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iren2701 [21]
2 years ago
11

Which fraction and decimal forms match the long division problem? 15) 8.000 75. 50 45, 50 45 5 O A. 8 and 0.53 15 O B. 8 15 and

0.53 O C. 15 8 and 0.53 O D. 15 8 and 0.533​

Mathematics
1 answer:
sveticcg [70]2 years ago
5 0

Answer: The answer is A

Step-by-step explanation:

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HELP PLEASE FAST I WILL MARK BRAINLIEST RANDOM ANSWERS WILL BE REPORTED
ElenaW [278]

Answer:

(-2,-4)

Step-by-step explanation:

Turning the point 180 degrees causes the point to be reversed, or upside down, therefore the numbers will become negative.

3 0
3 years ago
Find the value of x.
nevsk [136]

Answer:

<h2>x = 30°</h2>

Step-by-step explanation:

Since the figure above is a triangle, all it's angles sum up to 180°

To find x , add up all the angles and subtract them from 180°

That's

57 + 93 + x = 180

150 + x = 180

x = 180 - 50

We have the final answer as

<h3>x = 30°</h3>

Hope this helps you

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3 years ago
A jar contains 15 red marbles, 10 green marbles, and 11 blue marbles. What is the probability that a marble chosen at random fro
rosijanka [135]

Answer:

Why are you playing with marbles?

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
There is a Picture below
Veseljchak [2.6K]
1a. 0.05
1b. 100
1c. 0.5
1d. 50
2. Not sure if this is right but 0.5 divides by 0.05 to get the middle number
5 0
2 years ago
Read 2 more answers
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
2 years ago
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