The sale price of a pair of pants is 65% of the regular price of $25. How much do you save by buying the pants on sale?

A frequency table of grades has five classes (A, B, C, D,F) with frequencies of 4, 10, 17, 6, and respectively. Using percentage

fiasKO [112]

We have the frequencies for each of the grades. We can estimate the number of students graded by adding all those frequencies. Let's call N the total number of grades:

$\begin{gathered} N=4+10+17+6+2 \\ N=39 \end{gathered}$

We have then a total number of grades of 39.

The corresponding relative frequency for a grade is the ratio of the frequency to the total number of "samples", 39 in this case.

Then, for grade A, the relative frequency (RF) will be:

$RF_A=\frac{4}{39}\approx0.10256\approx10.26\text{ \%}$

This will be the fraction of the total grades that are A. Represented as a percentage will be 10.26%, rounded to two decimal places.

Now, to complete the table we do the same for the other frequencies:

For grade B:

$RF_B=\frac{10}{39}\approx0.2564\approx25.64\text{ \%}$

For grade C:

$RF_C=\frac{17}{39}\approx0.4359\approx43.59\text{ \%}$

For grade D:

$RF_D=\frac{6}{39}\approx0.1538\approx15.38\text{ \%}$

For grade F:

$RF_D=\frac{2}{39}\approx0.0513\approx5.13\text{ \%}$

4 0

1 year ago

Simplify the expression. p(p^-7 q^3)^-2q^-3

FinnZ [79.3K]

Answer:

Step-by-step explanation:

7323

4 0

3 years ago

49–2ax –a^2–x^2 factor

vampirchik [111]

49–2ax –a^2–x^2

Extract the negative sign

= 49 - (a^2 +2ax +x^2)

Factor using a^2 + 2ab + b^2 = (a+b)^2

= 49 - (a+x)^2

Factor using a^2 - b^2 = (a-b)(a+b)

= ( 7- (a + x)) * (7 + (a + x))

= ( 7- a - x) * (7 + a + x)

4 0

3 years ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

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:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$