A line goes through the point (–4,–2) and has an x-intercept of –1. Which point also lies on the line?

2 answers:

4 0

Im not getting exact answers but the closest was D. What i did was i graphed the points and i found the slope and then i found the y intercept. so the equation was y = 2/3x + .67. and i plugged in the points. so i would say the answer is D. because since i rounded the .67 it wasn't exact but it was close

Harman [31]3 years ago

3 0

Answer:

d. 5,4

Step-by-step explanation:

You might be interested in

A new clothing store is offering 30 free points to customers who sign up for their rewards card. Then for each shirt a customer

algol [13]

P = 3n + 30
p = 75
75 = 3n + 30
45 = 3n
n = 15
A customers earns 15 additional points on purchasing one shirt.

8 0

3 years ago

Read 2 more answers

Adele Garcia obtained a loan of 8,000$ at 12% interest for 24 months . The monthly payment is 367.80. after 15 payments the bala

Rama09 [41]

Answer:

Step-by-step explanation:

8 0

3 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

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".

The population proportion have the following distribution

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

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

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 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.