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4 years ago

8

the figure below is made of rectangles what is the perimeter of the figure below? note that not all the side lengths are labeled

.

1 answer:

Anit [1.1K]4 years ago

5 0

The perimeter is 30cm

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Find the circumferences of both circles to the nearest tenth.

Liula [17]

Answer:

The Orange circle is: 78.5 cm

And The Blue circle is 314.2 cm

Step-by-step explanation:

By taking the formula r^2(pi), and filling it in with the radius for the orange circle you will get 78.5. For the blue circle the radius would be the orange's radius combined with the rim given. Using the same formula you will get 314.2.

5 0

3 years ago

Read 2 more answers

I need the answer for #13

ira [324]

Is that the top or bottom? Assuming it's the top, when distributing the -2 to what's inside the parenthesis, they were correct in multiplying -2•7, but when they multiplied -2 and -y, the answer needs to be positive 2y because they are multiplying double negatives

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3 years ago

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Which is true about an equilateral triangle?

Keith_Richards [23]

Answer: all angle measures are equal

6 0

3 years ago

Guys, please help i can't move on

professor190 [17]

Answer:

15

Step-by-step explanation:

32-19 = 13 that have both

70-57= 13 that have both again from that sample

total number of people that have both : 13 + 13 = 26

11 kids don't have either one, so :

26-11 = 15

8 0

3 years ago

Fewer young people are driving. In year A, 67.9% of people under 20 years old who were eligible had a driver's license. Twenty y

aleksley [76]

Answer:

Case a Case b

margin of error 0.0216 0.0231

Interval estimate (0.7016 , 0.6795) (0.5031 , 0.4569)

margin of error is not same in both cases.

Step-by-step explanation:

a

At 95% confidence interval the interval estimate of number of 20 year old drivers in year A can be computed as

p' ± z $\sqrt{\frac{p'(1-p')}{n} }$

= 0.68 ± 1.96 $\sqrt{\frac{0.68(1-0.68)}{1800} }$

= 0.7016 , 0.6795

the margin of error can be written as

z $\sqrt{\frac{p'(1-p')}{n} }$

= 1.96 $\sqrt{\frac{0.68(1-0.68)}{1800} }$

= 0.0216

b

At 95% confidence interval the interval estimate of number of 20 year old drivers in year B can be computed as

p' ± z $\sqrt{\frac{p'(1-p')}{n} }$

= 0.48 ± 1.96 $\sqrt{\frac{0.48(1-0.48)}{1800} }$

= 0.5031 , 0.4569

the margin of error can be written as

z $\sqrt{\frac{p'(1-p')}{n} }$

= 1.96 $\sqrt{\frac{0.48(1-0.48)}{1800} }$

= 0.0231

c

Sample size is same in case A and B but proportion is different in both cases so margin of error is different in both cases

3 0

3 years ago