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VashaNatasha [74]
4 years ago
5

I don't know how to do this . Comment bellow to help my please

Mathematics
1 answer:
Lorico [155]4 years ago
4 0
4a. 99m + 81 = 130m. 4b. 40 + 30z = 70z

5a. 24v + 18 = 41v. 5b. 12c + 16 = 28c

6c. 44 + 48v= 92v. 6b. 40 + 12s = 52s.
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What is the slope of the line that includes points (36,-30) and (39,27)
DedPeter [7]

Answer:

The slope of the line is 19

Step-by-step explanation:

slope=(y2-y1)/(x2-x1)=(27-(-30))/(39-36)=57/3=19

Therefore, the slope of the line is 19

7 0
3 years ago
A sample of 250 people resulted in a confidence interval estimate for the proportion of people who believe that the federal gove
bagirrra123 [75]

Answer: Approximately 79 Percent

Step-by-step explanation:

The confidence level used in this estimation is approximately 79 percent.

3 0
3 years ago
6. A line that passes through the point (–1, 3) has a slope of 2. Find another point on the line. A. (0, 5) B. (–3, 2) C. (1, 4)
Bezzdna [24]
The answer is A because if we stare with (-1,3) and go up by 1x, since the slope is 2, then 3 + 2 is 5 and we have (0,5). Hope this helps!
4 0
3 years ago
Read 2 more answers
15 POINTS!<br> What is the correct answer for Dean's problem? Explain where he went wrong.
prohojiy [21]

Answer:

The slope is 5/-2.

Step-by-step explanation:

Slope is y1-y2 over m1-m2 (rise over run). The first ordered pair is -2 (m1) and 11 (y1). We then subtract the second ordered pair (4 (m2) and -4 (y2)) from the first.

11 - (-4) = 11 + 4 = 15

-2 - 4 = -6

Remember, slope is rise over run (y over x), so the slope is 15/-6. Now, we must simplify. 15/-6 = 5/-2

Dean went wrong because he thought that slope was run over rise (x over y). If he had switched the two numbers, his answer would have been correct.

8 0
4 years ago
You estimate that a box contains 141 envelopes the actual number of envelopes is 150 find the percent error
S_A_V [24]
0.0094% error



Divide 150/141. You get 0.94, then you need to find the percentage of that by clicking the percent button of your calculator
7 0
3 years ago
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