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enot [183]
3 years ago
14

Defining Terms

Mathematics
1 answer:
____ [38]3 years ago
7 0

Answers: line DB

H

D

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Solve this system by substitution. y = 5x - 15 y = 2x - 6 *Remember to write your answer as a coordinate point (x,y Solve this s
oksano4ka [1.4K]
5x - 15 = 2x - 6
3x = 9
x = 3
Y = 5(3) - 15
Y = 0
(3,0)
7 0
3 years ago
Thomas earns $2100 per month and then got a 15 raise .how much will Thomas make per month after his raise
adelina 88 [10]
So I guess you man 15% raise!
2100×0.15=315
now that's the actual increase in pay
so what he actually gets it 2100+315=2415 dollars per month more than previous month's
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3 years ago
The probability of winning a raffle is 2/5. What is the probability of not winning the raffle?
Karo-lina-s [1.5K]

Answer:

3/5

Step-by-step explanation:

the probability of winning + probability of not winning must = 1

these are the only 2 options

2/5 + not winning = 1

subtract 2/5 from each side

not winning = 5/5-2/5

not winning = 3/5

5 0
3 years ago
Read 2 more answers
Which if the following best describes this graph ?
Lelu [443]

Answer:

B

Step-by-step explanation:

4 0
3 years ago
A population has a known standard deviation of 1.27 and a sample space contains 85 values find the margin of error needed to cre
r-ruslan [8.4K]

Answer:

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

In this question:

\sigma = 1.27, n = 85

So

M = z*\frac{\sigma}{\sqrt{n}}

M = 2.575*\frac{1.27}{\sqrt{85}}

M = 0.3547

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

7 0
3 years ago
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