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andriy [413]
3 years ago
6

Camille bought 10 tickets to the play and 5 parking permits. She spent $145.50 on tickets to the play and $50.75 on parking perm

its. How much did Camille spend on two tickets to the play and 1 parking permit?
A.
Step 1: Calculate the price of 1 ticket to the play.
Step 2: Calculate the price of 1 parking permit.
Step 3: Add the 2 numbers together.

B.
Step 1: Calculate the price of 1 ticket to the play.
Step 2: Calculate the price of 1 parking permit.
Step 3: Add the cost of a play ticket to the cost of a parking permit.
Step 4: Divide the total by 2.

C.
Step 1: Calculate the price of 1 ticket to the play.
Step 2: Multiply that number by 2.
Step 3: Calculate the price of 1 parking permit.
Step 4: Add that number to the product found in Step 2.

D.
Step 1: Calculate the price of 1 ticket to the play.
Step 2: Calculate the price of 1 parking permit.
Step 3: Add the cost of a play ticket to the cost of a parking permit.
Step 4: Multiply the total by 2.
Mathematics
2 answers:
Dahasolnce [82]3 years ago
4 0
She would spend $29.10 on two tickets and $10.15 on parking permits.
Steps
1.) 145.50/ 10=14.55
2.) 14.55+14.55=29.10
3.) 50.75/5=10.15
That's how you get you answer.
Arturiano [62]3 years ago
4 0
Option C describes the correct way on how to solve this question.
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A police department reports that the probabilities that 0, 1, 2, and 3 burglaries will be reported in a given day are 0.53, 0.40
SpyIntel [72]
The mean of the probability distribution is given by the expected value given by

E(x)=\sum xp(x) \\  \\ =0(0.53)+1(0.40)+2(0.05)+3(0.02) \\  \\ =0.40+0.10+0.06=0.56

The variance of the probability distribution is given by

Var(x)=(x-\bar{x})^2p(x) \\  \\ =(0-0.56)^2(0.53)+(1-0.56)^2(0.40)+(2-0.56)^2(0.05)\\+(3-0.56)^2(0.02) \\  \\ =(-0.56)^2(0.53)+0.44^2(0.40)+1.44^2(0.05)+2.44^2(0.02) \\  \\ =0.3136(0.53)+0.1936(0.40)+2.0736(0.05)+5.9536(0.02) \\  \\ =0.166208+0.07744+0.10368+0.119072=0.4664

The standard deviation is given by

\sigma(x)=\sqrt{Var(x)} \\  \\ =\sqrt{0.4664}=0.683
5 0
3 years ago
Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gen
Aleksandr-060686 [28]

Let X be the number of boys in n selected births. Let p be the probability of getting baby boy on selected birth.

Here n=10. Also the male and female births are equally likely it means chance of baby boy or girl is 1/2

P(Boy) = P(girl) =0.5

p =0.5

From given information we have n =10 fixed number of trials, p is probability of success which is constant for each trial . And each trial is independent of each other.

So X follows Binomial distribution with n=10 and p=0.5

The probability function of Binomial distribution for k number of success, x=k is given as

P(X=k) = (10Ck) 0.5^{k} (1-0.5)^{10-k}

We have to find probability of getting 8 boys in n=10 births

P(X=8) = (10C8) 0.5^{8} (1-0.5)^{10-8}

= 45 * 0.0039 * 0.25

P(X = 8) = 0.0438

The probability of getting exactly 8 boys in selected 10 births is 0.044

8 0
3 years ago
Ryan has some nickels and dimes. altogether, he has 34 coins that total $2.60. write and solve a system of equations to determin
777dan777 [17]
N+d=34
.05n+.10d=2.60

Solve by substitution
n=34-d
.05(34-d)+.10d=2.60
1.7-.05d+.10d=2.60
.05d=.90
d=18 dimes
18+n=34
n=16 nickles

4 0
3 years ago
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