$b^{2}-15b-54=0. \newline
b^{2}-15b=54. \newline
b^{2}-15b+(\frac{15}{2})^{2}=54+(\frac{15}{2})^{2}. \newline
(b-\frac{15}{2})^{2}=\frac{441}{4}. \newline
b-\frac{15}{2}=\pm\frac{21}{2}. \newline
b=\pm\frac{21}{2}+\frac{15}{2}. \newline
b_1 = -3, b_2 = 18.$

3 0

3 years ago

Which represents the value of Point C on the number line?

postnew [5]

Answer:

Step-by-step explanation:

clue what this means LOLOLOL

3 0

3 years ago

Read 2 more answers

Consider a data set containing the following values: 70 65 71 78 89 68 50 75 The mean of the preceding values is 70.75. The devi

Leokris [45]

Answer:

If this is the sample data, the sample variance is 125.07 and the sample standard deviation is 11.18

If this is a population data, the population variance is 109.44 and the population standard deviation is 10.46

Step-by-step explanation:

We are given the following data-set:

70, 65, 71, 78, 89, 68, 50, 75

Deviations from the mean

–0.75, –5.75 , 0.25, 7.25, 18.25, –2.75, –20.75, 4.2

Sample size, n = 8

Sample:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences =

0.5625 + 33.0625 + 0.0625 + 52.5625 + 333.0625 + 7.5625 + 430.5625 + 18.0625 = 875.5

$s = \sqrt{\dfrac{875.5}{7}} = 11.18$

$\text{ Sample variance} = s^2 = 125.07$

Population:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences =

0.5625 + 33.0625 + 0.0625 + 52.5625 + 333.0625 + 7.5625 + 430.5625 + 18.0625 = 875.5

$\sigma = \sqrt{\dfrac{875.5}{8}} = 10.46$

$\text{ Population variance} = \sigma^2 = 109.44$

7 0

3 years ago