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Dmitry_Shevchenko [17]
4 years ago
12

Some easy questions please answer thank you

Mathematics
1 answer:
Umnica [9.8K]4 years ago
7 0

Answer:

Estimate \frac{5}{8}-\frac{1}{10}: B \frac{1}{2}

Common denominator of \frac{1}{4}and\frac{3}{5}: D 20

Step-by-step explanation:

In the first question, it is asking for just an estimate.  Looking at the first fraction of \frac{5}{8}, you can see that this the amount is very close to \frac{4}{8} which is \frac{1}{2}, but  \frac{5}{8} would be a little more.  Since you are subtracting \frac{1}{10}, which would be a very small amount, the best estimate is:

B \frac{1}{2}

In the second question, finding a common denominator between fractions means looking for the least common multiple:

4: 4, 8, 12, 16, 20

5:  5, 10, 15, 20

So, the lowest common denominator is 20 which is answer choice D.

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66 is what percent of 52.8?
Ivan

Answer:

125%

Step-by-step explanation:

If you take the question

66 is what percent of 52.8?

and convert it into an equation it would be

66=x*52.8

Using this information we divide 66 by 52.8 this then gives us our final answer of 1.25 or 125%

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Aneli [31]

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5 0
3 years ago
I need help!!!
Nataly_w [17]

Answer:

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6 0
4 years ago
Read 2 more answers
A music industry researcher wants to estimate, with a 90% confidence level, the proportion of young urban people (ages 21 to 35
Maslowich

Answer:

1,539

Step-by-step explanation:

Using Simple Random Sampling in an infinite population (this is such a large population that we do not know the exact number) we have that the sample size should be the nearest integer to

\large \frac{Z^2pq}{e^2}

where

<em>Z= the z-score corresponding to the confidence level, in this case 90%, so Z=1.645 (this means that the area under the Normal N(0,1) between [-1.645,1.645] is 90%=0.9) </em>

<em>p= the proportion of young urban people (ages 21 to 35 years) who go to at least 3 concerts a year= 35% = 0.35 </em>

<em>q = 1-p = 0.65 </em>

<em>e = the error proportion = 2% = 0.02 </em>

Making the calculations

\large \frac{Z^2pq}{e^2}=\frac{(1.645)^2*0.35*0.65}{(0.02)^2}=1,539.09

So, the sample size should be 1,539 young urban people (ages 21 to 35 years)

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Help view the picture find the value of p
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