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3 years ago

What is the solution for this inequality? -10x < 40

1 answer:

8 0

Let's solve your inequality step-by-step.−10x<40Step 1: Divide both sides by -10.−10x−10<40−10x>−4Answer:x>−4

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Help pleasssss Order the following numbers from least to greatest. 0.58 0.28 1/2

Tema [17]

Answer. 0.28, 1/2, 0.58
Explanation. If you convert all the numbers to decimal form, we find that this is their position from least to greatest. Hope this helps, let me know if it is correct so others can use it as well :)
Good luck.

8 0

2 years ago

Read 2 more answers

The director of a marching band asks the band members to line up in rows of four, but one is left over. Then she tries to line t

lions [1.4K]

I got 81 members. 4 x 20 = 80 81-80=1
6 x 13= 78 81-78= 3
7 x 11= 77 81-77= 4

6 0

3 years ago

Which of these is not an equation?

timurjin [86]

Answer:

Step-by-step explanation:

an equation always shows that sum is equal

in this case 1825026 will subtract 17 and the remainder will be used to find A

3 0

2 years ago

Using the letters suggested, construct a simple formula:

kompoz [17]

Answer:

Speed = distance / time

Time = distance/ Speed

Distance = Speed x time

8 0

2 years ago

The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,

strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

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".

$\bar X=400$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=80$ represent the population standard deviation

n=15 represent the sample size

2) Part a

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

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

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$400-2.58\frac{80}{\sqrt{15}}=346.708$

$400+2.58\frac{80}{\sqrt{15}}=453.292$

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

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 first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.98$

Now we have everything in order to replace into formula (1):

$400-2.98\frac{80}{\sqrt{15}}=338.445$

$400+2.98\frac{80}{\sqrt{15}}=461.555$

So on this case the 99% confidence interval would be given by (338.445;461.555)

4 0

2 years ago