38 < 4x + 3 +7 - 3x (original equation)
38 < 4x - 3x + 3 + 7 (combine like terms)
38 < x + 10 (simplify)
38 - 10 < x + 10 - 10 (subtract 10 from both sides to get (x) alone)
28 < x (simplify)
x > 28 (switch to get x on the left (its proper equation writing >.<) )
Answer:
the price of adult ticket and student ticket be $10.50 and $8.25 respectively
Step-by-step explanation:
Let us assume the price of adult ticket be x
And, the price of student ticket be y
Now according to the question
2x + 2y = $37.50
x + 3y = $35.25
x = $35.25 - 3y
Put the value of x in the first equation
2($35.25 - 3y) + 2y = $37.50
$70.50 - 6y + 2y = $37.50
-4y = $37.50 - $70.50
-4y = -$33
y = $8.25
Now x = $35.25 - 3($8.25)
= $10.5
Hence, the price of adult ticket and student ticket be $10.50 and $8.25 respectively
Area=πr^2
find the area then divide by 8
we know that diameter=2radius or diameter/2=radius so
12=diameter
12/2=radius=6
subsitute
area=π(6)^2
area=36π
divide 36π by 8
36π/8=18π/4=9π/2π=4.5π
area of one section is 4.5π square feet or if we aprox pito 3.14159 then we get
4.5(3.14159)=area=14.1372 square feet or 14.14 square feet
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
