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3 years ago

The system of equations is graphed in the figure. What is the solution of the system ?

Mathematics

1 answer:

Hunter-Best [27]3 years ago

7 0

The answer is (3,-2)

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PLS HELP WITH THIS MATH

laiz [17]

I think it’s 28.35

54/100 x 5 = 2.7
54 + 2.7 = 56.7
56.7/100 x 50 = 28.35
56.7 -28.35 = 28.35

6 0

3 years ago

Read 2 more answers

Results for: Timothy owns a t shirt printing buisnesses the expenses are $4 per t shirt Timothy sales each shirt for $10 he sold

topjm [15]

You have to do subtract 10 from 4 because 6 is how much we will profit from each shirt. Now, 6 times 327 is $1,962. That's how much Timothy earned of profit.

8 0

3 years ago

What is the constant of proportionality in this graph?

lana [24]

Answer:

one half or 1/2

Step-by-step explanation:

do your rise over your run on the graph.

7 0

3 years ago

Use a number line to solve 235-123

Tems11 [23]

The answer is 112 on the number line

7 0

3 years ago

A company receives shipments of a component used in the manufacture of a component for a high-end acoustic speaker system. When

Tanya [424]

Answer:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

Step-by-step explanation:

For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.

We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

$\hat p = \frac{24}{250}= 0.096$

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And the margin of error would be:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

4 0

4 years ago