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Solve the system of equations system

Korolek [52]

Answer:

(3, 3 )

Step-by-step explanation:

Given the 2 equations

3x - y = 6 → (1)

6x + y = 21 → (2)

Adding the 2 equations term by term will eliminate y, that is

(6x + 3x) + (y - y) = (21 + 6), that is

9x = 27 (divide both sides by 9 )

x = 3

Substitute x = 3 into either (1) or (2) and solve for y

Using (2), then

(6 × 3) + y = 21

18 + y = 21 ( subtract 18 from both sides )

y = 3

Solution is (3, 3 )

8 0

3 years ago

Read 2 more answers

The width of a large, rectangular-shaped tree farm is 1.9 kilometers. The length of the farm is 4.4 kilometers. What are the bes

Vinvika [58]

Answer:

9 square kilometers

Step-by-step explanation:

Let's round the width to 2.0 kilometers, and round the length to 4.5 kilometers.

We know that area is length times width, so

A = lw

A = 2.0*4.5

A = 9.0 square kilometers

If we do this on the calculator, it's around 9.36 square kilometers, so our estimate was good.

6 0

3 years ago

Solve the following system equation graphically on the set of axes below y= x+4 and y= -3/2 - 1 PLEASE HELP MEEE

gizmo_the_mogwai [7]

Answer:

y=4

y= -1

Step-by-step explanation:

8 0

2 years ago

Richard Lens receives a 3.5% commission for selling cleaning supplies. What is his commission on sales of $122,591?

LekaFEV [45]

Answer:

$4,290.69

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- To find the commission, just convert the percentage into a decimal (divide by 100) and multiply it by the sales.

3.5% = 0.035

122,591 x 0.035 = $4,290.69

5 0

2 years ago

Read 2 more answers

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