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4 years ago

Select all possible values for x if −3x < 15. Select one or more: A. -11 B. -7 C. -5 D. 2 E. 9

Mathematics

2 answers:

Archy [21]4 years ago

5 0

Answer:

D and C

Step-by-step explanation:

Since a negative times a negative is a positive, the only possible values (of the ones listed) for x are positive values that are equal to or higher than 15, and are not correct. Since a positive times a negative is a negative, the two positive answer choices, 2 and 9 are both correct.

Thepotemich [5.8K]4 years ago

4 0

-3 x -11 = 33
-3 x -7 = 21
-3 x -5 = 15
-3 x 2 = -6
-3 x 9= -27

the sign < means smaller than.
-6 and -27 are smaller than -15.

This means that the answer is D & E

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Dmitrij [34]

x is 53

angle PQD is equal to APQ. so its 62 degrees

PRC and PRD are supplement so their sum is 180.

PRD=115 PRC= 180 -115= 65

the sum of the angles of a triangle is equal to 180. so:

$x = 180 - (62 + 65) = 53$

good luck

7 0

3 years ago

Can you find the values of A and B please.

Tom [10]

Answer:

Step-by-step explanation:

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3 years ago

EXAMPLE:

schepotkina [342]

5 is 0.88888888888888888888 it can go on forever

6 0.06666666666666666666 still can go on forever

on number 6 if it was over the 0 then it would have be 0.060606060 repeating but if its only on the last number it just stays 0.0666

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4 years ago

F(x)=4x+7 Place the steps for finding f -1(x) in the correct order.

dybincka [34]

X=4x+7

X=4y+7

X-7=4y

X-7/4=y

F-1(x)=x-7/4

6 0

4 years ago

The mean weight of 10 randomly selected newborn babies at a local hospital is 7.14 lbs and the standard deviation is 0.87 lbs. A

sergejj [24]

Answer:

a) $ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$

b) 90% of all samples of size 10 have sample means within 0.5035 of the population mean.

c) The 90% confidence interval would be given by (6.636;7.643)

d) Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

e) If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

Step-by-step explanation:

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".

$\bar X=7.14$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.87 represent the sample standard deviation

n=10 represent the sample size

Part a

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

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

And the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that $t_{\alpha/2}=1.83$

$ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$

Part b

90% of all samples of size 10 have sample means within 0.5035 of the population mean.

Part c

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

$7.14-1.83\frac{0.87}{\sqrt{10}}=6.637$

$7.14+1.83\frac{0.87}{\sqrt{10}}=7.643$

So on this case the 90% confidence interval would be given by (6.636;7.643)

Part d

Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

Part e

If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

7 0

4 years ago