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saul85 [17]
3 years ago
12

PLS ANSWER THIS AS SOON AS POSSIBLE I RlY NEED HELP WITH IT AND IDK WHICH IS THE RIGHT ANSWER PLS HELP. ps the lines on graph yo

ur able to change to find the answer. And the photos are the same thing its just i couldnt fit it in one photo

Mathematics
1 answer:
Mariana [72]3 years ago
7 0

Answer:B

Step-by-step explanation:

Thats right

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Adrian's annual income changes every year because of the following three factors: On average, his salary is 1.2 times the previo
olya-2409 [2.1K]
In this question, Adrian initially has $52,000 and his salary is 1.2 times the previous year's salary. This means the income should look like this:52000*1.2^n 

30% of his income is budgeted for rent, then the equation should look like this: 30% * income= 30%*(52000*1.2^n)

Adrian's income increases by $2,300 each year as a result of gifts will make this equation:2300n

if you put the 3 equation together you will get
f(n)= 52000*1.2^n - 0.3(52000*1.2^n) +2300n
f(n)= 36400*1.2^n +2300n
7 0
3 years ago
It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other.
zheka24 [161]

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider  some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

P(X=x)=nCx*p^{x}*(1-p)^{(n-x)} with x = 0, 1, 2, 3,…, n

Where

n: number of independent trials

p: success probability  

x: number of successes in n trials

In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)} with x = 0, 1

So,

P(A)=P(X=0)+P(X=1)

P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}

P(A)=0.99^{10}+10*0.01*0.99^{9}

P(A)=0.9957

Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:

3*(1-P(A))*P(A)*P(A) = (1-0.9957)*0.9957*0.9957=0.0127

Finally, we have that the probability of returning exactly one of the three packages is 1.27%

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Step-by-step explanation:

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