Use intercepts to graph the linear equation −12x+3y=24.

Oh yes... Wonderful Proportions. There are ways to do proportions but this is how I learned how to do them :

PART %
=
WHOLE 100 That is the formula. I'm not sure of an example but I hope it helps

3 0

2 years ago

What is the solution to the system of equations?

yarga [219]

Answer:

(2, - 4 )

Step-by-step explanation:

Given the 2 equations

x + 2y = - 6 → (1)

3x + y = 2 → (2)

Multiplying (2) by - 2 and adding to (1) will eliminate the term in y

- 6x - 2y = - 4 → (3)

Add (1) and (3) term by term to eliminate y

- 5x = - 10 ( divide both sides by - 5 )

x = 2

Substitute x = 2 into either of the 2 equations and solve for y

Substituting into (1)

2 + 2y = - 6 ( subtract 2 from both sides )

2y = - 8 ( divide both sides by 2 )

y = - 4

Solution is (2, - 4 )

8 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

2 years ago