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sleet_krkn [62]

3 years ago

10

When a higher confidence level is used to estimate a proportion and all other factors involved are held constant

1 answer:

lbvjy [14]3 years ago

8 0

Options :

A) the confidence interval will be wider.

B) the confidence interval will be narrower.

C) there is not enough information to determine the effect on the confidence interval.

D) the confidence interval will be less likely to contain the parameter being estimated.

E) the confidence interval will not be affected.

Answer: A) The confidence interval will be wider

Step-by-step explanation: In statistics, When a higher confidence interval is used to estimate a proportion with other factors being held constant, the confidence interval will be wider. Since the confidence interval is aimed at tweaking the precision of our measurement or estimation, when we adopt a higher confidence level, we are trying g to reduce our chances of being wrong and hence increasing our precision and thus widening the level of confidence in the result obtained. That said, 95% confidence interval is wider than 90%. And 99% is wder than 95%.

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Use the power property to rewrite log3x9.

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Answer:

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Step-by-step explanation:

Given the expression

$\log _3\:x^9$

now rewriting the expression

$\log _3\:x^9$

$\mathrm{Apply\:log\:rule\:}\log _a\left(x^b\right)=b\cdot \log _a\left(x\right),\:\quad \mathrm{\:assuming\:}x\:\ge \:0$

$=9\log _3\left(x\right)$

Therefore,

$\log _3\left(x^9\right)=9\log _3\left(x\right)$

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If the measure of ZB is 40°, what measures can ZA and

lubasha [3.4K]

Answer: 30° and 110°

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70° and 70°

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Let a, b, c, and d be real numbers with a, c 6= 0. Prove that the lines y = ax+b and y = cx + d have the same x-intercept if and

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Step-by-step explanation:

We have got the lines :

$y=ax+b\\y=cx+d$

Both lines intercept the x-axis in the point :

$I = (i_{1} ,i_{2})$

In all point from x-axis the y-component is equal to 0.

$I=(i_{1},o)$

We replace the I point in the lines equations:

$0=a(i_{1})+b \\0=c(i_{1})+d$

From the first equation :

$0=a(i_{1})+b \\-b=a(i_{1})\\i_{1}=\frac{-b}{a}$

From the second equation :

$0=c(i_{1})+d\\ -d=c(i_{1})\\i_{1}=\frac{-d}{c}$

Then $i_{1}=i_{1}$

Finally :

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