Elements smaller than iron are made in supernovas, true or false?

If angle 2 and angle 3 are supplementary, what is the reason

kondaur [170]

They add up to 180 degrees

4 0

3 years ago

Mr. James has 53 paint brushes to share with 4 classes. Are there enough for each class to have the same amount? Are there any l

olga2289 [7]

Answer:

Mr. James can give each class 13 paint brushes and have one left over.

Step-by-step explanation:

Since he can't give a class a fraction of a paint brush, you don't divide any farther and get a remainder of 1. I attached a picture of the long division.

5 0

3 years ago

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple

polet [3.4K]

Answer:

95% confidence interval for the difference in the proportion is [-0.017 , 0.697].

Step-by-step explanation:

We are given that a simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 8 were "totaled," meaning that the cost of repairs is greater than the value of the car.

Another sample of 15 large cars were subjected to the same test, and 5 of them were totaled.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

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

where, $\hat p_1$ = sample proportion of small cars that were totaled = $\frac{8}{12}$ = 0.67

$\hat p_2$ = sample proportion of large cars that were totaled = $\frac{5}{15}$ = 0.33

$n_1$ = sample of small cars = 12

$n_2$ = sample of large cars = 15

$p_1$ = population proportion of small cars that are totaled

$p_2$ = population proportion of large cars that were totaled

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

So, 95% confidence interval for the difference between population population, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < $p_1-p_2$ < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

<u>95% confidence interval for</u> $p_1-p_2$ = [ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ]

= [ $(0.67-0.33)-1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ , $(0.67-0.33)+1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ ]