Subtract. (6x - 3) - (2x -2) What is the answer?

An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample

valentina_108 [34]

Answer:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Replacing we got:

$\bar X=58.875$ represent the mean

$s=5.083$ represent the sample standard deviation for the sample

$n=8$ sample size

$\mu_o =60$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis: $\mu \geq 60$

Alternative hypothesis: $\mu < 60$

The statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0

3 years ago

toni bought an old bike for 50$. She fixed it up and sold it for $80. What percent of the original value was toni’s sale price

ikadub [295]

For this case what you must do is the following rule of three:
50 ---> 100
80 ----> x
We clear x:
x = (80/50) * (100)
x = 160%
160-100 = 60%
Answer:
toni added 60% of the original value for his sale price

3 0

3 years ago

First line joins ordered pairs negative 4, 3 and 2, negative 3. Second line joins negative 4, negative 3 and 2, 3. Part A shaded

elena55 [62]

The part that represents the solution to the inequality will be Part B shaded below first line and above second line.

<h3>How to depict the inequality?</h3>

From the information given, the equation of the first line will be:

y - 3 = (-3 - 3/2 + 4)(x + 4)

y - 3 = -1(x + 4)

y + x = -4 + 3

x + y = -1

The equation of the second line will be:

y + 3 = -1(x + 4)

y = x + 4 - 3

y = x + 1

This is plotted on the graph attached.

From the systems of equations, the statement that is correct about the two systems of equations is that They will have the same solution because the first equation of System B is obtained by adding the first equation of System A to 3 times the second equation of System A.

Learn more about inequalities on:

brainly.com/question/12215820

#SPJ1

4 0

2 years ago

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter

Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and $\sqrt{8^{2} - x^{2} }$ or $\sqrt{64 - x^{2} }$

thus tan(arcsin(x/8)) = $\frac{x}{\sqrt{64 - x^{2} }}$ =

Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

B

Here we have cos = adjacent/hypotenuse where adjacent side is $\sqrt{64 - x^{2} }$ and hypothenuse = 8 we have $\sqrt{64 - x^{2} }$ /8

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=10%20-%206v%20%3D%20" id="TexFormula1" title="10 - 6v = " alt="10 - 6v = " align="absmiddle" c

MariettaO [177]

4v
You would subtract 10-6
Then you would add the v

6 0

3 years ago