For the answer to the question above asking <span>what is the probability that a randomly selected student will be a boy that brings his lunch to school? I'll provide a solution for the following problem below.
</span>P(boy |lunch) = P(boy and lunch)/P(lunch) = 0.40
P(boy and lunch) = P(lunch)*0.4 = 0.3*0.4 = 0.12
The value of a would be 3.5, as a=1 when b=10 means that a is worth 1/10 of b's value.
Answer:
0.2839 = 28.93%
Step-by-step explanation:
In order to find find the probability that one car chosen at random will have less than 68.5 tons of coal, we have to find z
Le x be tons of coal in a car
Z = (x – mean)/standard deviation
Z = (x – μ)/ σ
P (x<68.5) = P(z < (68.5-69)/0.9)
P (x<68.5) = P(z < -0.5555)
Then we use the z table to find the area under the curve.
P (x<68.5) = 0.2893
First, convert R percent to r a decimal
r = R/100
r = 7%/100
r = 0.07 per year,
Then, solve our equation for A
A = P(1 + r/n)nt
A = 200.00(1 + 0.005833333/12)(12)(5)
A = $ 283.53
Summary:
The total amount accrued, principal plus interest,
from compound interest on an original principal of
$ 200.00 at a rate of 7% per year
compounded 12 times per year
over 5 years is $ 283.53.
