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Irina18 [472]
3 years ago
5

Which numbers are necessary to solve this problem?

Mathematics
2 answers:
pochemuha3 years ago
5 0
D is the correct answer <span />
AlexFokin [52]3 years ago
3 0
D is the correct answer my dudeeess
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How do I do this ? factor the denominators
BARSIC [14]

time 2nd 5/5 it become 35/5x-20

2/5x-20+35/5x-20

=37/5x-20

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4 years ago
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7 0
3 years ago
In a chemistry lab, you have 2 vinegars. One is 5% acetic acid, and one is 6.5% acetic acid. You want to make 200 ml of a vinega
Iteru [2.4K]
We have 5% and 6.5% acetic acid solutions and we need 200 ml of 6% acetic acid.
Set up two equations:
f means 5% and s means 6.5%
A) f + s = 200
B) .05f + .065s = (.06 * 200)
Multiplying equation A by -.05
A) -.05f  -.05s = -10
B) .05f + .065s = 12 then adding both equations:
.015s = 2
<span> <span> <span> we need 133.33</span> ml of 6.5% acetic acid and
66.67 ml of 5% acetic acid solution.
</span> </span>
Source:
http://www.1728.org/mixture.htm


8 0
3 years ago
We are told 5 miles is 8 km. <br> Convert 32 km to miles.
sesenic [268]
19.884 miles, because 1km=0.621 mile
3 0
3 years ago
Read 2 more answers
A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop
attashe74 [19]

Answer:

0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286

0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562

The 99% confidence interval would be given by (0.286;0.562)

Step-by-step explanation:

Information given:

X= 36 represent the families owned at least one DVD player

n= 85 represent the total number of families

\hat p=\frac{36}{85}= 0.424 represent the estimated proportion of families owned at least one DVD player

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 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The confidence interval for the mean is given by the following formula:  

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

If we replace the values obtained we got:

0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286

0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562

The 99% confidence interval would be given by (0.286;0.562)

8 0
3 years ago
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