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3 years ago

The domain for f(x) and g(x) is the set of all real numbers.

Mathematics

1 answer:

Tom [10]3 years ago

3 0

Answer:

Option B

Step-by-step explanation:

=> $f(x) = 3x+5$

=> $g(x) = x^2$

Subtracting both

=> $(f-g)(x) = 3x+5-x^2$

Assembling

=> $(f-g)(x) = -x^2+3x+5$

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Find the quotient.<br> 52.06 divided by 1.9

Grace [21]

Answer:

27.4

Step-by-step explanation:

First multiply the numerator and denominator by 10

Then write the problem in long division format

Then divide 52 by 19 and get 2

Then multiply the quotient digit (2) by the divisor 19

Then subtract 38 from 52

Then bring down the next number of the dividend

Then divide 140 by 19 to get 7

Then multiply the quotient digit (7) by the divisor 19

Then subtract 133 from 140

Then add a decimal point to the solution

Then bring down the next number of the dividend

Then divide 76 by 19 to get 4

Then multiply the quotient digit (4) by the divisor 19

Then subtract 76 from 76

The solution for long division for 52.06./1.9 is 27.4

27.4

7 0

3 years ago

Read 2 more answers

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X 0 1 2

P(X) 42% 50% 8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c) P(win first game) = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X 0 1 2

P(X) 42% 50% 8%

P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value $\mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66$