360}=\frac{1}{18}[/tex]

Now the circumference is 6, so one eighteenth of that is [tex]\frac{1}{3}[/tex]

Answer:

144

Step-by-step explanation:

Answer:

116

Step-by-step explanation:

3x4=12

12x4=48

8x4=32

32-3=29

29x4=116

Hope it's clear

Answer:

[tex]$ \text {Sample mean} = \bar{x} = \mu = 500 \: hours $[/tex]

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.  

For the given scenario, it is known from numerous previous samples that when this service life is in control it is normally distributed with a mean of 500 hours and a standard deviation of 20 hours.

On three recent production batches, he tested service life on random samples of four headlamps.

We are asked to find the mean of the sampling distribution of sample means when the service life is in control.

Since we know that the population is normally distributed and a random sample is taken from the population then the mean of the sampling distribution of sample means would be equal to the population mean that is 500 hours.

[tex]$ \text {Sample mean} = \bar{x} = \mu = 500 \: hours $[/tex]

Whereas the standard deviation of the sampling distribution of sample means would be

[tex]\text {standard deviation} = s = \frac{\sigma}{\sqrt{n} } \\\\[/tex]

Where n is the sample size and σ is the population standard deviation.

[tex]\text {standard deviation} = s = \frac{20}{\sqrt{4} } \\\\ \text {standard deviation} = s = \frac{20}{2 } \\\\ \text {standard deviation} = s = 10 \: hours \\\\[/tex]

Answer:

Step-by-step explanation:

|x-1|=8

if (x-1) >= 0 meaning x >= 1

then |x-1| = x-1

and then the solution of the equation is

x-1=8

<=> x = 9

if (x-1) <= 0 meaning x <= 1

then |x-1| = -(x-1) = -x+1

so the solution of the equation is

-x+1=8

<=> -x = 7

<=> x = -7

so the solutions are -7 and 9

answer C

do no hesitate if you need further explanation

thank you

Answer:

8

Step-by-step explanation:

The other patterns go with the two factors on top (2 x 3 = 6 and 3 x 3 = 9).

So, 2 x 4 = 8

The probability that all 4 selected workers will be from the day shift is, = 0.0198

The probability that all 4  selected workers will be from the same shift is = 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5257

To solve this question properly, we will need to make use of the concept of combination along with set theory.

In mathematical concept, Combination is the grouping of subsets from a set without taking the order of selection into consideration.  

The formula for calculating combination can be expressed as:

[tex]\mathbf{(^n _r) =\dfrac{n!}{r!(n-r)! }}[/tex]

From the parameters given:

A quality control consultant is to select 4 of these workers for in-depth interviews:

Using the expression for calculating combination:

(a)

The number of selections results in all 4 workers coming from the day shift is :

[tex]\mathbf{(^n _r) = (^{10} _4)}[/tex]

[tex]\mathbf{=\dfrac{(10!)}{4!(10-4)!}}[/tex]

= 210

The probability that all 5 selected workers will be from the day shift is,

[tex]\begin{array}{c}\\P\left( {{\rm{all \ 4 \ selected \ workers\ will \ be \ from \ the \ day \ shift}}} \right) = \dfrac{{\left( \begin{array}{l}\\10\\\\4\\\end{array} \right)}}{{\left( \begin{array}{l}\\24\\\\4\\\end{array} \right)}}\\\end{array}[/tex]

[tex]\mathbf{= \dfrac{210}{10626}} \\ \\ \\ \mathbf{= 0.0198}[/tex]

(b) The probability that all 4 selected workers will be from the same shift is calculated as follows:

P( all 4 selected workers will be) [tex]\mathbf{= \dfrac{ \Big(^{10}_4\Big) }{\Big(^{24}_4\Big)}+\dfrac{ \Big(^{8}_4\Big) }{\Big(^{24}_4\Big)} + \dfrac{ \Big(^{6}_4\Big) }{\Big(^{24}_4\Big)}}[/tex]

where;

[tex]\mathbf{\Big(^{8}_4\Big) = \dfrac{8!}{4!(8-4)!} = 70}[/tex]

[tex]\mathbf{\Big(^{6}_4\Big) = \dfrac{6!}{4!(6-4)!} = 15}[/tex]

P( all 4 selected workers is:)

[tex]\mathbf{=\dfrac{210+70+15}{10626}}[/tex]

The probability that all 4 selected workers will be from the same shift is = 0.0278

(c)

The probability that at least two different shifts will be represented among the selected workers can be computed as:

 [tex]= 1-\dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}[/tex]

[tex]=1 - \dfrac{210+70+15}{10626}[/tex]

= 1 - 0.0278

=  0.9722

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

(d)

The probability that at least one of the shifts will be unrepresented in the sample of workers is:

[tex]P(AUBUC) = \dfrac{(^{6+8}_4)}{(^{24}_4)}+ \dfrac{(^{10+6}_4)}{(^{24}_4)}+ \dfrac{(^{10+8}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0[/tex]

[tex]P(AUBUC) = \dfrac{(^{14}_4)}{(^{24}_4)}+ \dfrac{(^{16}_4)}{(^{24}_4)}+ \dfrac{(^{18}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0[/tex]

[tex]P(AUBUC) = \dfrac{1001}{10626}+ \dfrac{1820}{10626}+ \dfrac{3060}{10626}-\dfrac{15}{10626}-\dfrac{70}{10626}-\dfrac{210}{10626} +0[/tex]

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5257

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Answer: The new price of  the car is $41856

Step-by-step explanation:

So we know the the original price as 43,600 which is 100% and is being dropped by 4%  so you would have to subtract 4% from a 100% and multiply it by the original price.

