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
Kimm since the function is g(x) = 3(x-2) and we are given g(x)=6 That means: 6 = 3(x-2) Then we need to know what is x? 6 = 3x - 6 (distribute the 3) 12 = 3x (add 6 to both sides) 4 = x (divide both sides by 3)You can check it this way. Is g(4) = 3(4-2) = 6 ? 3*2 = 6
Answer:
500
Step-by-step explanation:
First, 4✘5^3
Then, 4✘125
Last, =500
Answer:
400,000
Step-by-step explanation:
In total, you would be paying 400,000 for 2 years for the interest.
So, for one year, the interest would be 200,000. The monthly rate would be 16,666.6667.
Answer:
Check the explanation
Step-by-step explanation:
We have:
Null Hypothesis:
H o:μ = 215
Alternate Hypothesis:
Η α :μ> 215 This is a one tailed test
Standardized value to be tested:
== 
-= 1.93
Since this is a one tailed test, critical t value which occurs at 95% confidence level is 1.943.
Since 1.93 < 1.943, we can say that At α = 0.05, there is insufficient evidence to indicate that the mean price of all digital cameras exceeds $215.00.
To make it sufficient, we need value more than 1.943.
At α = 0.10, critical t value is 1.440 (less than 1.93) which makes it sufficient.
