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3 years ago

Given f(x) = x^3 - 5x^2 + kx + 4 and that x-2 is a factor of f(x), what is the value of k?

Mathematics

2 answers:

Flauer [41]3 years ago

5 0

Answer : k = 4
please refer to the solution below :)

kotykmax [81]3 years ago

4 0

Either I’m being really dumb or this doesn’t work, x=2 because x-2 is a factor... the answer seems to be a lot larger than these

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Caroline, Colin & Sarah share some money.

julia-pushkina [17]

Answer:

2/9

Step-by-step explanation:

After Caroline's share,

1 - 1/9 = 8/9 is left

Sarah gets:

1/(1+3) of 8/9

1/4 × 8/9

2/9

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3 years ago

A test is worth 30 points. Multiple-choice questions are worth 7 point and

kumpel [21]

Answer:cvg

Step-by-step explanation:

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3 years ago

julia is putting together bag of 45 blocks. if she wants the probability of choosing a red block to be 1/5 how many red blocks j

4vir4ik [10]

Answer:

Step-by-step explanation:

Given that the sample size is 45 blocks

Let number of red blocks in the bag to be x

Then the probability of chosing a red block will be x/45

Given that the probability of choosing a red block is 1/5 then you can form the expression;

x/45 = 1/5 ---------------find the value of x by performing cross multiplication

5x=45*1

5x/5=45/5

x=9

Answer

Julia put 9 red blocks in the bag.

7 0

3 years ago

The meat department of a local supermarket packages ground beef using meat trays of two sizes: 1 designed to hold 1 lb of meet a

r-ruslan [8.4K]

Answer:

c) 0.932

99% confidence interval for average weights of all packages sold in small meat trays.

(0.932 ,1.071)

Step-by-step explanation:

Explanation:-

Given random sample of 35 packages in small meat trays produced weight with an average of 1.01 lbs. and standard deviation of 0.18 lbs.

size of the sample 'n' = 35

mean of the sample x⁻= 1.01lbs

standard deviation of the sample 'S' = 0.18lbs

The 99% confidence intervals are given by

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )$

The degrees of freedom γ=n-1 =35-1=34

tₐ = 2.0322

99% confidence interval for average weights of all packages sold in small meat trays

$(1.01 - 2.0322 \frac{0.18}{\sqrt{35} } , 1.01+2.0322 \frac{0.18}{\sqrt{35} } )$

( 1.01 - 0.06183 , 1.01+0.06183)

(0.932 ,1.071)

Final answer:-

99% confidence interval for average weights of all packages sold in small meat trays.

(0.932 ,1.071)

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3 years ago

What is the measure of the angle

lozanna [386]

Answer:

140 degree

Step-by-step explanation:

You go anticlockwise, starting from the 0.

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3 years ago