What is 5x + 3y = 12 written in slope intercept form

Mathematics

2 answers:

Luba_88 [7]3 years ago

6 0

Answer:

y=4-5/3x

Step-by-step explanation:

Rearrange and simplify.

Reika [66]3 years ago

6 0

Y= -5/3x + 4 hope this helped you

You might be interested in

Need help ASAP its due soon

kicyunya [14]

Answer:

720

Step-by-step explanation:

When you multiply 12 by 6, you get 72, and in this case, it will end up being 720 to correspond with the way it is set up.

5 0

2 years ago

Select the correct answer.

laila [671]

The answer is B, hope this helps

5 0

3 years ago

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 16-inch pizzas.

erastova [34]

Answer:

n=50

Step-by-step explanation:

1) 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".

$\bar X=16.10$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=1.8 represent the sample standard deviation

n=25 represent the sample size

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

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

We need to find the degrees of freedom given by:

$df=n-1=25-1=24$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=\pm 1.96$

Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)