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elixir [45]
3 years ago
14

Can someone please help me ASAP

Mathematics
1 answer:
forsale [732]3 years ago
8 0

Answer:

its B

Step-by-step explanation:

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4 Student is 20% of ____ students
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4 Student is 20% of 20 students

Step-by-step explanation:

if 4 = 20% then 1 = 5% to get a hundred percent then you need to multiply by 20

so 20

4 Student is 20% of 20 students

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The table represents the metric volume measure of a liter. Each amount is represented by its power of 10.
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Answer:

10000

Step-by-step explanation:

What is 10 to the 6th power? That would be 1000000. 10 to the second power is 100.

1000000

100

There are 4 zeros left. Meaning that the answer is 10000.

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6. Rose used 0.9 meter of a piece of cloth for her doll's dress. Write 0.9 i
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Suppose that farmers grew an unexpectedly large number of tomatoes this year. How would this increase in production affect the p
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Then, the rate of tomatoes would decrease......
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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

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

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

8 0
3 years ago
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