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3 years ago

Evaluate the following algebraic expression: 2 x + y for x = 1 and y = 1

Mathematics

1 answer:

12345 [234]3 years ago

8 0

Answer:

Step-by-step explanation:

2*1+1

2+1

how this helps :)

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In a volleyball game, Caroline served the ball over the net 16 out of 20 times. What percent did she not make her serve over the

brilliants [131]

16/20. We reduce that and get 4/5.

After this it's much simpler to chage it.

100 dividied by 5 is 20. (20,40,60,80,100)

So the 4th number (4/100)

is 80. 80%.

20% she didn't and 80% she did :) (100-80)

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3 years ago

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How do i know when a set of ordered pairs that represents a function?

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Answer:

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Step-by-step explanation:

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3 years ago

A rectangular courtyard has a length of 24 m and a width of 15 m. Find the perimeter of the rectangle defined by this courtyard.

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2 years ago

acceleration is a(t) = 3t and velocity is v(t) = 1 at t = 1, what is the displacement from t = 2 to t = 4

lidiya [134]

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2 years ago

Let A and B be events with =PA0.7, =PB0.3, and =PA or B0.9. (a) Compute PA and B. (b) Are A and B mutually exclusive? Explain. (

kvasek [131]

Answer:

a) $P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

b) False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

c) False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

Step-by-step explanation:

For this case we have the following probabilities given:

$P(A) = 0.7, P(B) =0.7, P(A \cup B) =0.9$

Part a

We want to calculate the following probability: $P(A \cap B)$

And we can use the total probability rule given by:

$P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

Part b

False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

Part c

False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

4 0

3 years ago

Evaluate the following algebraic expression: 2 x + y for x = 1 and y = 1​

Evaluate the following algebraic expression: 2 x + y for x = 1 and y = 1