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otez555 [7]
2 years ago
6

The area of the rectangle below is 18 square units. What is the width of the rectangle? Write and solve a division equation to f

ind the width of the rectangle.
Mathematics
1 answer:
kherson [118]2 years ago
7 0

Answer:

24/5 or 4 and 4/5

Step-by-step explanation:

3 and 3/4=15/4

18 divided by 15/4=4 and 4/5

4 and 4/5=24/5

24/5 times 15/4=18

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Answers for the 2 boxes please ​
almond37 [142]

Answer:

(4,9)

Step-by-step explanation:

y = 2x + 1

y = x + 5

2x + 1 = x + 5

2x - x = 5 - 1

x = 4

y = x + 5

y = 4 + 5

y = 9

Check:

(4,9) ? Yep

3 0
2 years ago
What dose x represent
goldenfox [79]
A variable Because of the math question like let’s say 40•12*-* divided by x
5 0
3 years ago
Read 2 more answers
Celery<br> celery<br> celery<br> celery<br> celery
ad-work [718]

Answer:

d, 3 5/8

Step-by-step explanation:

3 0
2 years ago
PLZ HELP INSCRIBED ANGLES
den301095 [7]
54
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7 0
2 years ago
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.62 - 0.5}{0.05}

Z = 2.4

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

3 0
2 years ago
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