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3 years ago

I neeeeeed helpppppp

Mathematics

1 answer:

AysviL [449]3 years ago

4 0

Answer:

8.25y-14

Step-by-step explanation:

$4(1.75y-3.5)+1.25y$

$4 * 1.75y+4$ * $(-3.5)+1.25y$

$=7y-14+1.25y$

$(7y+1.25y)-14$

$8.25y-14$

The simplified expression is 8.25y-14

I hope this helps!

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

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The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .

Misha Larkins [42]

Answer:

0.1507 or 15.07%.

Step-by-step explanation:

We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.

First of all, we will find z-scores for data points using z-score formula.

$z=\frac{x-\mu}{\sigma}$ , where,

z = z-score,

x = Sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

$z=\frac{21.97-22}{0.016}$

$z=\frac{-0.03}{0.016}$

$z=-0.1875$

Let us find z-score of data point 22.03.

$z=\frac{22.03-22}{0.016}$

$z=\frac{0.03}{0.016}$

$z=0.1875$

Using probability formula $P(a$ , we will get:

$P(-0.1875$

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.

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3 years ago

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2 years ago

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Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

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