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4 years ago

-2xy(3xy^3 – 5xy+ 7y^2)

Mathematics

1 answer:

svetlana [45]4 years ago

3 0

Answer:

look below

Step-by-step explanation:

11x2 + 8xy + 5y

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((5xy+(7•(x2)))-3y)-(((0-22x2)-8y)-3xy)

Step 2 :

Equation at the end of step 2 :

((5xy + 7x2) - 3y) - (-4x2 - 3xy - 8y)

Step 3 :

Trying to factor a multi variable polynomial :

3.1 Factoring 11x2 + 8xy + 5y

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Value of x in the isosceles below

allochka39001 [22]

Being an isosceles, both sides would be equal.

5X - 32 = 2x +61

Add 32 to each side:

5X = 2X + 93

Subtract 2X from each side:

3X = 93

Divide both sides by 3:

X = 93/3

X = 31

3 0

3 years ago

The Highway Safety Department wants to study the driving habits of individuals. A sample of 121 cars traveling on the highway re

Ede4ka [16]

Answer:

{58.02007 , 61.97993]

Step-by-step explanation:

Data are given in the question

Sample of cars = n = 121

Average speed = sample mean = 60

Standard deviation = sd = 11

And we assume

95% confidence t-score = 1.97993

Therefore

Confidence interval is

$= [60 - \frac{1.97993 \times 11}{\sqrt{121} }] , [60 + \frac{1.97993 \times 11}{\sqrt{121} }]$

= {58.02007 , 61.97993]

Basically we applied the above formula to determine the confidence interval

4 0

3 years ago

A variable needs to be eliminated to solve the system of equations below. Choose the

notsponge [240]

Answer:

x=-2

y= 8

Step-by-step explanation:

Given data

10x – 2y = -36-----------1

7x – 2y = -30--------------2

-Subtract to eliminate y

3x-0= -6

3x= -6

x= -6/3

x= -2

Put x= -2 in

7(-2)– 2y = -30

-14-2y=-30

-14+30=2y

16=2y

y= 16/2

y= 8

8 0

3 years ago

The probability that a certain make of car will need repairs in the first seven months is 0.9. A dealer sells three such cars. W

aleksandr82 [10.1K]

Answer:

0.9990 = 99.90% probability that at least one of them will require repairs in the first seven months.

Step-by-step explanation:

For each car, there are only two possible outcomes. Either they will require repair in the first seven months, or they will not. The probability of a car requiring repair in the first seven months is independent of other cars. So we use the binomial probability distribution to solve this question.

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.

The probability that a certain make of car will need repairs in the first seven months is 0.9.

This means that $p = 0.9$ A dealer sells three such cars.

A dealer sells three such cars.

This means that $n = 3$

What is the probability that at least one of them will require repairs in the first seven months?

Either none will require repairs, or at least one will. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X > 0) = 1$

We want P(X > 0). So

$P(X > 0) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{3,0}.(0.9)^{0}.(0.1)^{3} = 0.001$

Then

$P(X > 0) = 1 - P(X = 0) = 1 - 0.001 = 0.9990$

0.9990 = 99.90% probability that at least one of them will require repairs in the first seven months.

4 0

3 years ago

Can someone please help

Bas_tet [7]

Number 6 is C, Number 7 is A. Do you need number 8? if so I will edit answer and add pic

7 0

3 years ago