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In-s [12.5K]
2 years ago
9

How would i do this ?????????????????

Mathematics
1 answer:
White raven [17]2 years ago
6 0

Answer:

f(x) = 2x + 7

f(8) = 2(8) + 7 = 16 + 7 = 23

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From a box containing 10 cards numbered 1 to 10, four cards are drawn together. The probability that their sum is even is 21 21
ankoles [38]

Answer:

Step-by-step explanation:

We know that between 1 to 10 there are 5 even and 5 odd numbers.

We could get 4 even cards , 4 odd cards or 2 odd and 2 even cards

Let´s check all this combinations

Case 1: When all 4 numbers are even:  

We are going to take 4 of the 5 even numbers in the box so we have

5C4=5

Case 2: When all 4 numbers are odd:  

We are going to take 4 of the 5 odd numbers in the box, so we have

5C4=5

Case 3: When 2 are even and 2 are odd:

We are giong to take 2 from 5 even and odd cards in the box so we have

 

5C2 * 5C2

Remember that we obtain the probability from

\frac{Number-of-favourable-Outcome}{Total-number-of-outcomes}

So we have the number of favourable outcomes but we need the Total cases for drawing four cards, so we have that:  

We are taking 4 of the 10 cards:

10C_4=210

Hence we have that the probability that their sum is even

\frac{5+5+100}{210}=\frac{11}{21}

8 0
3 years ago
Variable expression for 3x+7
evablogger [386]

Answer:

y=3x+7

Step-by-step explanation:

Where y is the dependent variable and x is the independent variable. When we change x , Y also changes.

3 0
3 years ago
If f(x)=3x-7 and g(x)=6x+4, what is f(x) + g(x)?
PSYCHO15rus [73]
F(x) + g(x)
3x-7+6x+4
=9x-3
8 0
4 years ago
What is the derivative of w=(t^2+1)^100<br><br> How is it 200t(t^2+1)^99?
GuDViN [60]
(t^2+1)^100

USE CHAIN RULE

Outside first (using power rule)

100*(t^2+1)^99 * derivative of the inside

100(t^2+1)^99 * d(t^2+1)

100(t^2+1)^99 * 2t

200t(t^2+1)^99
6 0
3 years ago
Read 2 more answers
Mr. Grant spent 8.40 to place an adin the newspapers.the price included a one time fee of 6.00 pulse 0.08 per word.which method
Sladkaya [172]
Your answer is 0048.00
5 0
3 years ago
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