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babunello [35]
3 years ago
6

CHECK MY ANSWER?!

Mathematics
2 answers:
MrRissso [65]3 years ago
4 0
First degree polynomial w two terms. a polynomial means there is more than one term and 2x is one term and 8 is the other :) hope this helped
Effectus [21]3 years ago
3 0
No, a second degree binomial is when the x is squared (to the 2nd power)
The correct answer is <span>first degree polynomial with two terms which is basically another way to say first degree binomial.
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Name a combination of coins that is 50% of the value of a dollar.
Luda [366]

Answer:

2 dimes and 1 nickel.

2 quaters

10 nickels

5 dimes

Step-by-step explanation:

Hope this helps.

4 0
3 years ago
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Help me out with this question ​
alisha [4.7K]
The answer is: Length
4 0
3 years ago
4/5z=16 solve equation (that's a fraction )
Assoli18 [71]
So we need to isolate the variable. In order to do that, we need to multiply both sides by 5/4.
Why? That's so that 4/5z will be canceled out when multiplied by 5/4 to get just z.
And so, we would get 20 as our answer.
6 0
3 years ago
Which two representations are considered to be only partial representations?
DaniilM [7]

Answer:

D) The graph and the table

Step-by-step explanation:

The equation and the verbal description fully represent the situation because their representation is valid for all possible x values. On the other hand, the table and the graph only represent certain x values.

4 0
3 years ago
Read 2 more answers
(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0
Goshia [24]

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

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

We know that

Maximum value of P(A\cup B)=1 and minimum value of P(A\cup B)=0

0\leq P(A\cup B )\leq 1

0\leq P(A)+P(B)-P(A\cap B)\leq 1

0\leq 0.4+0.7-P(A\cap B)\leq 1

0\leq 1.1-P(A\cap B)\leq 1

0\leq 1.1-P(A\cap B)

P(A\cap B)\leq 1.1

It is not possible that P(A\cap B) is equal to 1.1

1.1-P(A\cap B)\leq 1

-P(A\cap B)\leq 1-1.1=-0.1

Multiply by (-1) on both sides

P(A\cap B)\geq 0.1

Again, P(A\cup B)\geq P(B)

0.4+0.7-P(A\cap B)\geq 0.7

1.1-P(A\cap B)\geq 0.7

-P(A\cap B)\geq -1.1+0.7=-0.4

Multiply by (-1) on both sides

P(A\cap B)\leq 0.4

Hence, 0.1\leq P(A\cap B)\leq 0.4

3 0
3 years ago
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