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yulyashka [42]
3 years ago
8

Find the measurement of angle m.

Mathematics
1 answer:
Musya8 [376]3 years ago
8 0

Answer:

m = 35°

Step-by-step explanation:

Supplementary angles, sum = 180°

m + 145° = 180°

m = 180° -145°

m = 35°

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Part A: Solve A = (x + 23) for x. (4 points) Part B: Determine the value of x when A = 108. (2 points) Part C: Solve -np - 90 &g
Galina-37 [17]
<span>Part A: Solve A = (x + 23) for x.

A = x + 23

=> A - 23 = x + 23 - 23

=> A - 23 = x

=> x = A - 23 <------- answer

Part B: Determine the value of x when A = 108

Replace the value of A in the expression x = A - 23

x = 108 - 23 = 75

x = 75 <------- answer

Part C: Solve -np - 90 > 30 for n.

-np - 90 > 30

=> -np + np - 90 >30 +  np

=> - 90 > 30 + np

=> -90 - 30 > 30 - 30 + np

=> -120 > np

=> np < - 120 <----- answer
</span>
7 0
3 years ago
What is the solution of the equation? -6 = 2/3x<br> a. = -9<br> b. = -6<br> c. = 4<br> d. = -4
erastovalidia [21]
-6 = 2/3 x
x = -6*3 /2
x = -18/2
x = -9

Answer is Option A

Hope this helps!
8 0
3 years ago
What is the equation of the line that passes through the point 3,-1 and has a slope of 2
LenaWriter [7]

Answer:

y=2x-7

Step-by-step explanation:

so here's my work

I just plugged stuff in so:

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3 0
2 years ago
Show Your Work! <br> Will Mark Brainliest!
PSYCHO15rus [73]

Answer:

The answer is A.

Step-by-step explanation:

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6 0
2 years ago
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In a survey sample of 83 respondents, the mean number of hours worked per week is 39.04, with a standard deviation of 11.51. Cal
soldier1979 [14.2K]

Given that,

Sample size= 83

Mean number= 39.04

Standard deviation= 11.51

We know the critical t-value for  95% confidence interval which is equal to 1.989.

We also know the formula for confidence interval,

CI=( mean number - critical t-value*standard deviation/(sample size)^(1/2), mean number + critical t-value*standard deviation/(sample size)^(1/2))

So, we have

CI= (39.04 - 1.989*11.51/83^(1/2), 39.04 + 1.989*11.51/83^(1/2)

CI= (39.04 - 2.513,39.04 + 2.513)

CI= (36.527,41.553)

Therefore, 95% confidence interval for these data is (36.527,41.553), and this result interpret that the true value for this survey sample lie in the interval (36.527,41.553).

8 0
3 years ago
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