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agasfer [191]
3 years ago
13

In 2013 the population of the state of New York was approximately 19.65 million and the population of New York City was 8,406,00

0 in 2013 how many people in New York State did not live in New York City?
Mathematics
1 answer:
SVEN [57.7K]3 years ago
7 0

Answer: 11,244,000

Step-by-step explanation:

From the question, we are informed that In 2013 the population of New York state was approximately 19.65 million and the population of New York City was 8,406,000.

To know the number of people in New York State did not live in New York City, we subtract 8,406,000 from 19.65 million. This will be:

= 19,650,000 - 8,406,000

= 11,244,000

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A random sample of 25 boxes of cereal have a mean of 372.5grams and a standard deviation of 15 grams. Does an average box of cer
ss7ja [257]

Using the t-distribution, it is found that since the <u>test statistic is less than the critical value for the right-tailed test</u>, there is not enough evidence to conclude that an average box of cereal contain more than 368 grams of cereal.

At the null hypothesis, it is <u>tested if the average box of cereal does not contain more than 368 grams</u>, that is:

H_0: \mu \leq 368

At the alternative hypothesis, it is <u>tested if it contains</u>, that is:

H_1: \mu > 368

We have the <em>standard deviation for the sample</em>, hence, the t-distribution is used to solve this question.

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the values of the <em>parameters </em>are: \overline{x} = 372.5, \mu = 368, s = 15, n = 25.

Hence, the value of the <em>test statistic</em> is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{372.5 - 368}{\frac{15}{\sqrt{25}}}

t = 1.5

The critical value for a <u>right-tailed test</u>, as we are testing if the mean is greater than a value, with a <u>significance level of 0.05,</u> with 25 -1 = <u>24 df</u>, is of t^{\ast} = 1.71.

Since the <u>test statistic is less than the critical value for the right-tailed test</u>, there is not enough evidence to conclude that an average box of cereal contain more than 368 grams of cereal.

You can learn more about the use of the t-distribution to test an hypothesis at brainly.com/question/13873630

6 0
3 years ago
Select the gcf of these numbers. 25 · 5· 11 and 23· 52 · 7 a 2^2 · 3
White raven [17]
The greatest common factor of the numbers is that which is the fused factors of all the numbers without the repetition of the common factors present in the given.
                          25·5·11 = 5·5·5·11
                          23·52·7 =23·2·2·13·7
Since, there are no repeated factors, the answer would be,
                                        5·5·5·11·23·2·2·13·7
The answer unfortunately is not among the choices.
7 0
3 years ago
A researcher is interested in estimating the mean weight of a semi tracker truck to determine the potential load capacity. She t
FinnZ [79.3K]

Answer:

(19229.11 ,20770.89)

Step-by-step explanation:

We are given the following information:

Sample size, n = 17

Sample mean = 20,000 pounds

Sample standard deviation = 1,500 pounds

Confidence level = 95%

Significance level = 5% = 0.05

95% Confidence interval:  

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 16 and}~\alpha_{0.05} = \pm 2.119  

20000 \pm 2.119(\displaystyle\frac{1500}{\sqrt{17}} ) = 20000 \pm 770.89 = (19229.11 ,20770.89)  

4 0
3 years ago
We have a bag that contains 100 balls, 50 of them red and 50 blue. Select 5 balls at random. What is the probability that 3 are
MakcuM [25]

Answer:

Likely, I think I havent been in this for a long time so i may be a little rusty. But anyways, I hope this helped!

Step-by-step explanation:

5 0
2 years ago
Which expression is equivalent to StartFraction c squared minus 4 Over c + 3 EndFraction divided by StartFraction c + 2 Over 3 (
Firdavs [7]

Answer:

C

Step-by-step explanation:

Given:

\dfrac{c^2-4}{c+3}\div  \dfrac{c+2}{3(c^2-9)}

Changing the division to multiplication by taking the reciprocal of the second fraction.

\dfrac{c^2-4}{c+3}X  \dfrac{3(c^2-9)}{c+2}

<u>The correct option is C</u>

3 0
3 years ago
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