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ElenaW [278]
3 years ago
6

HELP WITH THESE SERIES OF QUESTIONS!!

Mathematics
1 answer:
Virty [35]3 years ago
5 0
1. Simplified already
2. 5x^2 x (3x+5)
3. 3x^3 x (3x^4 +11)
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Eight less than seven times a number is six find the number
Leokris [45]

Answer:

2

Step-by-step explanation:

let the number be n , then

7n - 8 = 6 ( add 8 to both sides )

7n = 14 ( divide both sides by 7 )

n = 2

The number is 2

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Simplify negative square root of 72
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2 years ago
Use the drop downs to determine which symbols would complete the inequality for the range.
Yuliya22 [10]

Answer:

(a) it is less then

(b) it is greater then

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it those because that the right answer

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2 years ago
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Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls on the skin, is essential for stro
tigry1 [53]

Answer:

a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

Step-by-step explanation:

Data given and notation  

n=2700 represent the random sample taken    

X represent the people in England who are deficient in Vitamin D

\hat p=0.2 estimated proportion of England people who are deficient in Vitamin D

\alpha=0.02 represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)    

p= population proportion of England people who are deficient in Vitamin D

Part a

The confidence interval would be given by this formula

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

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.33

And replacing into the confidence interval formula we got:

0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182

0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

3 0
3 years ago
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