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elena-14-01-66 [18.8K]
3 years ago
8

In constructing a confidence interval, suppose a student decides to decrease the level of confidence from 95% to 90 %. Assume al

l other quantities of interest remain unchanged. Which answer best describes the resulting effect (if any) on the confidence interval?Select one:a. The width of the confidence interval will be narrower.b. The width of the confidence interval will be unchanged.c. The width of the confidence interval will be wider.d. The upper and lower limits of the confidence interval will be shifted down by an equal distance.e. The confidence interval will no longer contain µ.
Mathematics
1 answer:
nydimaria [60]3 years ago
7 0

Answer:

Option A) The width of the confidence interval will be narrower.          

Step-by-step explanation:

We are given the following in the question:

A student decreases the level of confidence from 95% to 90%, while all other quantities remained unchanged.

The confidence level decreased from 95% to 90%.

Since the confidence level decreases, the corresponding statistic decreases.

This will decrease the marginal error.

Thus, the confidence interval will become narrower.

Option A) The width of the confidence interval will be narrower.

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Vadim26 [7]

The factored form of an equation is done by rewriting the equation in simpler forms.

The completely factored form of x^3 - 64x is x(x-8)(x + 8)

The expression is given as:

x^3 - 64x

Factor out x from the expression, x^3 - 64x

x^3 - 64x = x(x^2-64)

Express 64 as 8^2

x^3 - 64x = x(x^2-8^2)

Express the expression as a difference of two squares

x^3 - 64x = x(x-8)(x + 8)

Hence, the completely factored form of x^3 - 64x is x(x-8)(x + 8)

Read more about factorization at:

brainly.com/question/3024511

3 0
2 years ago
I’ve been stuck on this for a while!!!!!
maxonik [38]

Answer:

See explanation.

Step-by-step explanation:

Part A.

If two curves intersect, they have the same x- and y-coordinates.

So if we set the y-coordinates equal to each other, we are finding for what x-coordinate those y-coordinates will be the same. Therefore solving 2^(-x)=8^(x+4) gives us an x- so that both sides are the same (where both sides represent the corresponding y- coordinate for that x- in the curves given by y=2^(-x) and y=8^(x+4).

Part B.

x      | 2^(-x)                                | 8^(x+4)

-----------------------------------------------------------------------------

-3      2^(-(-3))=2^3=8                  8^(-3+4)=8^1=8

-2      2^(-(-2))=2^2=4                  8^(-2+4)=8^2=64

-1       2^(-(-1))=2^1=2                    8^(-1+4)=8^3=512

0       2^(-(0))=2^0=1                    8^(0+4)=8^4=4096

1        2^(-(1))=2^(-1)=1/2                8^(1+4)=8^5=32768

2       2^(-(2))=2^(-2)=1/4              8^(2+4)=8^6=262144

3       2^(-(3))=2^(-3)=1/8              8^(3+4)+8^7=2097152

So in the table we see both lists have the same y-coordinate when the x-coordinate is -3.

The solution to equation 2^(-x)=8^(x+4) is therefore x=-3 since the table tells us both sides is the same value, 8, for this x-coordinate.

Part C.

You can graph both curves given by y=2^(-x) and y=8^(x+4) on the same coordinate plane. Where they intersect is the solution.

I use the table in part B to help guide me in plotting the points to help graph the curves for each equation. I have included a visual in an attachment.

They cross at (-3,8) so this is the solution.

So this tells me 2^(-x)=8^(x+4) has solution at x=-3 since both sides will be 8 at that x-.

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3 years ago
What is the length of the hyptonenuse of the right triangle? the other sides are:<br>36 ft<br>15 f​
andreev551 [17]

Answer:

39

Step-by-step explanation:

We can use the Pythagorean theorem

a^2+b^2 = c^2 where a and b are the legs and c is the hypotenuse

15^2 + 36^2 = c^2

225+1296 = c^2

1521 = c^2

Take the square root of each side

39 = c

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Answer:

Isosceles Right Triangle

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I am having trouble with this could somebody give me a hand
MrRa [10]

check the picture below.


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+100}t\stackrel{\stackrel{c}{\downarrow }}{+12} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)


\bf \left( -\cfrac{100}{2(-16)}~~,~~12-\cfrac{100^2}{4(-16)} \right)\implies \left( \cfrac{25}{8}~~,~~12+\cfrac{625}{4}\right) \\\\\\ \left(\cfrac{25}{8}~,~\cfrac{673}{4} \right)\implies \left(\stackrel{\stackrel{\textit{took this long}}{\downarrow }}{3\frac{1}{8}}~~,~~\stackrel{\stackrel{\textit{went this high}}{\downarrow }}{168\frac{1}{4}} \right)

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4 years ago
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