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Vinil7 [7]
3 years ago
15

Which expression is equivalent to One-fifth (150 x minus 80 y + 50 minus 50 x minus 25 y + 20)?

Mathematics
2 answers:
V125BC [204]3 years ago
4 0

Answer:

a

Step-by-step explanation:

inna [77]3 years ago
3 0

Answer:

The answer to your question is  20x + 21y + 14        

Step-by-step explanation:

Data

Expression             1/5 (150x - 80y + 50 - 50x - 25y + 20)

Process

1.- Multiply 1/5 by each term

                                     150/5x - 80/5y + 50/5 - 50/5x - 25/5y + 20/5

2.- Simplify

                                      30x - 16y + 10 - 10x - 5y + 4

3.- Simplify like terms

                                      (30x - 10x) + (-16y - 5y) + (10 + 4)

4.- Result

                                        20x + 21y + 14                                        

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The average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 in
Murrr4er [49]

Answer:

Probability that average height would be shorter than 63 inches = 0.30854 .

Step-by-step explanation:

We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 64 inches

           \sigma = standard deviation = 4 inches

           n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{63-64}{\frac{4}{\sqrt{4} } } ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

                                                        = 1 - 0.69146 = 0.30854

Hence, it is 30.85%  likely that average height would be shorter than 63 inches.

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3 years ago
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