Answer:
3/4
Step-by-step explanation:
A: students present
B: students on time
P(all students present and on time) = P(A and B) = 3/10
P(all students present) = P(A) = 2/5
P(A and B) = P(A).P(B|A) where P(B|A) is the probability of everyone being on time given that everyone is present
So P(B|A) = P(A and B) /P(A) = 3/10 ÷ 2/5 = 3/10 * 5/2 = 15/20 which can be reduced to 3/4 by dividing numerator and denominator by 5
Answer:
In AMNO, n = 92 inches, N=119° and 20=20°. Find the length of m, to the nearest 10th of an inch In AJKL, j = 9.6 inches, k = 6.4 inches and I=8.4 inches. Find the measure of K to thelaearest degree. In AQRS. q = 49 cm, 7.1 cm and S=70°. Find the area of AQRS, to the nearest 10th of a square centimeter Find all angles, 0 = 0 < 360°, that satisfy the equation below, to the nearest 10th of a degree. 25 sino-1=0 Find all angles, 0 SC <360°, that satisfy the equation below, to the nearest tenth of a degree (if necessary). 12 sin C + 6 = 5 sin ( + 6
Step-by-step explanation:
Answer:
4.5
Step-by-step explanation:
4x+6 = 6x-3
3+6 = 6x-4x
9 = 2x
X =4.5
Total tickets sold = 159
Total sales = $1100.60
Child admission = $5.20
Adult admission = $8.90
Assumed all 159 tickets are Child admission tickets
Total sales = 159 x 5.2 = $826.80
Difference in amount = $1100.60 - $826.80 = $273.80
The difference must be contributed by the Adult Admission Tickets, which has a difference of $8.90 - $5.20 = $3.70
$273.80 ÷ $3.70 = 74 adult tickets
159 - 74 = 85 child tickets
Answer:
it's 10 cemetery in a two way too far and far from a lot for
