<h2>
Therefore he took 40 gram of 
type solution and 10 gram of 
type solution.</h2>
Step-by-step explanation:
Given that , A pharmacist 13% alcohol solution another 18% alcohol solution .
Let he took x gram solution of type solution
and he took (50-x) gram of type solution.
Total amount of alcohol =![[x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ]](https://tex.z-dn.net/?f=%5Bx%5Ctimes%5Cfrac%7B13%7D%7B100%7D%5D%20%2B%5B%2850%20-x%29%20%5Ctimes%5Cfrac%7B18%7D%7B100%7D%20%5D)
gram
Total amount of solution = 50 gram
According to problem
⇔![\frac{ [x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ]}{50}= \frac{14}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Bx%5Ctimes%5Cfrac%7B13%7D%7B100%7D%5D%20%2B%5B%2850%20-x%29%20%5Ctimes%5Cfrac%7B18%7D%7B100%7D%20%5D%7D%7B50%7D%3D%20%5Cfrac%7B14%7D%7B100%7D)
⇔
⇔- 5x= 700 - 900
⇔5x = 200
⇔x = 40 gram
Therefore he took 40 gram of type solution and (50 -40)gram = 10 gram of type solution.
1 file + 3 pens = $32.85 --------------- (1)
2 files + 8 pens = $83.50 ----------------(2)
(1) x 2 :
2 files + 6 pens = $65.70 ----------------(1a)
(2) - (1a) :
2 pens = $17.80
1 pen = $8.90
-------------------------------------------------------------------------------
Answer: One pen costs $8.90.
-------------------------------------------------------------------------------
(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.
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Answer:
<h2>14mph</h2>
Step-by-step explanation:
Given the gas mileage for a certain vehicle modeled by the equation m=−0.05x²+3.5x−49 where x is the speed of the vehicle in mph. In order to determine the speed(s) at which the car gets 9 mpg, we will substitute the value of m = 9 into the modeled equation and calculate x as shown;
m = −0.05x²+3.5x−49
when m= 9
9 = −0.05x²+3.5x−49
−0.05x²+3.5x−49 = 9
0.05x²-3.5x+49 = -9
Multiplying through by 100
5x²+350x−4900 = 900
Dividing through by 5;
x²+70x−980 = 180
x²+70x−980 - 180 = 0
x²+70x−1160 = 0
Using the general formula to get x;
a = 1, b = 70, c = -1160
x = -70±√70²-4(1)(-1160)/2
x = -70±√4900+4640)/2
x = -70±(√4900+4640)/2
x = -70±√9540/2
x = -70±97.7/2
x = -70+97.7/2
x = 27.7/2
x = 13.85mph
x ≈ 14 mph
Hence, the speed(s) at which the car gets 9 mpg to the nearest mph is 14mph
