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2 years ago

This is so important

Mathematics

1 answer:

Artist 52 [7]2 years ago

7 0

7/8 divided by 1/4 (or 2/8) = 3 1/8

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Can I have help with this!!!!!!!!!!

77julia77 [94]

Answer:

The answer is 19

Step-by-step explanation: 13 scored from 91-100, and 6 scored from 81-90, all you have to do is add them together.

3 0

2 years ago

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What is -10=10(x+9) and could you tell me how to do it cause I don't understand.

timofeeve [1]

-10=10x+90
80=10x
x=80/10
x=8

4 0

2 years ago

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The sum of the measures of the two angles of a linear pair must be

Pie

Answer: 180

Angles in a linear pair are always supplementary. This means they form a straight angle or a straight line.

6 0

2 years ago

Write the expression as the sine or cosine of an angle sin(pi/7) cos(x) + cos(pi/7) sin(x)

Natasha_Volkova [10]

Answer:

$\sin(\frac{\pi}{7}+x)$

Step-by-step explanation:

We are going to use the identity

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

because this identities right hand side matches your expression where

$a=\frac{\pi}{7}$ and $b=x$ .

So we have that $\sin(\frac{\pi}{7})\cos(x)+\cos(\frac{\pi}{7})\sin(x)$ is equal to $\sin(\frac{\pi}{7}+x)$ .

5 0

3 years ago

Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of

Alexxx [7]

Answer:

a) 0.018

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0

3 years ago

This is so important​

This is so important