Answer:
(b) 
Step-by-step explanation:
When two p and q events are independent then, by definition:
P (p and q) = P (p) * P (q)
Then, if q and r are independent events then:
P(q and r) = P(q)*P(r) = 1/4*1/5
P(q and r) = 1/20
P(q and r) = 0.05
In the question that is shown in the attached image, we have two separate urns. The amount of white balls that we take in the first urn does not affect the amount of white balls we could get in the second urn. This means that both events are independent.
In the first ballot box there are 9 balls, 3 white and 6 yellow.
Then the probability of obtaining a white ball from the first ballot box is:

In the second ballot box there are 10 balls, 7 white and 3 yellow.
Then the probability of obtaining a white ball from the second ballot box is:

We want to know the probability of obtaining a white ball in both urns. This is: P(
and
)
As the events are independent:
P(
and
) = P (
) * P (
)
P(
and
) = 
P(
and
) = 
Finally the correct option is (b) 
If you start with a 12x16 rectangle and cut square with side length x, when you bend the sides you'll have an inner rectangle with sides
and
, and a height of x.
So, the volume will be given by the product of the dimensions, i.e.

The derivative of this function is

and it equals zero if and only if

If we evaluate the volume function at these points, we have

So, the maximum volume is given if you cut a square with side length

Answer:
7 Decenas = 70 Unidades espero que te ayude
Step-by-step explanation:
Answer:
x=116°
Step-by-step explanation:
180° - 64° = 116°
Answer:
almost 0%
Step-by-step explanation:
Given that for an insurance company with 10000 automobile policy holders, the expected yearly claim per policyholder is $240 with a standard deaviation of 800
using normal approximation, the probability that the total yearly claim exceeds $2.7 million is calculated as follows:
Sea sumatoria de x = SUMX, tenemos que:


= P (z => 3.75)
= 1 - P ( z < 3.75)
P = 1 - 0.999912
P = 0.000088
Which means that the probability is almost 0%