For points (x1,y1) and (x2,y2)
slope=(y1-y2)/(x1-x2)
(-4,-13)
(19,11)
x1=-4
y1=-13
x2=19
y2=11
slope=(-13-11)/(-4-19)=-24/-23=24/23
B is answer
A) the probability it is brown would be 50%; the probability it is yellow or blue would be 35%; the probability it is not green is 95%; the probability it is striped is 0%.
B) the probability of all brown would be 12.5%; the probability that the third one is the first red one drawn is 8.1%; the probability that none are yellow is 61.4%; the probability that at least one is green is 14.3%.
Explanation:
A) The probability that it is brown is the percentage of brown we have. Brown is not listed, so we subtract what we are given from 100%:
100-(15+10+20+5) = 100-(50) = 50%. The probability that one drawn is yellow or blue would be the two percentages added together: 15+20 = 35%. The probability that it is not green would be the percentage of green subtracted from 100: 100-5=95%. Since there are no striped candies listed, the probability is 0%.
B) Since we have an infinite supply of candy, we will treat these as independent events. All 3 being brown is found by taking the probability that one is brown and multiplying it 3 times:
0.5*0.5*0.5 = 0.125 = 12.5%.
To find the probability that the first one that is red is the third one drawn, we take the probability that it is NOT red, 100-10 = 90% = 0.9, for the first two, and the probability that it IS red, 10% = 0.1, for the last:
0.9*0.9*0.1 = 0.081 = 8.1%.
The probability that none are yellow is found by raising the probability that the first one is not yellow, 100-15=85%=0.85, to the third power:
0.85^3 = 0.614 = 61.4%.
The probability that at least one is green is computed by subtracting 1-(probability of no green). We first find the probability that all three are NOT green:
0.95^3 = 0.857375
1-0.857375 = 0.143 = 14.3%.
9514 1404 393
Answer:
-3/4, 3
Step-by-step explanation:
The zeros are the values of x that make h(x)=0. Those are the values of x that make the factors of h(x) be zero.
-4x -3 = 0 ⇒ x = -3/4
x -3 = 0 ⇒ x = 3
The zeros of the function are -3/4 and 3.
I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
