The five numbers summary for a data set is shown below. What is the range of the data set? 3, 7, 11, 14, 16

Answers

Answer 1

Answer:

13

Step-by-step explanation:

The range of a set of data is the difference between the highest and lowest values in the set.

So, in order to find the range, you first order the data from least to greatest. Which it is already.

3, 7, 11, 14, 16

Then subtract the smallest value from the largest value in the set.

16 - 3 = 13

Hope this helps you out! : )


Related Questions

In the given figure, find AB, given thatAC = 14 andBC = 9.

Answers

Answer:

Given:

AC = 14      and       BC = 9

AB = ?

Solution:

From the fig:

AC = AB + BC

Putting the values

14 = AB + 9

AB = 14 - 9

AB = 5

(you can also take AB = x or any other variable)

Step-by-step explanation:

Examine the system of equations. y = 3 2 x − 6, y = −9 2 x + 21 Use substitution to solve the system of equations. What is the value of y? y =

Answers

Answer:

its 3/4

Step-by-step explanation: i got it right trust me

The solution to the system of equations will be x= 9 / 2 and y= 3 / 4.

What is a system of equations?

A finite set of equations for which common solutions are sought is referred to in mathematics as a set of simultaneous equations, often known as a system of equations or an equation system.

An equation is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

The given equations are y=(3/ 2 )x-6 and y=(-9/2)x+21 ​to calculate the values of x and y using the substitution method.

Since both equations are equated to y, you just need to use substitution to create the equation below:

(3/ 2 )x-6 =(-9/2)x+21

Solve the equation for x:

(3/ 2 )x + (9/2)x = 27

x = 27 / 6 = 9 / 2

Plug x into any one of the given equations to find the value of y:

y=(3/ 2 )x-6

Solve for the value of y.

y=(3/2) x (9 / 2)-6

y = ( 27 / 4 ) - 6

y = ( 27 - 24 ) / 4

y = 3 / 4

Hence, the solution for the equation will be x = 9 / 2 and y = 3 / 4.

To know more about the system of linear equations follow

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A data set is summarized in the frequency table below. Using the table, determine the number of values less than or equal to 6.

Answers

Answer:

18

Step-by-step explanation:

Given the above table of the data set, the number of values less than or equal to 6 would be the sum of the frequencies of all values that is equal to or less than 6.

From the table above, we would add up the frequencies of the values of 6 and below, which is:

2 + 3 + 6 + 4 + 3 = 18

Answer = 18

The number of values less than or equal to 6 is 18

Calculation of the number of values:

Here the number of values should be less than or equivalent to 6 represent the sum of the frequencies i.e. equal or less than 6

So, here the number of values should be

= 2 + 3 + 6 + 4 + 3

= 18

Hence, we can conclude that  The number of values less than or equal to 6 is 18

Learn more about frequency here: https://brainly.com/question/20875379

An object is moving at a speed of 7300 inches every 3 seconds. Express this speed in miles per day.

Answers

Answer:

≈3318 miles per day....

Step-by-step explanation:

VW=40in. The radius of the circle is 25 inches. Find the length of CT.

Answers

Answer:

The answer is B. 40 inches.

Step-by-step explanation:

The question starts by telling you that line VW is equal to 40 in. If you look at the picture you can see it is divided into 2 equal parts of 20 in each. If you look at line CT, you can see that there are the same marks meaning that those segments are also 20 in. That means that line CT and line VW are equal and that line CT is equal to 40 in.

. What is the percentage of VanArsdel's manufactured goods sold in Alberta? (to two decimal places in the format 00.00, without the % sign)

Answers

Answer:

Revenue : 47.77

Units Sold : 28.91

Step-by-step explanation:

The revenue is the amount that is received after selling the goods manufactured. VanArsdel's sold good of manufactured in Alberta. Goods manufactured by VanArsdel's is considered as 100 percent out of which it sold 28.91 % of units in Alberta. The revenue percentage is 47.77%.

determine the polynomial equivalent to this expression.
x^2-9/x-3

A. x-3

B. -3x-9

C. x+3

D. x^2+3x​

Answers

Answer:

[tex]\dfrac{x^2-9}{x-3}= \Large \boxed{x+3}[/tex]

Step-by-step explanation:

Hello,

We need to work a little bit of the expression to see if we can simplify.

Do you remember this formula?

   for any a and b reals, we can write

   [tex]a^2-b^2=(a-b)(a+b)[/tex]

We will apply it.

