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hodyreva [135]
3 years ago
5

Need help with number 10

Mathematics
1 answer:
katovenus [111]3 years ago
6 0
Multiply 62 by 3 and you get 186
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Is it ever possible for a triangle's altitude to lie entirely outside the triangle?
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Answer:

If all three of the latitudes lie entirely outside the triangle

its a because all the altitudes have to cross each other so when an obtuse angle is being measured the altitudes have to cross but the only way they can can cross is if they are outside of the angle to get  the right mesaurement.

so yes its possible.

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There are 20 wild pigs on an island and the number of pigs doubled each year for the past 5 years. The independent variable is:
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Step-by-step explanation:

This is because the wild pigs' fertilty rate increased for the past five years,  which caused them to double each year.

3 0
3 years ago
The diameters of a batch of ball bearings are known to follow a normal distribution with a mean 4.0 in and a standard deviation
8_murik_8 [283]

Answer:

a) 88.54% probability of a diameter between 3.8 in and 4.3 in

b) 25.14% probability of a diameter smaller than 3.9in

c) 90.82% probability of a diameter larger than 4.2 in

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 = 4, \sigma = 0.15

(a) a diameter between 3.8 in and 4.3 in,

This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.

X = 4.3

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

Z = \frac{4.3 - 4}{0.15}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 3.8

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

Z = \frac{3.8 - 4}{0.15}

Z = -1.33

Z = -1.33 has a pvalue of 0.0918

0.9772 - 0.0918 = 0.8854

88.54% probability of a diameter between 3.8 in and 4.3 in

(b) a diameter smaller than 3 9 in,

This is the pvalue of Z when X = 3.9. So

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

Z = \frac{3.9 - 4}{0.15}

Z = -0.67

Z = -0.67 has a pvalue of 0.2514

25.14% probability of a diameter smaller than 3.9in

(c) a diameter larger than 4.2 in

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

Z = \frac{4.2 - 4}{0.15}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

90.82% probability of a diameter larger than 4.2 in

6 0
3 years ago
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