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3 years ago

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the ages of students in a computer class are 14, 13, 14,15,14,35,14. Decide which measure of center(s) best describes the data s

et. Explain your reasoning

1 answer:

atroni [7]3 years ago

5 0

I think...Would it be 15?

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Find the exact value of the trigonometric function given that sinU=-7/25 and cosV=-4/5 (Both U and V are in Quadrant III)

Contact [7]

cosU:-

$\\ \rm\longmapsto \sqrt{1-sin^2U}=\sqrt{1-49/625}=24/25$

As U lies in Q3

cosU=-24/25

sinV

$\\ \rm\longmapsto \sqrt{1-cos^2V}=\sqrt{1-16/25}=3/4$

As V lies in Q3

sinV=-3/5

So

sin(V-U)=sinVcosU-cosVsinU=(-3/5)(-24/25)-(-4/5)(-7/25)=72/125-28/125=72-28/125=<u>4</u><u>4</u><u>/</u><u>1</u><u>2</u><u>5</u>
cos(U-V)=cosUcosV+sinUsinV=(-24/25)(-4/5)+(-7/25)(-3/5)=96/125+21/125=117/125

3 0

2 years ago

According to the graph, what is the value of the constant in the equation

BaLLatris [955]

Answer:

The answer is C: 75

Step-by-step explanation:

The constant is found by multiplying your height and width, so thats really all you do.

25x3=75

15x3=75

3x15=75

25x3=75

75 would be the constant.

7 0

3 years ago

Read 2 more answers

What are the possible values of x for the diagram

kaheart [24]

Answer:

I think its 21 but iam not sure about the rest

8 0

3 years ago

Read 2 more answers

NarStor, a computer disk drive manufacturer, claims that the median time until failure for their hard drives is more than 14,400

Ne4ueva [31]

Answer:

The test statistics is $t = -1.727$

Step-by-step explanation:

From the question we are told that

The data given is

330 620 1870 2410 4620 6396 7822 81028309 12882 14419 16092 18384 20916 23812 25814

The population mean is $\mu = 14400$

The sample size is n = 16

The null hypothesis is $\mu \le 14400$

The alternative hypothesis is $H_a : \mu > 14400$

The sample mean is mathematically evaluated as

$\= x = \frac{\sum x_i}{n}$

So

$\= x = \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}$

=> $\= x = 10799.9$

The standard deviation is mathematically represented as

$\sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}$

So

$\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}} \ ..$

$..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2 }{16}} \ ...$

$\ ... \sqrt{\frac{(23812 - 10799.9)^2 +(25814 - 10799.9)^2 }{16}}$

=> $\sigma = 8340$

Generally the test statistic is mathematically represented as

$t = \frac{10799.9- 14400}{ \frac{8340}{\sqrt{16} } }$

$t = -1.727$

From the z-table the p-value is

$p-value = P(Z > t) = P(Z > -1.727) = 0.95792$

From the values obtained we see that

$p-value > \alpha$ so we fail to reject the null hypothesis

Which implies that the claim of the NarStor is wrong

5 0

3 years ago

A publisher reports that 344% of their readers own a particular make of car. A marketing executive wants to test the claim that

inysia [295]

Answer:

No, there is not enough evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.

And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;

Null Hypothesis, $H_0$ : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}

Alternate Hypothesis, $H_1$ : p $\neq$ 0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}

The test statistics we will use here is;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, p = actual % of readers who own a particular make of car = 0.34

$\hat p$ = percentage of readers who own a particular make of car in a

sample of 220 = 0.30

n = sample size = 220

So, Test statistics = $\frac{0.30 -0.34}{\sqrt{\frac{0.30(1- 0.30)}{220} } }$

= -1.30

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.

3 0

3 years ago