For any x real number different from 3 (as dividing by 0 is not allowed)

[tex]\dfrac{x^2-9}{x-3}=\dfrac{x^2-3^2}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3[/tex]

So the winner is C !!

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole class is a population). A random sample of 16 students yielded an average of a 75 on the first exam (s=16). What is the z-score of the sample mean? Is this sample significantly different from the population? (Hint: Use the z-score formula for locating a sample mean)

Answers

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

WILL MARK AS BRAINLIEST!!! 5. A 2011 study by The National Safety Council estimated that there are nearly 5.7 million traffic accidents year. At least 28% of them involved distracted drivers using cell phones or texting. The data showed that 11% of drivers at any time are using cell phones . Car insurance companies base their policy rates on accident data that shows drivers have collisions approximately once every 19 years. That’s a 5.26% chance per year. Given what you know about probability, determine if cell phone use while driving and traffic accidents are related. Step A: Let DC = event that a randomly selected driver is using a cell phone. What is P(DC)? (1 point) Step B: Let TA = event that a randomly selected driver has a traffic accident. What is P(TA)? Hint: What is the probability on any given day? (1 point) Step C: How can you determine if cell phone use while driving and traffic accidents are related? (1 point) Step D: Given that the driver has an accident, what is the probability that the driver was distracted by a cell phone? Write this event with the correct conditional notation. (1 point) Step E: What is the probability that a randomly selected driver will be distracted by using a cell phone and have an accident? (2 points) Step F: For a randomly selected driver, are the events "driving while using a cell phone" and "having a traffic accident" independent events? Explain your answer. (2 points)

Answers

Answer:

Step-by-step explanation:

Hello!

Regarding the reasons that traffic accidents occur:

28% are caused by distracted drivers using cell phones or texting

11% of the drivers' user their phones at any time

The probability of a driver having an accident is 5.26%

a)

DC = event that a randomly selected driver is using a cell phone.

P(DC)= 0.11

b)

TA = event that a randomly selected driver has a traffic accident.

P(TA)= 0.0526

c) and f)

If both events are related, i.e. dependent, then you would expect that the occurrence of one of these events will affect the probability of the other one. If they are not related, i.e. independent events, then their probabilities will not be affected by the occurrence of one or another:

If both events are independent P(TA|DC)= P(TA)

If they are dependent, then:

P(TA|DC)≠ P(TA)

P(TA|DC)= 0.28

P(TA)= 0.0526

As you can see the probability of the driver having an accident given that he was using the cell phone is different from the probability of the driver having an accident. This means that both events are related.

d) and e)

You have to calculate the probability that "the driver was distracted with the phone given that he had an accident", symbolically P(DC|TA)

P(DC|TA) = [tex]\frac{P(DCnTA)}{P(TA)}[/tex]

[tex]P(TA|DC)= \frac{P(TAnDC}{P(DC)}[/tex] ⇒ P(DC∩TA)= P(TA|DC)*P(DC)= 0.28 *  0.11= 0.0308

P(DC|TA) = [tex]\frac{0.0308}{0.0526}= 0.585= 0.59[/tex]

I hope this helps!

Suppose that you expect SugarCane stock price to decline. So you decide to ask your broker to short sell 2000 shares. The current market price is $40. The proceeds from the short sale $80,000 is credited into your account. However, a few days later the market price of the stock jumps to $80 per share and your broker asks you close out your position immediately. What is your profit or loss from this transaction?

Answers

Answer:

Loss = $80000

Step-by-step explanation:

To determine if it's a profit or loss is simple.

He predicted the sugar cane stock to fall so he sold , but few days later the stock grew and went bullish.

He sold at$ 40 for 2000 shares

=$ 80000

But the stock went up to $80 per share that is gaining extra $40

So it was actually a loss.

The loss is =$40 * 2000

The loss = $80000

Which two points are on the graph of y=-x+ 3?
(-1,-2), (1,4)
(1, 2), (0, -3)
(0, 3), (4, -1)
(4, -1), (1, 3)

Answers

Answer:

(0, 3), (4, -1)

(1, 2)

Step-by-step explanation:

If the answers that have been provided to you are only in pairs then it'd just be the first answer I wrote. The points (1, 2) also are on the graph of y=x+3 but if the answers aren't individual than I'd just stick with the (0, 3), (4, -1). Does that make sense? I used a graphing calculator online called Desmos, it's very good. I highly recommend it for problems like these.

I hope this helps:) Select as brainliest because I actually put work into this and tried.

An insect population in a lab has 2 ¹² insect. If the population double how many insect will be there?

