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Gelneren [198K]

3 years ago

9

The reasures of two angles have a sum of 180°. The measures of the angles are in a ratio of 5:1. Determine the measures of both

angles.

1 answer:

Mademuasel [1]3 years ago

3 0

Answer:

150° : 30°

Step-by-step explanation:

total ratio

5 + 1 = 6

ratio 1

180 ÷ 6 = 30

ratio 5

30 × 5 = 150

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Answer: .20x+1.00

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3 years ago

Which is the correct coefficient matrix for this system? 3x - 4y = 12 x + 6y = 10

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2 years ago

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Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. 27% of the possible Z values are grea

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Answer:

27% of the possible Z values are greater than 0.613

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 0, \sigma = 1$

27% of the possible Z values are greater than

The 100 - 27 = 73rd percentile, which is X when Z has a pvalue of 0.73. So X when the z-score is 0.613.

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

$0.613 = \frac{X - 0}{1}$

$X = 0.613$

27% of the possible Z values are greater than 0.613

6 0

3 years ago

The mean per capita income is 20,90820,908 dollars per annum with a standard deviation of 407407 dollars per annum. What is the

LiRa [457]

Answer:

$z=\frac{20880-20908}{\frac{407}{\sqrt{55}}}= -0.51$

$z=\frac{20936-20908}{\frac{407}{\sqrt{55}}}= 0.51$

Then we can find the probability of interest with this difference:

$P(-0.51$

And using the normal standard distribution or excel we got:

$P(-0.51$

So then the probability that the sample mean would differ from the true mean by less than 28 dollars from the sample of 55 is approximately 0.390

Step-by-step explanation:

We define the variable of interest as the per capita income and we know the following properties for this variable:

$\mu=20908$ and $\sigma=407$

We want to find this probability:

$P(20908-28$

We select a sample size of n=55 and we define the z score formula given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

We can find the z score then for 20880 and 20936 and we got:

$z=\frac{20880-20908}{\frac{407}{\sqrt{55}}}= -0.51$

$z=\frac{20936-20908}{\frac{407}{\sqrt{55}}}= 0.51$

Then we can find the probability of interest with this difference:

$P(-0.51$

And using the normal standard distribution or excel we got:

$P(-0.51$

So then the probability that the sample mean would differ from the true mean by less than 28 dollars from the sample of 55 is approximately 0.390

7 0

3 years ago