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Alecsey [184]
3 years ago
12

Find the critical value necessary to form a confidence interval at the level of confidence shown below.

Mathematics
1 answer:
Rzqust [24]3 years ago
8 0

Answer:

the critical value for 0.87 level of confidence is 1.51

Step-by-step explanation:

Given the data in the question;

level of confidence = 0.87

now,

1 - c = 0.87

c = 1 - 0.87

c = 0.13

c/2 = 0.13/2 = 0.065

now

1 - c/2  = 1 - 0.065 = 0.935

so our level of significance ∝ = 0.935

Now, the critical value will be;

Z_{0.935 is obtained as follows;

P( Z < z ) = 0.935

so we find the value from ( standard normal distribution table )

P( Z < 1.51 ) = 0.935

Therefore, the critical value for 0.87 level of confidence is 1.51

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To add fractions, find the LCD and then combine.

Exact form:

9/2

Decimal form:

4.5

Mixed number form:

4 1/2

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4 1/2

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3 years ago
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X= r-h/y<br> solve for h and r
malfutka [58]
X = r - h/y ; Solve for h & r

Let's first solve for h :)

Add h/y to both sides.

h/y + x = r

So, r = h/y + x 

Now, let's solve for h.

x = r-h/y

Subtract r from both sides.

-r + x = -h/y

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In a certain Algebra 2 class of 21 students, 12 of them play basketball and 14 of them play baseball. There are 5 students who p
gizmo_the_mogwai [7]

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The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct
lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

y=\dfrac{x}{1+x}

Point P\left(1,\dfrac{1}{2}\right) lies on the curve.

Point Q\left(x,\dfrac{x}{1+x}\right) is an arbitrary point on the curve.

The slope of the secant line PQ is

\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]

1. If x=0.5, then the slope is

\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333

2. If x=0.9, then the slope is

\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632

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\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501

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\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381

7. If x=1.01, then the slope is

\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488

8. If x=1.001, then the slope is

\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499

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3 years ago
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Ludmilka [50]
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