We are given that one wants to estimate the mean PSLT for the population of all families in New York City with gross incomes in the range $35.000 to $40.000.

If sigma equals 2.0, we have to find that how many families should be surveyed if one wants to be 90% sure of being able to estimate the true mean PSLT within 0.5.

Here, we will use the concept of Margin of error as the statement "true mean PSLT within 0.5" represents the margin of error we want.

<u></u>

<u>SO, Margin of error formula is given by;</u>

Margin of error = $Z_(_\frac{\alpha}{2}_ ) \times \frac{\sigma}{\sqrt{n} }$

where, $\alpha$ = significance level = 10%

$\sigma$ = standard deviation = 2.0

n = number of families

Now, in the z table the critical value of x at 5% ( $\frac{0.10}{2} = 0.05$ ) level of significance is 1.645.

SO, Margin of error = $Z_(_\frac{\alpha}{2}_ ) \times \frac{\sigma}{\sqrt{n} }$

0.5 = $1.645 \times \frac{2}{\sqrt{n} }$

$\sqrt{n} =\frac{2\times 1.645 }{0.5}$

$\sqrt{n} =6.58$

n = $6.58^{2}$

= 43.3 ≈ 43

Therefore, number of families that should be surveyed if one wants to be 90% sure of being able to estimate the true mean PSLT within 0.5 is at least 43.

4 0

3 years ago

Calculate the total area of the shaded region.

LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$

7 0

2 years ago

Round 8 5/6 to the nearest whole number

kiruha [24]

Answer:

Step-by-step explanation:

8 5/6

5/6 is close to 1 so it will round up

8+1 = 9

8 5/6 rounds to 9

5 0

3 years ago

Read 2 more answers