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3 years ago

The length of a rectangle is 2 in longer than its width. if the perimeter of the rectangle is 44 in , find its area.

2 answers:

7 0

P = 2(L + W)
P = 44
L = W + 2

44 = 2(W + 2 + W)
44 = 2(2W + 2)
44 = 4W + 4
44 - 4 = 4W
40 = 4W
40/4 = W
10 = W

L = W + 2
L = 10 + 2
L = 12

A = L * W
L = 12
W = 10

A = 12 * 10
A = 120 square inches <===

Tpy6a [65]3 years ago

3 0

The are of the rectangle would be 120 bcause the perimeter is 44 and the lenght is 2in bigger then the width and 10+10+12+12=44 so if you do length times width you get 12*10=12

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Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.3. (Round your ans

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Answer:

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 1.3

Sample size, n = 12

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

Formula:

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

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{1.3}{\sqrt{12}} = 0.3753$

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$P( x \geq 51) = P( z \geq \displaystyle\frac{51 - 50}{0.3753}) = P(z \geq 2.6645)$

$= 1 - P(z < 2.6645)$

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

$P(x \geq 51) = 1 - 0.9961= 0.0039$

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.

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3 years ago

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