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3 years ago

What number gives you a result of -13 when 11 is subtracted from the quotient of the number and 5

Mathematics

1 answer:

Neporo4naja [7]3 years ago

3 0

5x-11=-13

add 11 on both sides

5x=-2

divide by five on both sides

-0.4

x=-0.4

I hope that's what you were asking

mark brainliest if correct

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3 years ago

A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

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 mean or variance 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.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=36.191$

$\chi^2_{1- \alpha/2}=7.633$

And replacing into the formula for the interval we got:

$\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}$

$8.400 \leq \sigma^2 \leq 39.827$

So the 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

7 0

3 years ago

The teachers of the two classes want to compare the heights of their students. Which statements about the data sets are accurate

aleksandr82 [10.1K]

This question is incomplete.

Complete question

The sets of data below show the heights, in inches, of students in two different preschool classes.

A) a box plot titled class 1.

The number line goes from 38 to 49. the whiskers range from 39 to 48, and the box ranges from 40 to 43. a line divides the box at 41.

class 1

B) A box plot titled class 2.

The number line goes from 38 to 49. the whiskers range from 38 to 49, and the box ranges from 39 to 42. a line divides the box at 41.

class 2

The teachers of the two classes want to compare the heights of their students. Which statements about the data sets are accurate? Select three options.

• Because the sets are symmetrical, the mean should be used to compare the data sets.

• Because the sets do not contain outliers, the MAD should be used to compare the data sets.

•Because the sets are not symmetrical, the IQR should be used to compare the data sets.

•Because the sets contain outliers, the median should be used to compare the data sets.

•The mean and mode cannot be accurately determined based on the type of data display.

Answer:

• Because the sets are not symmetrical, the IQR should be used to compare the data sets.

• Because the sets contain outliers, the median should be used to compare the data sets.

• The mean and mode cannot be accurately determined based on the type of data display.

Step-by-step explanation:

In the above question, we can see that a teacher intends to compare the heights of students from two different classes.

Because the data sets to be obtained would contain heights from different classes the results obtained would be non symmetrical.

Hence the best statistical analysis to use to solve for or to compare a set of non symmetrical data is known as either the Median or an Interquartile range.

It is important to note that a non symmetrical data would most likely contain or include the presence of outliers.

8 0

3 years ago