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koban [17]
2 years ago
8

Please solve this for me thank you!

Mathematics
2 answers:
melamori03 [73]2 years ago
6 0

Answer:

620.14

Step-by-step explanation:

Original Equation:

\frac{4y}{1.025^4}+y-2y(1.05)^2=1500

Calculate exponents

\frac{4y}{1.103812890625}+y-2y(1.1025)=1500\\

Simplify:

3.6238025791990193084270042653997y + y - 2.205y = 1500

Add like terms

2.4188025791990193084270042653997y \approx 1500

Divide both sides by 7.012

y\approx 213.91y\approx620.14

ryzh [129]2 years ago
5 0

Answer:

y = $620.14 (nearest cent)

Step-by-step explanation:

Given equation:

\dfrac{4y}{1.025^4}+y-2y(1.05)^2=\$1,500

Factor out y from the left side:

\implies y\left(\dfrac{4}{1.025^4}+1-2(1.05)^2\right)=\$1,500

Carry out the arithmetic operations inside the parentheses by following the <u>order of operations</u> PEMDAS:

Calculate the <u>exponents</u>:

\implies y\left(\dfrac{4}{1.103812891...}+1-2(1.1025)\right)=\$1,500

Carry out the <u>multiplication and division</u> from left to right:

\implies y\left(3.623802579...+1-2.205\right)=\$1,500

Carry out the <u>addition and subtraction</u> from left to right:

\implies y\left(4.623802579...-2.205\right)=\$1,500

\implies y\left(2.418802579...\right)=\$1,500

Finally, divide both sides by the coefficient of y to isolate y:

\implies \dfrac{y\left(2.418802579...\right)}{2.418802579...}=\dfrac{\$1,500}{2.418802579...}

\implies y=\$620.141558...

Therefore, y = $620.14 (nearest cent)

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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}}

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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:

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

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The confidence interval is:

p \pm zs

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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.

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

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Step-by-step explanation:

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