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Rudik [331]
3 years ago
15

Plzzzzz help for a brainly

Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0

Answer:

I think it would be C. 1/6. not for sure though

Step-by-step explanation:

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Ayuda porfavor es urgente​
san4es73 [151]
B for one and D for 2
Uno - B Dos - D
5 0
3 years ago
An article reports "attendance dropped 7% this year, to 6854." What was the attendance before the drop? (Round your answer to th
Masja [62]

Answer:

7370

Step-by-step explanation:

Let x represent attendance before drop.

We have been given that an article reports attendance dropped 7% this year, to 6854. We are asked to find the attendance before the drop.

Since attendance dropped by 7%, this means that attendance after drop was 93% of x.

We can represent this information in an equation as:

x-\frac{7}{100}x=6854

x-0.07x=6854

0.93x=6854

\frac{0.93x}{0.93}=\frac{6854}{0.93}

x=7369.89247

x\approx 7370

Therefore, the attendance before drop was 7370.

4 0
4 years ago
What is 3 divided by 11/12
Lena [83]

Answer:

0.02272

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Javier drove 299.6 miles on 14 gallons of gas. At
Bingel [31]

Answer:

545.7 Miles

Step-by-step explanation:

If Javier drave 299.6 miles on 14 gallons of gas then divide 299.6 by 14 to get 21.4. Then multiply 21.4 by 25.5 to get 545.7

7 0
3 years ago
A credit card company is about to send out a mailing to test the market for a new credit card. From that sample, they want to es
Sergeeva-Olga [200]

Answer:

n=\frac{0.005(1-0.005)}{(\frac{0.001}{1.96})^2}=19111.96  

And rounded up we have that n=19112

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

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".  

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.001 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.005(1-0.005)}{(\frac{0.001}{1.96})^2}=19111.96  

And rounded up we have that n=19112

3 0
3 years ago
Read 2 more answers
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