Answer:
Step-by-step explanation:
10(3)+2(10)(4)+2(3)(4)=30+80+24
=134
The probability that that a randomly selected student will buy a raffle ticket and win a prize = 0.03
Step-by-step explanation:
Step 1 :
Given,
The percentage of students buying the raffle ticket = 30%
The percentage that the student who bought the ticket wins the prize = 10%
We need to determined the probability that randomly selected student will buy a raffle ticket and win a prize.
Step 2 :
The probability that a student buys the raffle ticket = 
The probability that a student wins a prize =
= 
The probability that a student who buys the ticket wins a price can be computed by taking the product of the above 2 probabilities.
=
×
=
= 0.03
Step 3 :
Answer :
The probability that that a randomly selected student will buy a raffle ticket and win a prize = 0.03
Answer:
-3,6
Step-by-step explanation:
i think this is right
Answer:
75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.
Step-by-step explanation:
The question is missing. It is as follows:
Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64
Assume that the population of x values has an approximately normal distribution.
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)
75% Confidence Interval can be calculated using M±ME where
- M is the sample mean weight of the wild mountain lions (
)
- ME is the margin of error of the mean
And margin of error (ME) of the mean can be calculated using the formula
ME=
where
- t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
- s is the standard deviation of the sample(31.4)
Thus, ME=
≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5
I hope this helps you
when y=14 x=20
y=3,5 x=?
?.14=20.3,5
?=20.3,5/14
?=5