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jasenka [17]
3 years ago
7

The system of equations y = 2x + 5 and y = –3x – 15 is shown on the graph below.According to the graph, what is the solution to

this system of equations?

Mathematics
2 answers:
gtnhenbr [62]3 years ago
6 0

Answer:

(-4, -3)

Step-by-step explanation:

The graph of both lines will be shown below.

To find the solution, we need to find the point on the graph that both lines intersect on.

As we can see, both lines intersect on the point (-4, -3). Thus, that is the solution.

Best of Luck!

Naya [18.7K]3 years ago
5 0

Answer:

X=-4 y=-3

Step-by-step explanation:

Solve by substitution:

Step: Substitute

2x+5 for y in y = −3x−15 :

2x+5+3x=-3x-15+3x ( add 3x to both side)

5x+5=-15

5x+5+-5=-15+-5(add -5 to both sides)

5x=-20

5x/5=-20/5(divide both sides by 5)

X=-4

Step: substitute -4 for x in y=2x+5

Y=2x+5

Y=(2)(-4)+5

y=-3 (simplify both sides of the equation)

I HOPE ITS CORRECT IF IT SNOT IM SORRY!!

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A shirt costs $5 less than a sweater. Together they cost<br> $25. How much does the sweater cost?
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Answer:

Shirt: $7.5

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

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Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

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