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lubasha [3.4K]
3 years ago
12

Compute the standardized test statistic, χ 2, to test the claim σ 2 = 25.8 if n = 12, s 2 = 21.6, and α = 0.05.

Mathematics
1 answer:
ozzi3 years ago
8 0

Answer:

t^2=9.2

Step-by-step explanation:

From the question we are told that

Standard Deviation \sigma=28.8

Sample space n=12

s^2= 21.6

Generally the equation for test statistics is mathematically given by

t^2=frac{(n-1)*s^2}{\sigma^2}}

t^2=\frac{(12-1)*21.6}{25.8}

Therefore the standardized test statistics

t^2=9.2

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Answer:

Year 2004

Step-by-step explanation:

Population of a country in year 1995 = 286

Model that expresses the growth rate of the population is,

P(t) = 286(1.009)^{t-1995}

Here, t = year

a). For P = 311 million people

   311 = 286(1.009)^{t-1995}

   \text{log}(\frac{311}{286})=\text{log}(1.009)^{t-1995}

   log(311) - log(286) = (t - 1995)log(1.009)

   2.4928 - 2.4564 = (t - 1995)(0.00389)

   \frac{0.03639}{0.00389}=t-1995

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   t = 2004

In year 2004 population of the country will be 311.

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3 years ago
Can someone help me ?
insens350 [35]

Answer:

• changing an interior angle to a reflex angle

Step-by-step explanation:

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4 years ago
Aline has a rise of 6 and a slope of<br> 1/20 What is the run
Softa [21]

。☆✼★ ━━━━━━━━━━━━━━  ☾

The equation is:

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Rearrange for run:

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Substitute values in:

run = 6/(1/20)

Solve:

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<h3>Answer:  Choice A</h3>
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  • Range: y > 0

========================================================

Explanation:

We want to avoid having a negative number under the square root. Solving x-4 \ge 0 leads to x \ge 4

So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.

So in reality, the domain is x > 4.

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The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.

We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression \frac{3}{\sqrt{x-4}} is positive because positive/positive = positive.

Therefore, the range of the given equation is y > 0

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The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).

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