Answer:
6 * 3/4 = 9 2 = 412 = 4.5
Step-by-ste
6 * 3/4 = 6/1 3*4 = 18/4 = 9/2
2 · 2 = 9/2
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(18, 4) = 2. In the next intermediate step cancelling by a common factor of 2 gives 9
2
.
In words - six multiplied by three quarters = nine halfs.
Answer:
(3,20)
Step-by-step explanation:
so O must be 0,0 and A is a midpoint of OB
This means B must be (14,12)
B is the midpoint of TS.
This means that the distance of SB is equal to the distance of BT.
This must mean that the coordinate of T must be (3,20)
O: (0,0) x coordinate increases by 7. y coordinate increases by 6.
A: (7,6)
B: (14,12)
S: (25,4) x coordinate decrease by 11. y coordinate increases by 8.
B: (14,12)
T: (3,20)
<span>When you are sampling from a small finite lot, the hypergeometric distribution applies. The binomial is a poor approximation in this case.
The general equation for the hypergeometric where aCx means the number of combinations of a items selected x at-a-time.:
P(x) =[(aCx)(N-aCn-x)]/NCn
Where
N is the lot size = 20
a is the number of defectives in the lot = 3.
x is the number of defectives in the sample.
n is the sample size = 2.
A. The probability that the first item is defective is
P(x=1) = [(3C1)(17C1)]/(20C2)
= (3)(17)/190 = 0.268
The probability that the second item is defective is
P(x = 1) = [(2C1)(17C1)]/(19C2) = (2)(17)/171 = 0.199.
So the total probability is (0.268)(0.199) = 0.0532
B. The probability that the first item is good is:
P(x = 0) = (3C0)(16C2)]/20C2 = (1)(120)/190 = 0.632
The probability that the second item is defective is
P(x = 0) =[(3C0)(16C2)]/19C2
= (1)(120)/171 = 0.670.
The total probability is 0.632(0.670) = 0.4234</span>
1/2*5 = 2.5 ~ 3 (round-off)
1/3* 10 = 3.33 ~ 3