Answers

Answer:

8192

Step-by-step explanation:

2 ¹²= 4096

4096 x 2 = 8192

The mean of a normal distribution is 400 pounds. The standard deviation is 10 pounds. What is the probability of a weight between 415 pounds and the mean of 400 pounds

Answers

Answer:

The probability is [tex]P(x_1 \le X \le x_2 ) = 0.4332[/tex]

Step-by-step explanation:

From the question we are told that

   The population mean is  [tex]\mu = 400[/tex]

   The  standard deviation is  [tex]\sigma = 10[/tex]

   The considered values are  [tex]x_1 = 400 \to x_2 = 415[/tex]

Given that the weight follows a normal distribution

     i.e       [tex]\approx X (\mu , \sigma )[/tex]

Now the probability of a weight between 415 pounds and the mean of 400 pounds is mathematically as

     [tex]P(x_1 \le X \le x_2 ) = P(\frac{x_1 - \mu }{\sigma } \le \frac{X - \mu }{\sigma } \le \frac{x_2 - \mu }{\sigma } )[/tex]

So  [tex]\frac{X - \mu }{\sigma }[/tex] is equal to Z (the standardized value of  X  )

Hence we have  

     [tex]P(x_1 \le X \le x_2 ) = P(\frac{x_1 - \mu }{\sigma } \le Z \le \frac{x_2 - \mu }{\sigma } )[/tex]

substituting values

      [tex]P(x_1 \le X \le x_2 ) = P(\frac{400 - 400 }{10 } \le Z \le \frac{415 - 400}{415 } )[/tex]

      [tex]P(x_1 \le X \le x_2 ) = P(0\le Z \le 1.5 )[/tex]

      [tex]P(x_1 \le X \le x_2 ) = P( Z < 1.5) - P( Z < 0)[/tex]

From the standardized normal distribution table  [tex]P( Z< 1.5) = 0.9332[/tex] and

   [tex]P( Z < 0) = 0.5[/tex]

So

     [tex]P(x_1 \le X \le x_2 ) = 0.9332 - 0.5[/tex]

     [tex]P(x_1 \le X \le x_2 ) = 0.4332[/tex]

NOTE :  This above  values obtained from the standardized normal distribution table can also be obtained using the P(Z) calculator at  (calculator dot net).

Section 8
Find the mean of these numbers:
24 18
37
82 17
26​

Answers

Answer:

[tex]\boxed{Mean = 34.33}[/tex]

Step-by-step explanation:

Mean = Sum of Observations / No. Of Observations

Mean = (24+18+37+82+17+26)/6

Mean = 206 / 6

Mean = 34.33

WILL GIVE BRAINLIEST IF CORRECT!! Please help ! -50 POINTS -

Answers

Answer:

i think (d) one i think it will help you

The correct answer is c. 180 , 202

All the step by step is below

Hopefully this help you :)

Using traditional methods it takes 109 hours to receive an advanced flying license. A new training technique using Computer Aided Instruction (CAI) has been proposed. A researcher believes the new technique may lengthen training time and decides to perform a hypothesis test. After performing the test on 190 students, the researcher decides to reject the null hypothesis at a 0.02 level of significance.

What is the conclusion?

a. There is sufficient evidence at the 0.020 level of significance that the new technique reduces training time.
b. There is not sufficient evidence at the 0.02 level of significance that the new technique reduces training time.

Answers

I think the answer is option B.

Because while researchers research they believed that it will lengthen the time and it don't reduced the time.

Hope it's correct..

Find the probability of each event. A six-sided die is rolled seven times. What is the probability that the die will show an even number at most five times?

Answers

Answer:

[tex]\dfrac{15}{16}[/tex]

Step-by-step explanation:

When a six sided die is rolled, the possible outcomes can be:

{1, 2, 3, 4, 5, 6}

Even numbers are {2, 4, 6}

Odd Numbers are {1, 3, 5}

Probability of even numbers:

[tex]\dfrac{\text{Favorable cases}}{\text{Total cases }} = \dfrac{3}{6} = \dfrac{1}{2}[/tex]

This is binomial distribution.

where probability of even numbers,  [tex]p =\frac{1}{2}[/tex]

Probability of not getting even numbers (Getting odd numbers) [tex]q =\frac{1}{2}[/tex]

Probability of getting r successes out of n trials:

[tex]P(r) = _nC_r\times p^r q^{n-r}[/tex]

Probability of getting even numbers at most 5 times out of 7 is given as:

