I can give you the first half!
1. D2 - hours worked on monday, 3hrs
2. E2 - money earned on monday, $15
3. D3 - hours worked on tuesday, 3 hrs
4. E3 - money earned on tuesday, $15
5. D4 - hours worked on thursday, 4 hrs
6. E4 - money earned on thursday, $20
Answer:
D. (–3, –6) and (5, 10)
Step-by-step explanation:
The system has equations:
...(1)
and
...(2)
Equate both equations;
Rewrite in standard form:
We factor to obtain:
By the zero product principle;
When x=-3, y=2(-3)=-6
This yields the ordered pair (-3,-6).
When x=5, y=2(5)=10
This yields the ordered pair (5,10).
The correct choice is D.
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:
Step-by-step explanation:
52.5 mi(8 km/5mi) = 84 km
