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Alik [6]
3 years ago
10

The point of a square pyramid is cut off, making each lateral face of the pyramid a trapezoid with the dimensions shown. A trape

zoid has a base of 3 inches, height of 1 inches, and top side length of 1 inch:
What is the area of one trapezoidal face of the figure?

Mathematics
1 answer:
aksik [14]3 years ago
5 0
Use the trapezium formula 1/2( sum of parallel lines)( height )

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Step-by-step explanation:

this is the slope and b

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I count 10 triangles.....

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What is 3 1/2 - 2 5/7 and show the work
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3 \frac{1}{2}  -2 \frac{5}{7} = \frac{(3\cdot2)+1}{2} -  \frac{(2\cdot7)+5}{7} =  \frac{7}{2} -  \frac{19}{7} \\\\= \frac{7\times 7}{2\times 7} -  \frac{19\times 2}{7\times 2} =  \frac{49}{14} - \frac{38}{14} \\\\= \frac{49-38}{14} =\boxed{\bf{\frac{11}{14}}}
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3 years ago
(3 marks) A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year. The battery live
Ksenya-84 [330]

Answer:

0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years

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.

A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year

This means that \mu = 3, \sigma = 0.5

What is the probability that a given battery will last between 2.3 and 3.6 years?

This is the p-value of Z when X = 3.6 subtracted by the p-value of Z when X = 2.3. So

X = 3.6

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

Z = \frac{3.6 - 3}{0.5}

Z = 1.2

Z = 1.2 has a p-value of 0.8849

X = 2.3

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

Z = \frac{2.3 - 3}{0.5}

Z = -1.4

Z = -1.4 has a p-value of 0.0808

0.8849 - 0.0808 = 0.8041

0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years

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The profit a company earns every month depends on the amount of product sold,p, for $855 each and the amount spent in rent, util
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the CEO earns about <em><u>31045.5</u></em> in a given month

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