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Nuetrik [128]
3 years ago
11

The business college computing center wants to determine the proportion of business students who have personal computers (PC's)

at home. If the proportion exceeds 30%, then the lab will scale back a proposed enlargement of its facilities. Suppose 250 business students were randomly sampled and 75 have PC's at home. Find the rejection region for this test using a = .05
- reject h is z is greater than 1.645




reject h is z= 1.645




reject h if z is less than -1.645




reject h if z is greater than 1.96 or z is less than -1.96
Mathematics
1 answer:
Anastasy [175]3 years ago
6 0

Answer:

Option A) reject null hypothesis if z is greater than 1.645

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 250

p = 30% = 0.3

Alpha, α = 0.05

Number of women belonging to union , x = 75

First, we design the null and the alternate hypothesis  

H_{0}: p = 0.3\\H_A: p > 0.3

This is a one-tailed(right) test.

Rejection Region:

z_{critical} \text{ at 0.05 level of significance } = 1.645

So, the rejection region will be

z > 1.64

That is we will reject the null hypothesis if the calculated z-statistic is greater than 1.645

Option A) reject null hypothesis if z is greater than 1.645

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What’s the question??????
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What is the equation of a line perpendicularto y=-2/5x-1 that passes through(2,-8)​
Andrews [41]
<h2>Answer: y =  ⁵/₂ x - 13   OR  y + 8 =  ⁵/₂ x - 5 </h2>

<h3>Step-by-step explanation:</h3>

<u>Find the slope of the perpendicular line</u>

When two lines are perpendicular, the product of their slopes is -1. This means that the slopes are <em>negative-reciprocal</em>s of each other.

⇒  if the slope of this line = - ²/₅

      then the slope of the perpendicular line (m) = ⁵/₂

<u>Determine the equation</u>

We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line:

                                ⇒  y - (-8) =  ⁵/₂ (x - 2)

                                   ∴  y + 8 = ⁵/₂ (x - 2)

We can also write the equation in the slope-intercept form by making y the subject of the equation and expanding the bracket to simplify:

                  since   y + 8 =  ⁵/₂ (x - 2)

                                    y =  ⁵/₂ x - 13

7 0
2 years ago
Now suppose that bigger cups are ordered and the machine’s mean amount dispensed is set at μ=12. Assuming we can precisely adjus
Elina [12.6K]

Answer:

σ should be adjusted at 0.5.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean 12.

Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?

13 should be 2 standard deviations above the mean of 12, and 11 should be two standard deviations below the mean.

So 1 should be worth two standard deviations. Then

2\sigma = 1

\sigma = \frac{1}{2}

\sigma = 0.5

σ should be adjusted at 0.5.

4 0
2 years ago
During second period, Edna completed a grammar worksheet. She got 57 questions correct and 18 questions incorrect. What percenta
irinina [24]
57+18=75
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hope this helps! :)
5 0
3 years ago
f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int
eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

as

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

So we must have

(C1 + (1/2)C3) = 0...........................(1)

and

(C2 + (1/2)C3) = 0..........................(2)

From (1)

2C1 + C3 = 0

=> C3 = -2C1.................................(3)

From (2)

2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

3 0
2 years ago
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