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3 years ago

8

Lao needs to be at school at 8:25 am. It takes her 15 minutes to walk to school, 3/12 hour to eat breakfast, 0.75 hours to get d

ressed, and 0.2 hours to shower. At what time should Lao wake up in the morning to be at school 10 minutes early?

1 answer:

Sati [7]3 years ago

8 0

Answer:

6:15

Step-by-step explanation:

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X+7/(7×+35)*(×^2-3×-40)/×-8

Leona [35]

I think it would be (x+7)/7

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3 years ago

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there is a beaker of 3.5% acid solution and a beaker of 6% acid solution in the science lab. Mr. Larson needs 200ml of 4.5% acid

disa [49]

Below are the choices:

A. 80 mL of the 3.5% solution and 120 mL of the 6% solution

<span>B. 120 mL of the 3.5% solution and 80 mL of the 6% solution </span>

<span>C. 140 mL of the 3.5% solution and 60 mL of the 6% solution </span>

<span>D. 120 mL of the 3.5% solution and 80 mL of the 6% solution
</span>
Let fraction of 3.5% in final solution be p.

<span>p * 3.5 + (1 - p) * 6 = 4.5 </span>

<span>3.5p + 6 - 6p = 4.5 </span>

<span>2.5p = 1.5 </span>

<span>p = 3/5 </span>

<span>3/5 * 200 = 120 </span>

<span>Therefore the answer is B. 120 ml of 3.5% and 80 ml of 6%.</span>

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3 years ago

Graph the line that represents this equation<br> Y=-5x+2

defon

Answer:

i cant answer this in brainly because of lack of display but if you plug in y=5x+2 into mathaway it will show you on a graph

Step-by-step explanation:

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3 years ago

B is 30 more than a. B is 6 times as many as a. What is the value of a and b ?

12345 [234]

Answer:

180 is the answer you multiple 6 times 30

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3 years ago

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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