<span>Using the information we have
3x+4=40
Do the same to each side of the equation to eliminate for x.
3x+4=40 Minus 4 from each side
3x=40-4
3x=36
Divide 3 from each side
x=36/3
x=12
AC=3x+4
insert the value of x
3(12)+4=40
AC=40
AD=20</span>
Answer:
6x + 3.25y = 42.50
x + y = 8
Step-by-step explanation:
x is the amount of drinks, and y represents the amount of candies that are bought. The first equation is used to figure out the amount of candies and drinks needed to be bought to add up to 42.50, and the second equation is used to make sure that the quantity of drinks and candies add up to 8.
This is the following condition in order to get the specific output for this specific problem: if is_a_prime(n):<span> is_prime = True</span> <span><span>Now all you have to do is write is_a_prime().
For the hard code for this problem:
</span>if n == 2:<span>
is_prime = True
elif n % 2 == 0:
is_prime = False
else:
is_prime = True
for m in range (3, int (n * 0.5) + 1, 2):
if n % m == 0:
is_prime = False
<span>break.</span></span></span>
<span>
To add, a high-level programming language that is widely used for general-purpose programming<span>, created by Guido van Rossum and first released in 1991 is called Python.</span></span>
(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.
