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DedPeter [7]
2 years ago
12

the length of a sandbox is three feet longer than it’s width. Write the expression that would represent the area of the sandbox.

Mathematics
1 answer:
Rashid [163]2 years ago
6 0

Part A : D.)

Part B : Length of the sandbox is 10 feet.

Step-by-step explanation:

Given,

Perimeter = 29 ft

We need to find the equation for the perimeter and also the length of the sandbox.

Solution,

Let the width of the sandbox be 'w'.

Now as per question said;

The length of the sandbox is 1 foot longer than twice the width of the sandbox.

So we can say that;

Length =

Now we know that the perimeter is equal to the sum of twice of length and width.

framing in equation form, we get;

Perimeter =

we have given the perimeter, so on substituting the value, we get;

Hence The equation used to find the width is  .

Now we solve for 'w'.

Applying distributive property, we get;

Subtracting both side by '2' we get

Dividing both side by 6 we get;

Width of the sandbox = 4.5 ft

Length of the sandbox =

Hence Length of the sandbox is 10 feet.

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The sketch is drawn at the end.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 0°C and a standard deviation of 1.00°C.

This means that \mu = 0, \sigma = 1

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

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

Z = \frac{-1.69 - 0}{1}

Z = -1.69

Z = -1.69 has a p-value of 0.0455

X = -2.23

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

Z = \frac{-2.23 - 0}{1}

Z = -2.23

Z = -2.23 has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

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