Find the cost for one ticket for the science center.
36.75 ÷ 3 = 12.25
Multiply the product with 12 to get the cost of 12 tickets.
12.25 × 12 = 147
Find the cost of 1 zoo ticket.
51 ÷ 4 = 12.75
Find the cost of 12 tickets by multiplying the product and 12
12.75 × 12 = $153
12 Science Center Tickets cost $147
12 zoo tickets cost $153
The Zoo costs more than the Science Center.
To find out how much more the Zoo costed, subtract 147 from 153.
153 - 147 = 6
The zoo only costs $6 more for 12 students.
So, you're answers are Zoo and $6
Answer:
a) p=0.2
b) probability of passing is 0.01696
.
c) The expected value of correct questions is 1.2
Step-by-step explanation:
a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.
b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6
.
Recall that
In this case, the student passes if X is at least four correct questions, then

c)The expected value of a binomial random variable with parameters n and p is
. IN our case, n=6 and p =0.2. Then the expected value of correct answers is 
Answer:
Check the explanation
Step-by-step explanation:
let the money on bet is X.
probability of winning =18/38=9/19
probability of losing =(1-9/19)=10/19
expected outcome =
probability *return =(
Expected value of return after one bet is =(9/19*x)-(10/19*x)=-1x/19
it is negative which is obvious cause casinos are there to earn money.
a) Our best strategy in this case as probability of winning is near by 50 %, we should place a bet of 1 $ each,and when we lose one bet consecutively we should bet twice the amount..
Cause two consecutive losses on black has less probability.
c) In case we have to reach 30 $ we have to use the same strategy as above.
Answer:
13
Step-by-step explanation:
Option B is the answer because angle A is greater than C and A