Use an appropriate procedure to show that x-4 is a factor of the function f(x)=2x^3-5x^2-11x-4. Show work pleassseeee

Solve 6 1/8 ÷ 3 5/8 ( meaning six and one eighth divided by three and five eighths )

Stells [14]

The answer is 1 20/29.

3 0

2 years ago

Create a factorable polynomial with a GCF of 6. Rewrite that polynomial in two other equivalent forms. Explain how each form was

mel-nik [20]

We are asked in the problem to devise a polynomial equation that has a GCF of 6 which means each of the terms can be divided to 6. For example: 6*(x^2 + x+1) = 6x^2 + 6x +6. This polynomial is created by multiplying each terms by the number 6 which is distinguished by factoring.

8 0

2 years ago

Read 2 more answers

Suppose the test scores for a college entrance exam are normally distributed with a mean of 450 and a s. d. of 100. a. What is t

svet-max [94.6K]

Answer:

a) 68.26% probability that a student scores between 350 and 550

b) A score of 638(or higher).

c) The 60th percentile of test scores is 475.3.

d) The middle 30% of the test scores is between 411.5 and 488.5.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 450, \sigma = 100$

a. What is the probability that a student scores between 350 and 550?

This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So

X = 550

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{550 - 450}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 350

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{350 - 450}{100}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a student scores between 350 and 550

b. If the upper 3% scholarship, what score must a student receive to get a scholarship?

100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88

$Z = \frac{X - \mu}{\sigma}$

$1.88 = \frac{X - 450}{100}$

$X - 450 = 1.88*100$

$X = 638$

A score of 638(or higher).

c. Find the 60th percentile of the test scores.

X when Z has a pvalue of 0.60. So it is X when Z = 0.253

$Z = \frac{X - \mu}{\sigma}$

$0.253 = \frac{X - 450}{100}$

$X - 450 = 0.253*100$

$X = 475.3$

The 60th percentile of test scores is 475.3.

d. Find the middle 30% of the test scores.

50 - (30/2) = 35th percentile

50 + (30/2) = 65th percentile.

35th percentile:

X when Z has a pvalue of 0.35. So X when Z = -0.385.

$Z = \frac{X - \mu}{\sigma}$

$-0.385 = \frac{X - 450}{100}$

$X - 450 = -0.385*100$

$X = 411.5$

65th percentile:

X when Z has a pvalue of 0.35. So X when Z = 0.385.

$Z = \frac{X - \mu}{\sigma}$

$0.385 = \frac{X - 450}{100}$

$X - 450 = 0.385*100$

$X = 488.5$

The middle 30% of the test scores is between 411.5 and 488.5.

7 0

2 years ago

PLEASE HELP ME I BEG

Kay [80]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

• 1 + cot²x = csc²x

Consider the left side

$csc^{4}$ x - $cot^{4}$ x

Factorise as a difference of squares

= (csc²x - cot²x)(csc²x + cot²x)

= (1 + cot²x - cot²x)(csc²x + cot²x)

= csc²x + cot²x

= csc²x + csc²x - 1

= 2csc²x - 1 = right side ⇒ verified

8 0

3 years ago

Find

shepuryov [24]

Answer:

The two odd numbers are 15 and 17

Step-by-step explanation:

Given

Let the odd numbers be represented with x and y

Let x be the greater number

$x + 5y = 92$

Required

Find x and y

Since x and y are consecutive odd numbers and x is greater, then

$x = y + 2$

Substitute y + 2 for x in $x + 5y = 92$

$y + 2 + 5y = 92$

Collect Like Terms

$y + 5y = 92 - 2$

$6y = 90$

Divide both sides by 6

$\frac{6y}{6} = \frac{90}{6}$

$y = \frac{90}{6}$

$y = 15$

Substitute 15 for y in $x = y + 2$

$x = 15 + 2$

$x = 17$

Hence; the two odd numbers are 15 and 17

4 0

2 years ago