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3241004551 [841]
3 years ago
6

in a class of students, the following data table summarizes how many students have a brother or sister. What is the probability

that a student chosen from randomly from the class does not have a sister? ​

Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
6 0
Answer: 13/28
Explanation:
melamori03 [73]3 years ago
4 0

Answer:

Step-by-step explanation:

The fact that the student does not have a brother either only means that you have chosen an only child. So there are 13 people that do not have a sister. The 13 comes from the bottom row. 9 + 4 = 13

What is the total number of people in the class? There must be 15 more who do have a sister.

15 + 13 = 28

So the answer should be 13 / 28 or 0.464

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) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa
asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

<u><em>Let </em></u>p_1<u><em> = population proportion of drivers 16-24 years old who buckle up .</em></u>

<u><em /></u>p_2<u><em> = population proportion of drivers 25-69 years old who buckle up .</em></u>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no significant difference in seat belt use}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a difference in seat belt use}

The test statistics that would be used here <u>Two-sample z proportion statistics;</u>

                     T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of drivers 16-24 years old who buckle up = 79%

\hat p_2 = sample proportion of drivers 25-69 years old who buckle up = \frac{924}{1100} = 84%

n_1 = sample of 16-24 years old drivers = 1000

n_2 = sample of 25-69 years old drivers = 1100

So, <u><em>test statistics</em></u>  =  \frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }  

                              =  -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u><em>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

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

Therefore, we conclude that there is a difference in seat belt use.

4 0
3 years ago
Given that g(x)=x+3 calculate g(-5)
hjlf

Note that -5 = x, plug in -5 for x

g(x) = x + 3

g(-5) = (-5) + 3

Simplify

g(-5) = -5 + 3

g(-5) = -2

g(-5) = -2 is your answer

~

6 0
3 years ago
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Tamara took guitar lessons for 5.8 years. Then she took trumpet lessons for the next 3.4 years.
BlackZzzverrR [31]
The answer is 9.2 hope this helps
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2/3,2/6,2/8
23 is the greatest because it has larger pieces
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Answer: The measurement of angle B is 34. The measurement of angle C is 108. The measurement of angle F is 72.

Step-by-step explanation:

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