Mike had thirty seven dollars to spend on eight books. After buying them

Jennifer is saving money to buy a bike The buce costs $343 She has $115 saved, and each week she adds $19 to her savings How lon

satela [25.4K]

Answer: 12 weeks

Step-by-step explanation:

343-115=228

228/19=12

5 0

3 years ago

6.Find the equation of<br> the line:<br> through: (-2, 3) and<br> (1,5)

prohojiy [21]

Answer:

y=2/3x+4.3

Step-by-step explanation:

y=mx+c

(5-3)/(1--2)

2/3= gradient

y=2/3x +c

5=2/3(1)+c

-2/3

4.3=c

6 0

3 years ago

When the mountain was last measured, its height was 3,570 meters. Now it is 3,927 meters. What is the percent of increase in the

Elina [12.6K]

Answer:

110%

Step-by-step explanation:

3,927/3,570

=1.1

=110%

6 0

3 years ago

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A ran

mart [117]

Answer:

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

$X=48$ number of vehicles classified as sports utility.

$n=500$ random sample taken

$\hat p=\frac{48}{500}=0.096$ estimated proportion of vehicles classified as sports utility vehicles.

$p$ true population proportion of vehicles classified as sports utility vehicles.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

The confidence interval would be given by this formula :

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070$

$0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121$

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

7 0

3 years ago

What is the difference? startfraction 2 x 5 over x squared minus 3 x endfraction minus startfraction 3 x 5 over x cubed minus 9

weeeeeb [17]

The difference of the expression $\frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x}$ is $\frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}$