100% - 4% = 96%

Now 96% of the original price is the new price.

96% * 43,600= ?

0.96 * 43,600 = 41856

Answer:

20.58

Step-by-step explanation:

Force A = 16 pounds

direction = N75E = 75° in the first quadrant

Force B = 20 pounds

direction = S20E = 180° - 20° = 160° in the fourth quadrants

we resolve the forces into their x and y resultant forces.

For the y axis forces,

-(16 x sin 75°) + (20 x sin 160°) = Fy

-15.45 + 6.84 = Fy

Fy = -8.61 N

For the x axis forces'

(16 x cos 75°) + (20 x cos 160°) = Fx

-4.14 - 18.79 = Fx

Fx = -22.93 N

Resultant force = [tex]\sqrt{(Fx)^{2} + (Fy)^{2} }[/tex] =  [tex]\sqrt{(-8.61)^{2} + (-22.93)^{2} }[/tex]

Resultant force = 24.49 N

angle made by the resultant force is,

∅ = [tex]tan^{-1}[/tex] [tex]\frac{Fy}{Fx}}[/tex]

∅  =  [tex]tan^{-1}[/tex] [tex]\frac{-8.61}{-22.93}}[/tex]

∅ =  [tex]tan^{-1}[/tex] 0.3755

∅ = 20.58

Answer:

The angles measured between the shadow and the wall trim and the shadow and the top or bottom of the picture should be equal if the picture is level

Step-by-step explanation:

Whereby the picture is 12 inches above the wall trim and the Sun is casting a shadow on the wall, therefore, the edges of the shadow of a straight edged object is straight

If the picture is level, the shadow cast by the sun is transversal to the wall trim and the line formed by extending the bottom or top of the picture in the direction of the shadow. That is, if the picture is parallel, the shadow cast by the Sun on the wall is transversal to the picture and the wall trim if or when the shadow eventually crosses the picture

With the protractor, the angle between the shadow and the wall trim and the shadow and the top or bottom is measured

The angles measured should be the same if the picture is level.

Answer:

79.91% of loaves are between 26.94 and 32.18 centimeters

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

[tex]\mu = 30, \sigma = 2[/tex]

What percentage of loaves are between 26.94 and 32.18 centimeters

This is the pvalue of Z when X = 32.18 subtracted by the pvalue of Z when X = 26.94.

X = 32.18:

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

[tex]Z = \frac{32.18 - 30}{2}[/tex]

[tex]Z = 1.09[/tex]

[tex]Z = 1.09[/tex] has a pvalue of 0.8621

X = 26.94:

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

[tex]Z = \frac{26.94 - 30}{2}[/tex]

[tex]Z = -1.53[/tex]

[tex]Z = -1.53[/tex] has a pvalue of 0.0630

0.8621 - 0.0630 = 0.7991

79.91% of loaves are between 26.94 and 32.18 centimeters

There is a missing content in the question.

After the statements and before the the options given; there is an omitted content which says:

Referring to Table 16-5, in testing the coefficient of X in the regression equation (0.117) the results were a t-statistic of 9.08 and an associated p-value of 0.0000. Which of the following is the best interpretation of this result?

Answer:

C. The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05).

Step-by-step explanation:

From the given question:

The resulting regression equation can be represented as:

[tex]\hat Y = 3.37 + 0.117 X - 0.083 Q_1 + 1.28 Q_2 + 0.617Q_3[/tex]

where;

the estimated number of contracts in a quarter X is the coded quarterly value with X = 0

the first quarter of 2010 Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise

Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise

Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise

Our null and alternative hypothesis can be stated as;

Null hypothesis :

[tex]H_0 :[/tex]  The quarterly growth rate in the number of contracts is not  significantly different from 0% (? = 0.05)

[tex]H_a:[/tex]  The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05)

The decision rule is to reject the null hypothesis if the p-value is less than 0.05.

From the missing omitted part we added above; we can see that the   t-statistics value = 9.08 and the p-value = 0.000 .

Conclusion:

Thus; we reject the null hypothesis and accept the alternative hypothesis. i.e

The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05)

Answer:

No

Step-by-step explanation:

Use the vertical line test. If the line intercepts more than one point, it is not a function. Since there are two points where the value of 'x' is two, the line will pass both points. The graph is not a function.

Answer:

A.

Step-by-step explanation:

Add 4:

-5x ≤ 10

Divide by -5:

x ≥ -2

Answer:

a) 38

Step-by-step explanation:

The normal distribution can be applied if:

[tex]np \geq 5[/tex] and [tex]n(1-p) \geq 5[/tex]

In this question:

[tex]p = 0.27[/tex]

Then

a) 38

n = 38.

Then

38*0.27 = 10.26

38*0.73 = 27.74

Satisfies. But is this the smallest sample of the options which satisfies.

b) 14

n = 14

Then

14*0.23 = 3.22

14*0.77 = 10.78

Does not satisfy

c) 10

Smaller than 14, which also does not satisfy, so 10 does not satisfy.

d) 48

Greater than 38, which already satisfies. So the answer is a)