P(0) + P(1) +P(2) + P(3) +P(4) + P(5)

[tex]\Rightarrow _7C_0\times \frac{1}{2}^0 \frac{1}{2}^{7}+_7C_1\times \frac{1}{2}^1 \frac{1}{2}^{6}+_7C_2\times \frac{1}{2}^2 \frac{1}{2}^{5}+_7C_3\times \frac{1}{2}^3 \frac{1}{2}^{4}+_7C_4\times \frac{1}{2}^4 \frac{1}{2}^{3}+_7C_5\times \frac{1}{2}^5 \frac{1}{2}^{2}[/tex]

[tex]\Rightarrow (\dfrac{1}{2})^7 (_7C_0+_7C_1+_7C_2+_7C_3+_7C_4+_7C_5)\\[/tex]

[tex]\Rightarrow (\dfrac{1}{2})^7 (1+7+\dfrac{7 \times 6}{2}+\dfrac{7 \times 6 \times 5}{3\times 2}+\dfrac{7 \times 6 \times 5}{3\times 2}+\dfrac{7 \times 6}{2})\\\Rightarrow \dfrac{120}{128} \\\Rightarrow \dfrac{15}{16}[/tex]

A manufacturer knows that on average 20% of the electric toasters produced require repairs within 1 year after they are sold. When 20 toasters are randomly selected, find appropriate numbers x and y such that (a) the probability that at least x of them will require repairs is less than 0.5; (b) the probability that at least y of them will not require repairs is greater than 0.8

Answers

Answer:

(a) The value of x is 5.

(b) The value of y is 15.

Step-by-step explanation:

Let the random variable X represent the number of electric toasters produced that require repairs within 1 year.

And the let the random variable Y represent the number of electric toasters produced that does not require repairs within 1 year.

The probability of the random variables are:

P (X) = 0.20

P (Y) = 1 - P (X) = 1 - 0.20 = 0.80

The event that a randomly selected electric toaster requires repair is independent of the other electric toasters.

A random sample of n = 20 toasters are selected.

The random variable X and Y thus, follows binomial distribution.

The probability mass function of X and Y are:

[tex]P(X=x)={20\choose x}(0.20)^{x}(1-0.20)^{20-x}[/tex]

[tex]P(Y=y)={20\choose y}(0.20)^{20-y}(1-0.20)^{y}[/tex]

(a)

Compute the value of x such that P (X ≥ x) < 0.50:

[tex]P (X \geq x) < 0.50\\\\1-P(X\leq x-1)<0.50\\\\0.50<P(X\leq x-1)\\\\0.50<\sum\limits^{x-1}_{0}[{20\choose x}(0.20)^{x}(1-0.20)^{20-x}][/tex]

Use the Binomial table for n = 20 and p = 0.20.

[tex]0.411=\sum\limits^{3}_{x=0}[b(x,20,0.20)]<0.50<\sum\limits^{4}_{x=0}[b(x,20,0.20)]=0.630[/tex]

The least value of x that satisfies the inequality P (X ≥ x) < 0.50 is:

x - 1 = 4

x = 5

Thus, the value of x is 5.

(b)

Compute the value of y such that P (Y ≥ y) > 0.80:

[tex]P (Y \geq y) >0.80\\\\P(Y\leq 20-y)>0.80\\\\P(Y\leq 20-y)>0.80\\\\\sum\limits^{20-y}_{y=0}[{20\choose y}(0.20)^{20-y}(1-0.20)^{y}]>0.80[/tex]

Use the Binomial table for n = 20 and p = 0.20.

[tex]0.630=\sum\limits^{4}_{y=0}[b(y,20,0.20)]<0.50<\sum\limits^{5}_{y=0}[b(y,20,0.20)]=0.804[/tex]

The least value of y that satisfies the inequality P (Y ≥ y) > 0.80 is:

20 - y = 5

y = 15

Thus, the value of y is 15.

