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3 years ago

I am a solid figure its two congruent polygons are bases, connected with lateral faces that are rectangles

1 answer:

8 0

A rectangular prism is a figure with congruent polygon bases, connected with lateral faces that are rectangles, so your answer would be rectangular prism!

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In a survey of women in a certain country the mean height was 62.9 inches with a standard deviation of 2.81 inches answer the fo

Natasha2012 [34]

The question is incomplete. The complete question is :

In a survey of women in a certain country ( ages 20-29), the mean height was 62.9 inches with a standard deviation of 2.81 inches. Answer the following questions about the specified normal distribution. (a) What height represents the 99th percentile? (b) What height represents the first quartile? (Round to two decimal places as needed)

Solution :

Let the random variable X represents the height of women in a country.

Given :

X is normal with mean, μ = $62.9$ inches and the standard deviation, σ = $2.81$ inches

Let,

$Z=\frac{X - 62.9}{2.81}$ , then Z is a standard normal

a). Let the $99th$ percentile is = a

The point a is such that,

$$P(X$

$P \left( Z < \frac{a-62.9}{2.81} \right) = 0.99$

From standard table, we get : $P( Z < 2.3263) =0.99$

∴ $\frac{(a-62.9)}{281} = 2.3263$

$a= (2.3263 \times 2.81 ) +62.9$

= 6.536903 + 62.9

= 69.436903

= 69.5 (rounding off)

Therefore, the height represents the $99th$ percentile = 69.5 inches.

b). Let b = height represents the first quartile.

It is given by :

$P( X < b) =0.25$

$P \left( Z < \frac{(b-62.9)}{2.81} \right) = 0.25$

From the standard normal table,

$P( Z < -0.6745) =0.99$

∴ $\frac{(b-62.9)}{2.81}= 0.6745$

$b=(0.6745 \times 2.81) +62.9$

= 1.895345 + 62.9

= 64.795345

= 64.8 (rounding off)

Therefore, the height represents the 1st quartile is 64.8 inches.

8 0

3 years ago

Oscar is trying to incorporate more exercise into his busy schedule. He has several short

katovenus [111]

Answer:

The arm routine lasts 10 minutes, and the abs routine lasts 15 minutes.

Step-by-step explanation:

Since Oscar is trying to incorporate more exercise into his busy schedule, and he has several short exercise routines he can complete at home, and last week, he worked out for a total of 35 minutes by doing 2 arm routines and 1 abdominal routine while That this week, he has completed 4 arm routines and 1 abdominal routine and spent a total of 55 minutes exercising, to determine how long does each routine last the following calculation must be performed:

2ARM + 1ABS = 35

4ARM + 1ABS = 55

4ARM + 1ABS - 2ARM - 1ABS = 55 - 35

2ARM = 20

ARM = 20/2

ARM = 10

2 x 10 + ABS = 35

ABS = 15

Therefore, the arm routine lasts 10 minutes, and the abs routine lasts 15 minutes.

7 0

3 years ago

Use a system of equations to solve this problem. Dr. Graham currently has two acid solutions. She only has a 60% acid solution a

Natalka [10]

Dr. Graham currently has two acid solutions.
60% acid AND 20% acid 
Dr. Graham needs 30 L of a 50% acid solution
We set up 2 equations in which s = 60% acid and t = 20% acid
A) s + t = 30
B) .60s + .20t = (.50 * 30)
We multiply equation A by -.20
A) = -.20s -.20t = -6 then we add it to B)
B) .60s + .20t = 15
.40s = 9
s = 22.5
t = 7.5
So, she needs to mix 22.5 liters of 60% acid with 7.5 liters of 20% acid.

Source:
http://1728.org/mixture.htm

4 0

3 years ago

7-(5-4)/3 A)7 B)1 C)4 D)2

Tpy6a [65]

The answer is around 6.66 but rounded up would be 7 so the answer would be A

3 0

4 years ago

Simplify the square root of 32/5

elixir [45]

Remember $\sqrt{ \frac{x}{y} } = \frac{ \sqrt{x} }{ \sqrt{y} }$
and
$\sqrt{ab^2} =b \sqrt{a}$
$\sqrt{ \frac{32}{5} } = \frac{ \sqrt{(4^2)(2)} }{ \sqrt{5} } = \frac{ 4\sqrt{2} }{ \sqrt{5} }$
times whoe thing by $\frac{ \sqrt{5} }{ \sqrt{5} }$
$\frac{4 \sqrt{10} }{5}$

3 0

3 years ago