C it’s I think lol!!!!!^^*
Answer:
One number is 15 and 3 less than that is 12 so that is your other number
Using the z-table, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.
For a normally distributed set of data, given the mean and standard deviation, the probability can be determined by solving the z-score and using the z-table.
First, solve for the z-score using the formula below.
z-score = (x – μ) / σ
where x = individual data value = 100
μ = mean = 125
σ = standard deviation = 18
z-score = (100 – 125) / 18
z-score = (-25) / 18
z-score = -1.39
Find the probability that corresponds to the z-score in the z-table. (see attached images)
-1.39 - (-1.3) : -1.4 - (-1.3) = x - 0.0968 : 0.0808 - 0.0968
-0.09 : -0.1 = x - 0.0968 : -0.016
x - 0.0968 = -0.09(-0.016)/-0.1
x = -0.0144 + 0.0968
x = 0.0824
x = 0.0824
Hence, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.
Learn more about probability here: brainly.com/question/26822684
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Answer:

Step-by-step explanation:
let
denote grams of
formed in
mins.
For
of
we have:
of A and
of B
Amounts of A,B remaining at any given time is expressed as:
of A and
of B
Rate at which C is formed satisfies:

Apply the initial condition,
,to the expression above

Now at
:

Substitute in X(t) to get

Answer:
a

b

Step-by-step explanation:
From the question we are told that
The proportion that has outstanding balance is p = 0.20
The sample size is n = 15
Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card

Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as

=> 
Here C stand for combination
=>
Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as
![P(X \le 4) = [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]](https://tex.z-dn.net/?f=P%28X%20%5Cle%204%29%20%3D%20%20%5B%20%5E%7B15%7DC_0%20%2A%20%280.20%29%5E0%20%2A%20%281%20-%200.20%29%5E%7B15-0%7D%5D%2B%5B%20%5E%7B15%7DC_1%20%2A%20%280.20%29%5E1%20%2A%20%281%20-%200.20%29%5E%7B15-1%7D%5D%2B%5Ccdots%2B%5B%20%5E%7B15%7DC_4%20%2A%20%280.20%29%5E4%20%2A%20%281%20-%200.20%29%5E%7B15-4%7D%5D)
=> 