Evaluate the expression

Answers

Answer: C)  tan(pi/56)

=============================================

Explanation:

I recommend using a trig identity reference sheet. The specific identity we will be using is [tex]\frac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)} = \tan(A-B)[/tex]

What we are given is in the form [tex]\frac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}[/tex] with A = pi/7 and B = pi/8

A-B = (pi/7)-(pi/8)

A-B = pi(1/7-1/8)

A-B = pi(8/56 - 7/56)

A-B = pi*(1/56)

A-B = pi/56

Therefore,

[tex]\frac{\tan\left(\pi/7\right)-\tan(\pi/8)}{1+\tan(\pi/7)\tan(\pi/8)} = \tan\left(\pi/56\right)[/tex]

Given that 243√3 =3^a, find the value of a

Answers

Answer:

a=11/5 OR 5.5

Step-by-step explanation:

Please answer this correctly without making mistakes

Answers

Answer:

Centerville is 13 kilometers away from Manchester

Step-by-step explanation:

26.1 - 13.1 = 13

Which is the value of this expression when p = 3 and q = negative 9? ((p Superscript negative 5 Baseline) (p Superscript negative 4 Baseline) (q cubed)) Superscript 0 Negative one-third Negative StartFraction 1 Over 27 EndFraction StartFraction 1 Over 27 EndFraction One-third Edge 2020

Answers

Answer:

I am pretty sure that the answer is D. The value should be 1.

Step-by-step explanation:

Answer:

Answer is D

Step-by-step explanation:

On Edge 2020

Line AB and Line CD are parallel lines. Which translation of the plane can we use to prove angles x and y are congruent, and why?

Answers

Answer:

Option C.

Step-by-step explanation:

In the given figure we have two parallel lines AB and CD.

A transversal line FB intersect the parallel lines at point B and C.

We know that the if a transversal line intersect two parallel lines, then corresponding angles are congruent.

[tex]\angle ABC=\anle ECF[/tex]

[tex]x=y[/tex]

To prove this by translation, we need a translation along the directed line segment CB maps ine CD onto line AB and angle y onto angle x.

Therefore, the correct option is C.

The circle graph shows the percentage of numbered tiles in a box. If each numbered tile is equally likely to be pulled from the box, what is the probability of pulling out a tile with a 6 on it? (Hint: Remember that percents are based out of 100% and probability is represented as a fraction of 100%)

Answers

Answer: [tex]\dfrac{1}{5}[/tex]

Step-by-step explanation:

From, the circle graph in the attachment below,

The percentage of portion taken by 6 (dark blue) = 20%  

So, the probability of pulling out a tile with a 6 on it = percentage of portion taken by 6 (dark blue) = 20%     [Probability can also be written as a percentage]

[tex]=\dfrac{20}{100}\\\\=\dfrac{1}{5}[/tex]  [we divide a percentage by 100 to convert it into fraction]

Hence, the probability of pulling out a tile with a 6 on it = [tex]\dfrac{1}{5}[/tex]

A candidate for political office wants to determine if there is a difference in his popularity between men and women. To test the claim of this difference, he conducts a survey of voters. The sample contains 250 men and 250 women, of which 44% of the men and 52% of the women favor his candidacy. Do these values indicate a difference in popularity?Use a 0.01 significance level.
What are the hypothesis statements?
a) H0:pm=pw
HA:pm b) H0:pm=pw
HA:pm>pw
c) H0:pm=pw
HA:pm≠pw

Answers

Answer:

c) H0:pm=pw

HA:pm≠pw

Step-by-step explanation:

We formulate our hypothesis as

H0: pm = pw  " probability of men = probabilityof women" meaning there's no difference in the probabilityof the men and women in favor of his candidacy.

Alternate Hypothesis HA :pm≠pw " probability of men ≠ probabilityof women" meaning there's a difference in the probability of the men and women in favor of his candidacy.

the significance level α= 0.01

The test statistic under H0 is

Z = pm- pw/ √p`q` ( 1/n.m + 1/n.w)

pm= probability of men= 0.44

pw= probability of women = 0.52

p`= n.m pm+ n.w pw/ n.m + n.w

p`= 250 *0.44 + 250 *0.52/ 250 + 250

p`= 110 + 130 /500 = 240 /500 = 0.48

q`= 1- p`= 1-0.48= 0.52

Putting the values

Z= 0.44- 0.52/ √ 0.48 * 0.52

z= 0.08 / √0.2496

z=  0.08/ 0.4995

z= 0.1601

The critical region for α= 0.01 is Z= ± 2.58

Conclusion: Since the calculated z = 0.1601 does not fall in the critical region , so we accept the null hypothesis H0:pm=pw and conclude that the data does not appear to indicate that the tow probabilities are different.

Using the z-distribution, it is found that since the absolute value of the test statistic is less than the critical value, there values do not indicate a difference in popularity.

