The 300th customer would be the first to recieve both. They would have given away 5 of the 20$ gift cards and 12 of the 10$ ones for a total of 17 gift cards.
Answer:
1.06 x
(3sf)
Step-by-step explanation:
using compound interest formula A = P
from 95 to 08 is 13 yrs
Total Revenue = (5.02 x
)
= 1.057 x 
Answer:
The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.
Step-by-step explanation:
With the weekly average we can estimate the daily average for customers, assuming 7 days a week:

We can model this situation with a Poisson distribution, with parameter λ=108. But because the number of events is large, we use the normal aproximation:

Then we can calculate the z value for x=100:

Now we calculate the probability of x>100 as:

The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.
<h2>Answer</h2>
0.43
<h2>Explanation</h2>
Remember that 
Since the problem is telling us "Among tenth graders", we must focus on the 10th graders row only. From the row, we can infer that the frequency is the number of 10th graders who prefer going to sporting events, so
. Now, the sum of all frequencies will be the sum of all the 10th graders, so
. Let's replace the values:



And rounded to the nearest hundredth:

Answer:
p-value (0.0208) is less than alpha = 0.05 reject H0.
Step-by-step explanation:
we have the following data:
sample size = n = 75
x, the number to evaluate is 45
the sample proportion would be: x / n = 45/75
p * = 0.6
Now, the null and alternative hypotheses are:
H0: P = 0.72
Ha: P no 72
two tailed test
statistic tes = z = (p * - p) / [(p * (1-p) / n)] ^ (1/2)
replacing we have:
z = (0.6 - 0.72) / [(0.72 * (1-0.72) / 75)] ^ (1/2)
z = -2.31
p-vaule = 2 * p (z <-2.31)
using z table, we get:
p-vaule = 2 * (0.0104)
p-vaule = 0.0208
Therefore, p-value (0.0208) is less than alpha = 0.05 reject H0.