Find the range of f(x) = -x + 4 for the domain (-3,-2,-1,1).

1 answer:

4 0

The domain: D={-3;-2;-1; 1}

f(x) = -x + 4

subtitute

f(-3) = -(-3) + 4 = 3 + 4 = 7
f(-2) = -(-2) + 4 = 2 + 4 = 6
f(-1) = -(-1) + 4 = 1 + 4 = 5
f(1) = -1 + 4 = 3

Answer: The range: {3; 5; 6; 7}

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If the confidence level is 90%, find the Margin of sampling error. The population is normally distributed, the sample size is 15

charle [14.2K]

Answer:

$ME = 1.761* \frac{5}{\sqrt{15}}=2.273$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X= 75$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=5 represent the sample standard deviation

n=15 represent the sample size

Solution to the problem

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,14)".And we see that $t_{\alpha/2}=1.761$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we replace the values we got:

$ME = 1.761* \frac{5}{\sqrt{15}}=2.273$

6 0

3 years ago

est scores for a statistics class had a mean of 79 with a standard deviation of 4.5. Test scores for a calculus class had a mean

Arte-miy333 [17]

Answer:

The z-score for the statistics test grade is of 1.11.

The z-score for the calculus test grade is 7.3.

Due to the higher z-score, the student performed better on the calculus test relative to the other students in each class

Step-by-step explanation:

Z-score:

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.

In this question:

The grade with the higher z-score is better relative to the other students in each class.

Statistics:

Mean of 79 and standard deviation of 4.5, so $\mu = 79, \sigma = 4.5$