Given :
Brianna is a waitress in an upscale restaurant. On Friday she waited on 12 groups of customers who spent an average of $186 per group.
Her tips averaged 18% of the total amount spent by all the groups of customers that day.
To Find :
Brianna's tip income earnings on Friday.
Solution :
Total amount of money spent by 12 groups :
P = $ 12×186 = $2232 .
She got tip of 18% of the total amount spent by all the groups of customers.
Brianna's tip income earnings on Friday is $446.4 .
Hence, this is the required solution.
Step-by-step explanation:
0.4 x + 3.9 = 5.7
hope 0.4 x = 5.78 - 3.9
0.4x = 1.88
x = 1.88/0.4
x = 4.7
Answer:
its c because parallel lines are lines that are always the same distance apart all of these are the same distance apart
hoped this helped let me know if it did
Answer:
The p value for this case would be given by:
For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42
Step-by-step explanation:
Information given
n=1045 represent the random sample selected
X=502 represent the college graduates with a mentor
estimated proportion of college graduates with a mentor
is the value that we want to test
z would represent the statistic
represent the p value
Hypothesis to test
We want to test if the true proportion is higher than 0.42, the system of hypothesis are.:
Null hypothesis:
Alternative hypothesis:
The statistic is given by:
(1)
Replacing the info we got:
The p value for this case would be given by:
For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42
Answer:
0.0143
Step-by-step explanation:
In this question, we are asked to use the binomial distribution to calculate the probability that 10 or fewer passengers from a sample of MIT data project sample were on American airline flights.
We proceed as follows;
The probability that a passenger was an American flight is 15.5%= 15.55/100 = 0.155
Let’s call this probability p
The probability that he/she isn’t on the flight, let’s call this q
q =1 - p= 0.845
Sample size, n = 155
P(X < A) = P(Z < (A - mean)/standard deviation)
Mean = np
= 125 x 0.155
= 19.375
Standard deviation = √npq
= √ (125 x 0.155x 0.845)
= 4.0462
P(10 or fewer passengers were on American Airline flights) = P(X \leq 10)
= P(Z < (10.5 - 19.375)/4.0462)
= P(Z < -2.19)
= 0.0143
