Write an equation of the line passing theough the point (4,-1) and perpendicular to the line y=2x-5.

What measure of central tendency is most affected by outliers?.

Elden [556K]

Answer: Mean

Mean is the only measure of central tendency that is always affected by an outlier.

hope this helps!

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8 0

2 years ago

2x-7=117 what is the value of x

Tcecarenko [31]

$2x - 7 = 117$

Add 7 to both sides

$7 + 2x - 7 = 117 + 7$

At the left hand side 7 will cancel 7

At the right hand side 7 will add up

$2x = 124$

Divide through by 2

$x = \frac{124}{2}$

simplify the fraction

$x = 62$

4 0

3 years ago

Please help quickly this is timed

liberstina [14]

Answer:

the one you have selected is correct

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

9. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of

netineya [11]

Answer:

9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 12, \sigma = 3$

What is the probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated?

This is 1 subtracted by the pvalue of Z when X = 16. So

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

$Z = \frac{16 - 12}{3}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

1 - 0.9082 = 0.0918

9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated

5 0

3 years ago

The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to in

oksano4ka [1.4K]

Answer:

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

$p_v =P(Z>2)=0.0228$

Step-by-step explanation:

1) Data given and notation

$\bar X=8300$ represent the sample mean

$\sigma=1200$ represent the population standard deviation

$n=64$ sample size

$\mu_o =800$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

4) P-value

Since is a one-side upper test the p value would given by:

$p_v =P(Z>2)=0.0228$

5) Conclusion

If we compare the p value and the significance level assumed, for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.

4 0

3 years ago