8|7+(-10) Simplify: 2-32

Solve the inequality for x. –4x + 6 < 50 A) x > –11 B) x > –14 C) x < –11 D) x < –14

Natali [406]

Answer:

x > -11

Step-by-step explanation:

Step 1: Subtract 6 from both sides

-4x + 6 - 6 < 50 - 6

-4x < 44

Step 2: Divide both sides by -4

-4x / -4 < 44 / -4

If you divide by a negative number, you flip the sign.

x > -11

Answer: x > -11

6 0

3 years ago

Help me, please!! Estoy desesperada y no hablo inglês!! Help

Aleksandr-060686 [28]

$D:x>0\\
\log x+\log x^2+\log x^3=-6\\
\log (x\cdot x^2\cdot x^3)=\log 10^{-6}\\
x^6=10^{-6}\\
x^6=\left(\frac{1}{10}\right)^6\\ \boxed{x=\frac{1}{10}}$

8 0

3 years ago

Read 2 more answers

25 out of 27 as a percentage

Basile [38]

25/27 = 0.925925925...

It's almost equal to 1 and it turns out that 1 is 100%.

So, we'd say that it's equal to:

92.59% (to 2 decimal places)

We've multiplied the value above (25/27) by 100.

8 0

3 years ago

Read 2 more answers

How to write 21/6 as a mixed number

NNADVOKAT [17]

Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction.

3 and 3 over 6

5 0

3 years ago

You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$

And if we round up the answer we see that the value of n to ensure the margin of error required $Me=\pm 60$ $ is n=28.

5 0

3 years ago