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Pavlova-9 [17]
4 years ago
13

A survey of magazine subscribers shows that 19% rented a car during the past twelve months for business reasons, 52% rented a ca

r during the last twelve months for personal reasons, and 3% rented a car during the last twelve months for both business and personal reasons.
a. What is the probability that a subscriber rented a car during the past 12 months for
business or personal reasons?
b. What is the probability that a subscriber did not rent a car during the past 12 months
for either business or personal reasons?
Mathematics
1 answer:
irinina [24]4 years ago
4 0

Answer:

a. 0.68 or 68%

b. 0.32 or 32%

Step-by-step explanation:

a. The probability that a subscriber rented a car during the past 12 months for  business or personal reasons (P(R)) is given by the probability that they rented a car for business reasons (P(B)=0.19), added to the probability that they rented for personal reasons (P(P)=0.52), subtracted by the probability that they rented for both reasons (P(B and P) = 0.03):

P(R)=P(B)+P(P)-P(B\cap P)\\P(R) = 0.19+0.52-0.03\\P(R)=0.68

b. The probability that a subscriber did not rent a car during the past 12 months  for either business or personal reasons (P(N)) is 100% minus the probability that they rented a car (P(R) = 0.68).

P(N) = 1-P(R) = 1-0.68\\P(N) =0.32

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Hey~freckledspots!\\----------------------

We~will~solve~for~i^{425}!

Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\

Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------

Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------

Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------

Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------

Answer:\\\large\boxed{-1+i}\\----------------------

Hope~This~Helped!~Good~Luck!

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Read 2 more answers
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