All categories

3 years ago

Find the measures of the interior angles.

Mathematics

1 answer:

NARA [144]3 years ago

6 0

Answer:

angle A: 59 degrees

angle B: 48 degrees

angle C: 73 degrees

Step-by-step explanation:

in every triangle, if you add up all of the degrees, you get 180

x + 73 + (x - 11) = 180

2x + 62 = 180

2x = 118

x = 59

then, you can just insert x into the respected expressions to solve for their angles

A. x = 59 degrees

B. x - 11

59 - 11

48 degrees

C. 73 degrees

You might be interested in

Anybody wanna help me on this?

igomit [66]

Answer:

(0,-2) , (2,1) , (4,4)

Step-by-step explanation:

those are the points, now connect then with a line

7 0

3 years ago

The diagram below represents a street intersection. The dashed lines are new streets that are being proposed. If the turn repres

Katarina [22]

Answer:

Step-by-step explanation:

5 0

3 years ago

1) Which expression represents the volume<br> of a shipping box?

crimeas [40]

Answer:

(s+2)^3

Step-by-step explanation:

I think this is referring to a square as a box, so the length of a jewelry box would be s. We learn that a shipping box is 2 inches longer, so s+2 = the length of the shipping box.

Volume = Length*Width*Height, but if square, all dimensions are the same, so s+2 to the power of 3 would be your answer.

4 0

4 years ago

Which set of side lengths forms a right triangle?

sergij07 [2.7K]

Answer:

My guess is 9,12 ,18

Step-by-step explanation:

5 0

3 years ago

A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds. Unfortunately, the company has a problem

blsea [12.9K]

Answer: $0.2461$

Step-by-step explanation:

Given : A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds.

Sample size : n=11

Degree of freedom =n-1=10

Sample standard deviation : s= 0.71 pounds

Significance level for 95% confidence interval : $\alpha=1-0.95=0.05$

We assume that the bowling ball weight is normally distributed.

Using chi-square distribution table, the required critical values are :-

$\chi^2_{df, \alpha/2}=20.48$

$\chi^2_{df, 1-\alpha/2}=3.25$

Then, the 95% confidence interval for the variance of the bowling ball weight will be :

$\dfrac{s^2(n-1)}{\chi_{\alpha/2}}$

$=\dfrac{(0.71)^2(10)}{20.48}$

∴ The 95% confidence interval for the variance of the bowling ball weight will be : $0.2461$

7 0

3 years ago