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4 years ago

Help me please, estimate it too please everybody forgets that part. (ಥ_ಥ)

Mathematics

2 answers:

Maslowich4 years ago

5 0

I hope this helps you 1) 2/5+2/3 2.3/5.3+2.5/3.5 6/15+10/15 16/15 2) 4.8+1/8-3.2/4.2 33/8-6/8 27/8 3) 1.3+1/3-1/4 4/3-1/4 4.4/3.4-1.3/4.3 16/12-3/12 13/12

Elis [28]4 years ago

4 0

It is 3 if you estimate because it is less then 3 and a half'

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Write an equation in slope-intercept form of the line that passes through (7, 2) and (2, 12)

Effectus [21]

Y=ax+b
for x=7, y=2
7a+b=2

for x=2, y=12
2a+b=12

7a-2a+b-b=2-12
5a= -10
a= -10/5
a=-2
2a+b=12
2*(-2)+b=12
-4+b=12
b=12+4
b=16
y=-2x+16

6 0

4 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> $p_1-p_2$ = [ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ]

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

= [-0.017 , 0.697]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].

6 0

4 years ago

What is the equation of a line with a y-intercept of -3 and a slope of - 4/5?

ElenaW [278]

I think it’s D not sure good luck though!

7 0

3 years ago

The width of a rectangle is 13 units less than the length. If the area is 90 square units, then find the dimensions of the recta

Artemon [7]

If the area is 90 square units. Then the dimension of the rectangle will be 5 units and 18 units.

<h3>What is the area of the rectangle?</h3>

Let L be the length and W be the width of the rectangle.

Then the area of the rectangle will be

Area of the rectangle = L×W square units

The width of a rectangle is 13 units less than the length.

If the area is 90 square units.

W = L – 13

Then we have

90 = L(L – 13)

L² – 13L – 90 = 0

L² – 18L + 5L – 90 = 0

L(L – 18) + 5(L – 18) = 0

(L – 18)(L + 5) = 0

L = -5, 18

Then the width will be

W = 18 – 13

W = 5 units

Then the dimension of the rectangle will be 5 units and 18 units.

More about the area of the rectangle link is given below.

brainly.com/question/20693059

#SPJ1

4 0

2 years ago

How to change the decimal 4.5 into a fraction

alukav5142 [94]

Answer:

9/2

Step-by-step explanation:

8 0

3 years ago

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