Determine the value of r so that line through. -2,5 and 6,r has a slope of -3,2

the mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5 %. assume

zhuklara [117]

Given:
Population proportion, $\mu _{p}$ = 57% = 0.57
Population standard deviation, σ = 3.5% = 0.035
Sample size, n = 40
Confidence level = 95%

The standard error is
$SE_{p} = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.57*0.43}{40} } =0.0783$

The confidence interval is
$\hat{p} \pm z^{*}SE_{p}$
where
$\hat{p}$ = sample proportion
z* = 1.96 at the 95% confidence lvvel

The sample proportion lies in the interval
(0.57-1.96*0.0783, 0.57+1.96*0.0783) = (0.4165, 0.7235)

Answer: Between 0.417 and 72.4), or between (41% and 72%)

8 0

3 years ago

Compare discrete domains and continuous domains.

shtirl [24]

Discrete domains and continuous domains are both sets. However a discrete domain contains a finite number of elements and continuous can contain and infinite number of elements.

6 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago

What is x^2+15x+7 factored?

Serga [27]

Answer:

x^2+15x+7

Step-by-step explanation:

There are no like terms.

4 0

3 years ago

A baseball team played 25 games and won 10 of them. What percent of the games did they win?

GalinKa [24]

Answer: 40%

Step-by-step explanation:

Setting up an equation to solve,

25 as the denominator representing the total number of baseball games played to represent the "whole."

10 as the numerator representing the number of games the baseball games won to represent the "part" you want to calculate as a percentage.

The equation, looking like this,

10/25 = ___

after calculating the equation you get 2/5 as reduced, evaluating that in terms of percentage you get 40% as 5 represents 100% (the whole) and 2 representing 40% (the number of baseball games won)

3 0

3 years ago