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3 years ago

Начало сказки "Снежная королева" на 20 странице, а конец на 50.

1 answer:

7 0

If you start on page 20 and end on Page 50, you will have read 31 pages. I just counted from 20 to 50, including the number 20 in it.

OR: You already know there are 50-20=30 pages. But that's without counting the 20 in it. So when you add the extra 20: 30 + 1 = 31

Translated version:

Если вы начнете со страницы 20 и закончите со страницы 50, вы прочтете 31 страницу. Я только что насчитал от 20 до 50, включая число 20 в нем.

ИЛИ: Вы уже знаете, что 50-20 = 30 страниц. Но это без учета 20 в нем. Поэтому, когда вы добавляете дополнительные 20: 30 + 1 = 31

Я использовал Google Translate, поэтому результаты могут быть неточными.

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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.

Let $p_1$ = population proportion of drivers 16-24 years old who buckle up .

$p_2$ = population proportion of drivers 25-69 years old who buckle up .

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 Two-sample z proportion statistics;

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, test statistics = $\frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }$

= -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%. Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

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 we reject our null hypothesis.

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

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3 years ago

I got two different answers and ended up just confusing myself.

Vadim26 [7]

Answer:

x=16

Step-by-step explanation:

multiply both sides of the equation by 2

multiply the number

reduce the number

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