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Semmy [17]
3 years ago
6

H e l p ; - ; idk how to manipulate it ; - ; .

Mathematics
1 answer:
Allushta [10]3 years ago
3 0

Answer:

Step-by-step explanation:

Part A

To keep it simple, I will use v instead of vi, x instead of triangle x and t instead of triangle t.

The equation becomes: x = vt + 1/2at^2

Solve for a:

x = vt + 1/2at^2

x - vt = 1/2at^2

(x - vt)*(2/t^2) = 1/2at^2*(2/t^2) = a

a = 2(x-vt)/t^2

or writing in the right format

a = 2(Δx - viΔt) / Δt^2

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Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbic
jolli1 [7]

Answer:

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

The randomly sampled 400 dogs from homes where an herbicide was used on a regular basis, diagnosing lymphoma in 230 of them.

This means that:

p_h = \frac{230}{400} = 0.575, s_h = \sqrt{\frac{0.575*0.425}{400}} = 0.0247

Of 200 dogs randomly sampled from homes where no herbicides were used, only 25 were found to have lymphoma.

This means that:

p_n = \frac{25}{200} = 0.125, s_n = \sqrt{\frac{0.125*0.875}{200}} = 0.0234

Distribution of the difference:

p = p_h - p_n = 0.575 - 0.125 = 0.45

s = \sqrt{s_h^2+s_n^2} = \sqrt{0.0247^2 + 0.0234^2} = 0.034

Confidence interval:

The confidence interval is:

p \pm zs

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower bound is 0.45 - 1.96(0.034) = 0.3834

The upper bound is 0.45 + 1.96(0.034) = 0.5166

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

3 0
3 years ago
What is the supplement angle of 100?​
aleksley [76]

Answer:

An angle measuring 100 degrees would be supplementary to any angle measuring 80 degrees, but it cannot be called supplementary by itself. "Complementary" and "supplementary" are terms that describe the relationship between two angles. They are comparative words like "larger."

Step-by-step explanation:

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3 years ago
The triangles shown below must be congruent
Leokris [45]

Yes; they  have the same side lengths and angles

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A number increased by 15 is 38 as an equation
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Answer:

15 + x = 38

KEY:

<u>Increased = addition</u>

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3 0
3 years ago
Consider a Poisson distribution with μ = 6.
bearhunter [10]

Answer:

a) P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b) f(2) = 0.04462

c) f(1) = 0.01487

d) P(X \geq 2) = 0.93803

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of successes

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

In this question:

\mu = 6

a. Write the appropriate Poisson probability function.

Considering \mu = 6

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b. Compute f (2).

This is P(X = 2). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

P(X \geq 2) = 1 - P(X < 2)

In which:

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

Then

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803

So

P(X \geq 2) = 0.93803

5 0
2 years ago
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