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Vera_Pavlovna [14]
3 years ago
10

PLS HURRY FIRST ONE TO DO IT GETS BRAINLEIST Maria and her classmates are standing with other people in a line of 35 people wait

ing to buy tickets. Maria is last in this line of classmates and 20th from the front of the whole line. Maria's classmate leading the line of classmates is 23rd from the end of the whole line. How many classmates does Maria have?
Mathematics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

7 classmates

Step-by-step explanation:

The last person in the line is in position 35.

The Maria's leading classmate is in the position 35-23+1 = 13.

So, Maria's classmates are from position 13 to 19 (not including herself), and their number is 19-12 = 7 persons.

ANSWER.  Maria has 7 classmates  (not including herself).

Hope this helps.

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Hypothesis Testing
Yuri [45]

Answer:

<u>Problem 1</u>: We conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

<u>Problem 2</u>: We conclude that the volume of Google stock has changed.

Step-by-step explanation:

<u>Problem 1:</u>

We are given that in a recent survey conducted by Pew Research, it was found that 156 of 295 adult Americans without a high school diploma were worried about having enough saved for retirement.

Let p = <em>proportion of adult Americans without a high school diploma who are worried about having enough saved for retirement</em>

So, Null Hypothesis, H_0 : p \leq 50%    {means that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement}

Alternate Hypothesis, H_A : p > 50%     {means that a majority of adult Americans without a high school diploma are worried about having enough saved for retirement}

This is a right-tailed test.

The test statistics that would be used here is <u>One-sample z-test</u> for proportions;

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

where, \hat p = sample proportion of adult Americans who were worried about having enough saved for retirement = \frac{156}{295} = 0.53

           n = sample of adult Americans = 295

So, <u><em>the test statistics</em></u> =  \frac{0.53-0.50}{\sqrt{\frac{0.50(1-0.50)}{295} } }

                                    =  1.03

The value of z-test statistics is 1.03.

<u>Also, the P-value of the test statistics is given by;</u>

              P-value = P(Z > 1.03) = 1 - P(Z \leq 1.03)

                           = 1 - 0.8485 = <u>0.1515</u>

Now, at a 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.03 < 1.645, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

<u>Problem 2:</u>

We are given that a random sample of 35 trading days in 2014 resulted in a  sample mean of 3.28 million shares with a standard deviation of 1.68 million shares.

Let \mu = <em>mean daily volume in Google stock</em>

So, Null Hypothesis, H_0 : \mu = 5.44 million shares    {means that the volume of Google stock has not changed}

Alternate Hypothesis, H_A : \mu \neq 5.44 million shares     {means that the volume of Google stock has changed}

This is a two-tailed test.

The test statistics that would be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                       T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean volume in Google stock = 3.28 million shares

            s = sample standard deviation = 1.68 million shares

           n = sample of trading days = 35

So, <u><em>the test statistics</em></u> =  \frac{3.28-5.44}{\frac{1.68}{\sqrt{35} } }  ~ t_3_4

                                    =  -7.606

The value of t-test statistics is -7.606.

<u>Also, the P-value of the test statistics is given by;</u>

              P-value = P(t_3_4 < -7.606) = Less than 0.05%

Now, at a 0.05 level of significance, the t table gives a critical value of -2.032 and 2.032 at 34 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, <u><em>so we sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the volume of Google stock has changed.

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3 years ago
Can some one help me i need it done by today
lapo4ka [179]

Answer:


Number 9: 1.25 miles

Step-by-step explanation:

In total Brian hiked 16 miles

In total Christine hiked 14.75

So, 16 - 14.75 = 1.25

Overall, Brian hiked 1.25 more miles than Christine.

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A graphing calculator shows it will take about 5.5 seconds.

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Is the number 526 a multiple of 6?
Allisa [31]
No!!! because 526 is not perfectly divisible by 6
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A gardener is planting two types of trees. Type A is 2 feet tall and grows at a rate of 22 inches per year. Type B is 3 feet tal
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