How do i solve this problem

Answer this please

Gelneren [198K]

Answer:

x = 48

Step-by-step explanation:

We now that supplementary angles (as is the case for a straight line) add up to 180 degrees.

3x - 12 + x = 180

4x = 192

x = 48

6 0

2 years ago

What is the difference between no solution and no real solution? When do you get multiple solutions? This is for solving absolut

Dmitriy789 [7]

No solution would mean that there is no answer to the equation. It is impossible for the equation to be true no matter what value we assign to the variable. Infinitesolutions would mean that any value for the variable would make the equation true.No Solution Equations.

In other words, it "discriminates" between the possible solutions. The discriminant is the expression found under the square root part of the quadratic formula (that is, . The value of tells how many solutions, roots, or x-intercepts the quadratic equation will have. If , there are two real solutions.

8 0

2 years ago

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

garri49 [273]

Answer:

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

The statistic is given by:

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

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0

3 years ago

What is the answer to this expression when x=5 3x+15=?

Nikitich [7]

3(5)+15= 30
because BIDMAS so we would multiply 3 by 5 then add the answer which would be 15 to 15 and we would get 30.

5 0

3 years ago

Read 2 more answers

What’s the value of the ratio 21:45