a rectangle swimming pool is 9 meters wide with a surface area of 3 square meters. what is the length of the pool?

What is the average rate of change for this function for the interval from x= 1 to x = 3?

MArishka [77]

Answer:

The average rate of change is 12x=12.0x.

Description:

Function: x= 1x = 3 convert to short form: x 1x 3

Interval: x= 1 , x 3

Steps:

Input: Find the average rate of change of f(x)=3x2 on the interval [x,3x].

We have that a=x, b=3x, f(x)=3x2

Thus, f(b)−f(a)b−a=3((3x))2−(3(x)2)3x−(x)=12x.

Answer: the average rate of change is 12x=12.0x.

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2 years ago

A survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the resul

pav-90 [236]

Answer:

95% confidence interval for the proportion of the adults who were opposed to the death penalty is (0.668, 0.704).

Step-by-step explanation:

We are given that a survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the results below, where X refers to the number of the respondents who were in favor.

X = 1,790

N = 2,610

$\hat p$ = Sample proportion = X/N = 0.6858

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = 0.6858

n = sample of respondents = 2,610

p = population proportion

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at

2.5% level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]