Ask question

Ask question

All categories

2 years ago

6

An app developer projects that he will earn $230.00 for every 100 apps downloaded. Which of the following equations can be used

to represent the proportional relationship between the number of apps, x, and the total earnings, y? WILL MARK BRAINLIEST!

1 answer:

rjkz [21]2 years ago

7 0

The answer is 100x+230=y
Sorry if I’m late :/ hope it helped

You might be interested in

Solve x2 + 6x = 7 by completing the square. Which is the solution set of the equation?

Ber [7]

So first you have to find the perfect square that matches up with x^2 + 6x

so half of 6, and square it. your perfect square is 9

x^2 + 6x + 9 = 7 + 9

then, condense the left side of the equation into a squared binomial:

(x + 3)^2 = 16

take the square root of both sides:

x + 3 = ± √16

therefore:

x + 3 = ± 4

x = - 3 ± 4

so your solution set is:

x = 1, -7

7 0

2 years ago

Read 2 more answers

a scientist wants to study 24 stars he studies 1/3 of them one week and 1/4 of them next week how many of them are left to study

zvonat [6]

6 because 1/3 of 24 is 8 and 1/4 of 8 is 6

4 0

2 years ago

Circle G and X connect at point Y. The length of G Y is 10 centimeters and the length of Y X is 16 centimeters. Point G is the c

poizon [28]

Answer:

its b

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A sample of size n is selected at random from an infinite population. as n increases, the standard error of the sample mean incr

andrew11 [14]

FALSE!

big time false

hope it helped

7 0

2 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

Step-by-step explanation:

Previous concepts

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".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by:

$z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

4 0

2 years ago