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3 years ago

Which of the following shows the complete factorization x3+3x2

Mathematics

2 answers:

Ksju [112]3 years ago

8 0

X^3 + 3x^2...factor out x^2

x^2 (x + 3) <==

SIZIF [17.4K]3 years ago

5 0

Hello!

x^3 + 3x^2
x (x^2 + 3x)
x^2 (x + 3)

Therefore, The Correct Answer is "x2(x + 3)".

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If the critical z-value for a hypothesis test... If the critical z-value for a hypothesis test equals 2.45, what value of the te

schepotkina [342]

Answer: z score = 0.00714

Step-by-step explanation: the value of test statistics is gotten using the standard normal distribution table.

Z= 2.45 has area to the left (z<2.45) and area to the right (z>2.45).

Level of significance α is the probability of committing a type 1 error. The area under the distribution is known as the rejection region and it is the area towards the right of the distribution.

The table I'm using is towards the left of the distribution.

But z>2.45 + z<2.45 = 1

z> 2.45 = 1 - z<2.45

But z < 2.45 = 0.99286

z > 2.45 = 1 - 0.99286

z >2.45 = 0.00714

Hence the test statistics that would produce the least type 1 error is 0.00714

8 0

3 years ago

Write the equation of the line perpendicular to the line x - 3y = 9 and passing through the point (3,5).

PilotLPTM [1.2K]

Here is a sample problem
has to be perpendicular to y=(4/9)*x-2 and pass through 4,3,

y-b=m(x-a)
slope of -9/4 is perpendicular to slope of 4/9, as y=4/9x-2 has slope 4/9. Our point is (4,3) therefore.
y-3=-9/4*(x-4)
y-3=-9/4x+9
y=-9/4x+9+3
y=-9/4x+12

5 0

3 years ago

Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w

Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0

4 years ago

The diameter of a circle is two times its radius. Which

atroni [7]

I can’t type it out but the equation is Pi(radius)^2

3 0

3 years ago

.................................

Ivan

Its a pull and the location has to be less than 400N

4 0

3 years ago