At the null hypothesis, it is tested if the proportions are equal, that is, their subtraction is of 0, hence:

[tex]H_0: p_w - p_m = 0[/tex]

At the alternative hypothesis, it is tested if they are different, that is, their subtraction is different of 0, hence:

[tex]H_1: p_w - p_m \neq 0[/tex]

The proportions and standard errors are:

[tex]p_m = 0.44, s_m = \sqrt{\frac{0.44(0.56)}{250}} = 0.0314[/tex]

[tex]p_w = 0.52, s_w = \sqrt{\frac{0.52(0.48)}{250}} = 0.0316[/tex]

For the distribution of the differences, the mean and the standard error are given by:

[tex]\overline{p} = p_w - p_m = 0.52 - 0.44 = 0.08[/tex]

[tex]s = \sqrt{s_m^2 + s_w^2} = \sqrt{0.0314^2 + 0.0316^2} = 0.0445[/tex]

The test statistic is given by:

[tex]z = \frac{\overline{p} - p}{s}[/tex]

In which p = 0 is the value tested at the null hypothesis.

Hence:

[tex]z = \frac{0.08}{0.0445}[/tex]

[tex]z = 1.795[/tex]

The critical value, for a two-tailed test, as we are testing if the mean is different of a value, with a significance level of 0.01, is of [tex]|z^{\ast}| = 2.5758[/tex]

Since the absolute value of the test statistic is less than the critical value, there values do not indicate a difference in popularity.

A similar problem, also involving an hypothesis test for a proportion, is given at https://brainly.com/question/24302053

Use z scores to compare the given values. The tallest living man at one time had a height of 249 cm. The shortest living man at that time had a height of 120.2 cm. Heights of men at that time had a mean of 176.55 cm and a standard deviation of 7.23 cm. Which of these two men had the height that was more​ extreme?

Answers

Answer:

Step-by-step explanation:

Average height = 176.55 cm

Height of tallest man = 249 cm

Standard deviation = 7.23

z score of tallest man

= (249 - 176.55) / 7.23

= 10.02

Average height = 176.55 cm

Height of shortest  man = 120.2 cm

Standard deviation = 7.23

z score of smallest  man

= ( 176.55 - 120.2 )  / 7.23

= 7.79

Since Z - score of tallest man is more , his height was more extreme .  

The number of cars sold annually by used car salespeople is normally distributed with a standard deviation of 17. A random sample of 470 salespeople was taken and the mean number of cars sold annually was found to be 69. Find the 95% confidence interval estimate of the population mean

Answers

Answer: Estimate mean is between 67.463 and 70.537

Step-by-step explanation: A 95% Confidence interval of a sample mean:

mean ± [tex]z.\frac{s}{\sqrt{n} }[/tex]

α = 1 - 0.95

α = 0.05

α/2 = 0.025

z-score of α/2 = 1.96

Knowing that mean = 69, sd = 17 and there were 470 salespeople in the sample:

69 ± [tex]1.96.\frac{17}{\sqrt{470} }[/tex]

69 ± [tex]1.96.\frac{17}{21.68}[/tex]

69 ± [tex]1.96.0.78[/tex]

69 ± 1.537

lower limit: 69 - 1.537 = 67.463

upper limit: 69 + 1.537 = 70.537

With a confidence of 95%, the estimate mean number of cars sold is between 67.463 and 70.537

Translate the sentence into an equation seven times the sum of a number and 5 is 4

Answers

Answer:

7(x+5) = 4

Step-by-step explanation:

"A number" refers to a variable. Seven is multiplied to the sum of a number and 5, meaning they must be in parentheses, indicating it is the sum.

Find the distance between the points (-3, -2) and (-1, -2). 2 √6 4

Answers

Answer:

Let the distance be AB.

So, by using distance formula, we get

AB=√(x^2-x^1)^2+(y^2-y^1)^2

AB=√[-1-(-3)]^2+[-2-(-2)]^2

AB=√(-1+3)²+(-2+2)²

AB=√2²+0²

AB=√4

AB=2 units

hope it helps u...

plz mark as brainliest...

Answer: The distance between the points (-3, -2) and (-1, -2). is 2

Simplify the expression.
16 • 4^-4
A. 256
B. -256
C. 1/16
D. -4,096

Answers

Answer:

C. 1/16

Step-by-step explanation:

[tex]16 * 4^{-4}[/tex]

16 can be written as a power of 4.

[tex]4^2 * 4^{-4}[/tex]

The bases are same, add exponents.

[tex]4^{2+-4}[/tex]

[tex]4^{-2}[/tex]

Simplify negative exponent.

[tex]\frac{1}{4^2 }[/tex]

[tex]\frac{1}{16}[/tex]

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