Binomial distribution formula: P(x) = (n k) p^k * (1 - p)^n - k
a) Probability that four parts are defective = 0.01374
P(4 defective) = (25 4) (0.04)^4 * (0.96)^21
P(4 defective) = 0.01374
b) Probability that at least one part is defective = 0.6396
Find the probability that 0 parts are defective and subtract that probability from 1.
P(0 defective) = (25 0) (0.04)^0 * (0.96)^25
P(0 defective) = 0.3604
1 - 0.3604 = 0.6396
c) Probability that 25 parts are defective = approximately 0
P(25 defective) = (25 25) (0.04)^25 * (0.96)^0
P(25 defective) = approximately 0
d) Probability that at most 1 part is defective = 0.7358
Find the probability that 0 and 1 parts are defective and add them together.
P(0 defective) = 0.3604 (from above)
P(1 defective) = (25 1) (0.04)^1 * (0.96)^24
P(1 defective) = 0.3754
P(at most 1 defective) = 0.3604 + 0.3754 = 0.7358
e) Mean = 1 | Standard Deviation = 0.9798
mean = n * p
mean = 25 * 0.04 = 1
stdev = 
stdev =
= 0.9798
Hope this helps!! :)
Answer:
$5.61
Step-by-step explanation:
Since David only has $9.39 but he needs $15.00 for his purchase. So, you would subtract the $9.39 from the $15.00 to see how much more money David needs to make this purchase.
15.00 - 9.39
When you carry out this problem, it equals $5.61.
Therefore, David needs $5.61 to be able to complete his purchase
Answer:
5.61×10^8 ton/day
Step-by-step explanation:
(2.08×10^5 ft³/s)·(8.64×10^4 s/day)·(6.24×10^1 lb/ft³)/(2×10^3 lb/ton)
= (2.08·8.64·6.24/2)×10^(5+4+1-3) ton/day
≈ 56.07×10^7 ton/day
≈ 5.61×10^8 ton/day
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We have expressed the answer using 3 significant digits, to match the precision of the rate and weight.
2(3x-4) = 5x+6
First, expand the bracket.
6x-8=5x+6
Next, move the variables to the same side.
x-8=6
Move the -8 to the other side, change to +
x=6+8
x=14
Charlie’s solution is incorrect. When expanding the bracket, he changed the - to a +.