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sp2606 [1]
3 years ago
5

-950-?= -379 Help Will give brainliest!!

Mathematics
1 answer:
Karolina [17]3 years ago
8 0
Answer: -571

Step-by-step-explanation:
-950 - (-571) = -379
-950 + 571 = -379
Hope this helps :)
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Solve 3/4x+1/4y=5 for y​
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y=3x+20

Step-by-step explanation:

So first, I don't like fractions so I would multiply the whole equation by 4. This will make the equation 3x+y=20. Then you have to isolate the y to find the answer so you should subtract 3x from both sides. This will make the equation look like this y=-3x+20. Therefore, y=3x+20

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Goofy's fast food center wishes to estimate the proportion of people in its city that will purchase its products. Suppose the tr
Westkost [7]

Answer:

The probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.0143.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em>  ≥ 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

The sample selected consists of <em>n</em> = 254 individuals. The sample is quite large, i.e. <em>n</em> = 254 > 30. So the central limit theorem can be applied to approximate the distribution of sample proportion of  people in the city that will purchase the products of Goofy's fast food.

The mean is:

\mu_{\hat p}=p=0.05

And the standard deviation is:

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.05(1-0.05)}{254}}=0.0137

Now, we need to compute the probability that the sample proportion will differ from the population proportion by greater than 0.03.

That is:

\hat p-p>0.03\\\hat p -0.05>0.03\\\hat p>0.08

Compute the value of P(\hat p>0.08) as follows:

P(\hat p>0.08)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.08-0.05}{0.0137})

                   =P(Z>2.19)\\=1-P(Z

*Use a <em>z</em>-tale for the probability.

Thus, the probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.0143.

8 0
3 years ago
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