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madreJ [45]
2 years ago
6

10. The continuously compounded annual return on a stock is normally distributed with a mean of 20% and standard deviation of 30

%. With 95.44% confidence, we should expect its actual return in any particular year to be between which pair of values
Mathematics
1 answer:
ehidna [41]2 years ago
7 0

Answer: We should expect its actual return in any particular year to be between<u> -40%</u> and<u> 80%</u>.

Step-by-step explanation:

Given : The continuously compounded annual return on a stock is normally distributed with a mean 20% and standard deviation of 30%.

From normal z-table, the z-value corresponds to 95.44 confidence is 2.

Therefore , the interval limits for 95.44 confidence level will be :

Lower limit = Mean -2(Standard deviation) = 20% -2(30%)= 20%-60%=-40%

Upper limit =  Mean +2(Standard deviation)=20% +2(30%)= 20%+60%=80%

Hence, we should expect its actual return in any particular year to be between<u> -40%</u> and<u> 80%</u>.

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A recipe calls for One-fourth cup of broth. The largest volume-measuring tool that Tabitha has is a teaspoon. She knows that the
djverab [1.8K]

<u>Answer:</u>

Tabitha will use 12 teaspoons.

<u>Explanation: </u>

According to the question, Tabitha has a largest volume-measuring tool teaspoon, and she wants to use one-fourth cup of broth.

Given that 1 cup consists of 16 tablespoons and 1 tablespoon has 3 teaspoons.

From the data, we can calculate that;

16 tablespoons will be equal to (3*16) = 48 teaspoons.

Therefore, 1 cup = 48 teaspoons.

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1 year ago
The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. If a sample
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Given:
standard deviation = 4.3%
sample size = 15
mean = 26.4%

error = σ/√n

error = 4.3% / √15

error = 4.3% / 3.873

error = 0.0111

0.0111 x 100% = 1.11%
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2 years ago
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Danika concludes that the following functions are inverses of each other because f(g(x)) = x. Do you agree with Danika? Explain
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Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ
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y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

\ell_i=\dfrac{i-1}2

r_i=\dfrac i2

with 1\le i\le10.

These subintervals have midpoints given by

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

Over each subinterval, we approximate f(x) with the quadratic polynomial

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that the integral we want to find can be estimated as

\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that

\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6

so that the arc length is approximately

\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

5 0
2 years ago
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