PLEASE HELP PLEASE PLEASE PLEASE PLZZZZZZZZZZZ

If mZ1 = (4x + 2)° and m/3 = (5x – 15)", what is the value of a?

posledela

Answer:

17

Step-by-step explanation:

angle one and angle three are equal to each other

so if you set the equations equal to each other

(4x+2)=( 5x-15)

move the 4x the right by subtracting it giving you 1x

then add 15 to 2 giving you 17

17=1x

8 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

See attachment for result of 1-VarStats

8 0

3 years ago

Simplify the equations below

IRISSAK [1]

Number 2 is 4x^5/3y^5 and 14 is 3125x^10

5 0

3 years ago

A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decre

Dmitry [639]

Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is less than the standard significance level of 0.05, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

At the null hypothesis, it is tested if the consumption is not different, that is, if the subtraction of the means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, it is tested if the consumption is different, that is, if the subtraction of the means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

Two groups of 22 patients, hence, the standard errors are:

$s_1 = \frac{45.1}{\sqrt{22}} = 9.6154$

$s_2 = \frac{26.4}{\sqrt{22}} = 5.6285$

The distribution of the differences is has:

$\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{25 - 0}{11.14}$

$t = 2.2438$

The p-value of the test is found using a two-tailed test, as we are testing if the mean is different of a value, with t = 2.2438 and 22 + 22 - 2 = 42 df.

Using a t-distribution calculator, this p-value is of 0.0302.

Since the p-value of the test is of 0.0302, which is less than the standard significance level of 0.05, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

A similar problem is given at brainly.com/question/25600813

4 0

3 years ago

EASY BRAINLIEST PLEASE HELP!!

DedPeter [7]

Answer:

C. AB = 5.8 cm

Step-by-step explanation:

In trapezoid ABCD, EF is the Midsegment.

.:. EF = {(AB+CD)

.:: 9.2 = {(AB + 12.6)

:: 9.2 x 2 = AB + 12.6

.. 18.4 = AB + 12.6

.. 18.4-12.6 = AB

:: AB = 5.8 cm

6 0

3 years ago

Read 2 more answers