How do I solve 3-5x=2

Simplify and reduce to the lowest terms.<br> <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B-7%7D%7B12%7D%20" id="TexFormula1" t

Paladinen [302]

When you divide two fractions, you're actually multiplying one of them by the reciprocal of the other. First, find the reciprocal of the second fraction by flipping it upside down. Then, multiply it by the first fraction. (Numerator x numerator and denominator x denominator)

$\frac{-7}{12}$ ÷ $\frac{2}{3}$ Replace the second fraction with it's reciprocal
$\frac{-7}{12}$ x $\frac{3}{2}$ Multiply (-7 x 3 and 12 x 2)
$\frac{-21}{24}$ Both 21 and 24 are divisible by three, so divide them by 3
$\frac{-7}{8}$

6 0

2 years ago

Y=3/2(-1)-1<br> Linear Equation

levacccp [35]

X = -5/2
X is a single horizontal line across the graph, because x has a set value

5 0

3 years ago

Ill give brainliest to who ever answers these questions first

svetlana [45]

High blood pressure, high cholesterol, and smoking.

6 0

2 years ago

Read 2 more answers

Use the quadratic formula to find the solutions to the

nadezda [96]

Answer:

D

Step-by-step explanation:

The quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where a is the coefficient of $x^2$ , b being the coefficient of $x$ and c being the constant.

Plug in the values:

$\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}$

$\frac{5\pm\sqrt{-15} }{4}$

Since $\sqrt{-15} =\sqrt{-1} \sqrt{15} =i \sqrt{15}$

The answer is:

$\frac{5\pm i \sqrt{15} }{4}$

6 0

2 years ago

Read 2 more answers

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago