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Korolek [52]
3 years ago
8

Solve equations with more than two unknowns.. The linear combination technique by solving the following system of two equations

with two unknowns... Need help solving this

Mathematics
1 answer:
Iteru [2.4K]3 years ago
5 0
You can do it using substitution too

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7. y = x2 + 3; D:{-2, -1, 0, 1, 2}
Andrei [34K]

Answer:

-1

1

3

5

7

Step-by-step explanation:

5 0
3 years ago
Suppose that you invest $6000 into a high yield savings account that earns 2% annual interest and is compounded monthly. If you
Dmitry_Shevchenko [17]

Answer:


Step-by-step explanation:

FV   =   P (1  +  r / n)Yn

n=12

Y=20

P =6000


7 0
3 years ago
A farmer sells 8.2 kilograms of apples and pears at the farmer's market. 1/4 of this weight is
Bumek [7]

Answer:

6.15 kilograms of pears

Step-by-step explanation:

The total is 8.2 kilograms. Since 1/4 of the weight is apples, we know that 3/4 of the weight is pears. So we multiply 8.2*0.75.

8.2*0.75=6.15

Therefore, the farmer sold 6.15 kilograms of pears at the farmer's market.

5 0
2 years ago
Begin collecting them. You buy the same number of cards once each
topjm [15]

Answer:

y=-2

Step-by-step explanation:

you have to subtract a positive by a negative and do to the to the 2

7 0
2 years ago
Read 2 more answers
In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that
leonid [27]

Given:

Number of students who has a cat and a dog = 5

Number of students who has a cat but do not have a dog = 11

Number of students who has a dog but do not have a cat = 3

Number of students who neither have a cat nor a dog = 2

To find:

The probability that a student has a cat given that they do not have a dog.

Solution:

Let the following events:

A = Student has a cat

B = Do not have a dog

Total number of outcomes is:

5+3+11+2=21

The probability that a student has a cat but do not have a dog is:

P(A\cap B)=\dfrac{11}{21}

The probability that a student do not have a dog is:

P(B)=\dfrac{11+2}{21}

P(B)=\dfrac{13}{21}

The conditional probability is:

P\left(\dfrac{A}{B}\right)=\dfrac{P(A\cap B)}{P(B)}

P\left(\dfrac{A}{B}\right)=\dfrac{\dfrac{11}{21}}{\dfrac{13}{21}}

P\left(\dfrac{A}{B}\right)=\dfrac{11}{13}

Therefore, the probability that a student has a cat given that they do not have a dog is \dfrac{11}{13}.

5 0
3 years ago
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