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2 years ago

Please help 20 points

Mathematics

1 answer:

S_A_V [24]2 years ago

6 0

Answer:

can you post a clearer picture ?

Step-by-step explanation:

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PLEASE HELP ILL GIVE 77 POINTS!!!!!!!!!!!!

Vikki [24]

Answer:

i think its twelve nine and five

Step-by-step explanation:

not one hundred percent sure

8 0

3 years ago

Had 2 pages of algebra ! :C my brain is not working anymore..

AleksandrR [38]

Step-by-step explanation:

given,

$2k \div 3 = 6 \\ 2k \div3 \times 3 = 6 \times 3 \\ 2k = 18 \\ k = 18 \div 2 \\ = 9$

7 0

3 years ago

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).

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3 years ago

Several students at a university were surveyed about how they

Anarel [89]

Okay so the answer would be

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3 years ago

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Shalnov [3]

Ok so the consent rate is 5 and 8

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3 years ago

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