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zloy xaker [14]
3 years ago
9

Theo is building a garden against his house. He bought two pieces of wood that are each five feet long. He wants to create a tri

angular garden against his house using the two pieces of wood, without cutting them. His house will be the third side of the triangle. He also wants the perimeter of his garden to be a whole number.
There are different triangles he can create that fit these conditions. Of these triangles, there are (fill in) isosceles, (fill in) scalene, and (fill in) equilateral.
Mathematics
1 answer:
liubo4ka [24]3 years ago
7 0
Given:
two pieces of wood that are 5 feet long each.
form a triangle with a perimeter of a whole number.
third side the triangle will be the house 

Since two sides are already of the same length, he can create an isosceles triangle. 

An isosceles triangle has two equal sides and two equal angles.

Assuming that the 3rd side is also 5 feet long, then, he can create an equilateral triangle.

An equilateral triangle has three equal sides and three equal angles, measuring 60° each angle.
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What number is 40% of 307<br> Input only whole numbers.
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A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
lara [203]

Answer:

7.3% of the bearings produced will not be acceptable

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}

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 0.499, \sigma = 0.002

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

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

Z = \frac{0.504 - 0.494}{0.002}

Z = 2.5

Z = 2.5 has a pvalue of 0.9938

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pvalue of Z when X = -1.5

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

Z = \frac{0.496 - 0.494}{0.002}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.0668 + 0.0062 = 0.073

7.3% of the bearings produced will not be acceptable

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3 years ago
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