This problem is a combination of the Poisson distribution and binomial distribution.
First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961
For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
=6.18101*10^(-5) or approximately
=0.00006181
Answer:
<h3>Figure 1</h3>
- Perimeter of base = 5 + 5 + 8 = 18 ft
- Base area = 1/2(8)(3) = 12 ft²
<u>Surface area:</u>
- S = 18*7 + 2*12 = 150 ft²
<h3>Figure 2</h3>
<u>Surface area of cube:</u>
- S = 6a² = 6(2.5)² = 37.5 m²
<u>Surface area of prism:</u>
- S = 2(11 + 9)(7) = 280 m²
<u>Overlapping area:</u>
<u>Surface area of composite figure:</u>
- S = 280 + 37.5 - 2(6.25) = 305 m²
(-$11) x 12. That would be the equation because there are 12 months, so you would multiply -11 from 12 to get -132 every year for the tennis club membership.
The formula for a circumference of a circle is 2πr
2 × π × 25 = 157
so your answer is A 157in.
I hope I've helped!
Solution :
To claim to be tested is whether "the mean salary is higher than 48,734".
i.e. μ > 48,734
Therefore the null and the alternative hypothesis are

and 
Here, n = 50

s = 3600
We take , α = 0.05
The test statistics t is given by


t = 2.15
Now the ">" sign in the
sign indicates that the right tailed test
Now degree of freedom, df = n - 1
= 50 - 1
= 49
Therefore, the p value = 0.02
The observed p value is less than α = 0.05, therefore we reject
. Hence the mean salary that the accounting graduates are offered from the university is more than the average salary of 48,734 dollar.