Find the complete factored form of the polynomial 48m^5+8n^2

Which values of k are solutions to the inequality StartAbsoluteValue negative k minus 2 EndAbsoluteValue less-than 18? Check all

Stella [2.4K]

Answer:

You could just replace all the given possible values of k in the inequality and see which ones are solutions, but let's solve this in a more interesting way:

First, remember how the absolute value works:

IxI = x if x ≥ 0

IxI = -x if x ≤ 0

Then if we have something like:

IxI < B

We can rewrite this as

-B < x < B

Now let's answer the question, here we have the inequality:

I-k -2I < 18

Then we can rewrite this as:

-18 < (-k - 2) < 18

Now let's isolate k:

first, we can add 2 in the 3 parts of the inequality:

-18 + 2 < -k - 2 + 2 < 18 + 2

-16 < -k < 20

Now we can multiply all sides by -1, remember that this also changes the direction of the signs, then:

-1*-16 > -1*-k > -1*20

16 > k > -20

Then k can be any value between these two limits.

So the correct options (from the given ones) are:

k = -16

k = -8

k = 0

6 0

2 years ago

Read 2 more answers

The slope of a line is -1/3, and the y intercept is 10/3. What is the equation of the line written in slope-intercept form

REY [17]

Should be y=-1/3x+10/3

Hope this helps :)

7 0

3 years ago

Please help pleasehelp

grandymaker [24]

Answer:

≥ and ≤ will both be solid dots, as it includes the number

x ≥ -7 is everything greater than -7; x ≤ 4 is everything less than 4.

to include both, you'd click -7, drag right, and stop at 4

7 0

2 years ago

A sample of 11001100 computer chips revealed that 62b% of the chips fail in the first 10001000 hours of their use. The company's

STALIN [3.7K]

Answer:

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

z score = 1.35

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

Question; A sample of 1100 computer chips revealed that 62% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 60% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.10 level.

Step-by-step explanation:

Given;

n=1100 represent the random sample taken

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 1100

po = Null hypothesized value = 0.60

p^ = Observed proportion = 0.62

Substituting the values we have

z = (0.62-0.60)/√{0.60(1-0.60)/1100}

z = 1.354

z = 1.35

To determine the p value (test statistic) at 0.01 significance level, using a two tailed hypothesis.

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

3 0

3 years ago

Which of the following are roots of the polynomial function? Check all that apply.

lutik1710 [3]

You know 1 is not a root because the sum of the coeffcients of the equation is 14, not zero.

It is fairly easy to try 3 by synthetic division (see attachment), which tells you that 3 is a root and the remaining quadratic factor is x²-3x-5. The quadratic formula tells you the roots of that factor are

... x = (-b±√(b²-4ac))/(2a) = (3±√29)/2

The appropriate choices are

... C. (3-√29)/2

... D. (3+√29)/2

... F. 3

_____

The quadratic formula tells you the solution to

... ax²+bx+c=0

is x = (-b±√(b²-4ac))/(2a)

We have a=1, b=-3, c=-5.

8 0

3 years ago