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MA_775_DIABLO [31]
1 year ago
15

According to the Rational Root Theorem, the following are potential roots of f(x) = 6x4 + 5x3 – 33x2 – 12x + 20. -5/2,-2, 1, 10/

3 Which is an actual root of f(x)?
Mathematics
2 answers:
stiks02 [169]1 year ago
8 0
-5/2 is the answer. I just took the test and got it right.
Arisa [49]1 year ago
4 0

Answer:

The actual root of f(x) is -5/2 (first option)

Step-by-step explanation:

A root of f(x) is a value of the variable "x" that makes f(x)=0

1) With x=-5/2

f(-5/2)=6(-5/2)^4+5(-5/2)^3-33(-5/2)^2-12(-5/2)+20

f(-5/2)=6(5^4/2^4)+5(-5^3/2^3)-33(5^2/2^2)+6(5)+20

f(-5/2)=6(625/16)-5(125/8)-33(25/4)+30+20

f(-5/2)=3(625/8)-625/8-825/4+50

f(-5/2)=1,875/8-625/8-825/4+50

f(-5/2)=[1,875-625-2(825)+8(50)]/8

f(-5/2)=(1,875-625-1,650+400)/8

f(-5/2)=0/8

f(-5/2)=0 then x=-5/2 is a root of f(x)


2) With x=-2

f(-2)=6(-2)^4+5(-2)^3-33(-2)^2-12(-2)+20

f(-2)=6(2^4)+5(-2^3)-33(2^2)+24+20

f(-2)=6(16)-5(8)-33(4)+44

f(-2)=96-40-132+44

f(-2)=-32 different of 0, then x=-2 is not a root of f(x)


3) With x=1

f(1)=6(1)^4+5(1)^3-33(1)^2-12(1)+20

f(1)=6(1)+5(1)-33(1)-12+20

f(1)=6+5-33+8

f(1)=-14 different of 0, then x=1 is not a root of f(x)


4) With x=10/3

f(10/3)=6(10/3)^4+5(10/3)^3-33(10/3)^2-12(10/3)+20

f(10/3)=6(10^4/3^4)+5(10^3/3^3)-33(10^2/3^2)-4(10)+20

f(10/3)=6(10,000/81)+5(1,000/27)-33(100/9)-40+20

f(10/3)=2(10,000/27)+5,000/27-11(100/3)-20

f(10/3)=20,000/27+5,000/27-1,100/3-20

f(10/3)=[20,000+5,000-9(1,100)-27(20)]/27

f(10/3)=(20,000+5,000-9,900-540)/27

f(10/3)=14,560/27 different of 0, then x=10/3 is not a root of f(x)

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We have the following function:

f(x)=-3x^{2} +168x-1920

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In order to calculate the maxium profit for the company, and how many cakes should be prepared in order to reach it, we have to calculate where the parabola's vertex is ubicated. To do so, we use the following formula:

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the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of
soldi70 [24.7K]

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = \sigma = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

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We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

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Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266

Since,

P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

z = -0.496

Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

-0.496=\frac{x-22}{4}\\\\ x=4(-0.496) + 22\\\\ x=20.02\\\\ x \approx 20

Thus, a test score of 20 represent the 31st percentile of the distribution.

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