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yarga [219]
3 years ago
6

Can someone help me with this?!?

Mathematics
2 answers:
Oksi-84 [34.3K]3 years ago
8 0
The answer is D, 9.
**
If c is 9, the points are (9,3) and (27,9).
See how dividing the first number by three gives the second number?
In other words, x is directly proportional to y.
This means the line goes through the origin.
y=x/3
0=0/3
lora16 [44]3 years ago
6 0

Answer:

D] 9

Step-by-step explanation:

the answer is would be 9...

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area=length*width

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Ashlyn is selling boxes of cookies for a fundraiser. If she raised $26 selling eight boxes, how much will she raise if she sells
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PLEASE HELP ASAP THERE ARE 20 POINTS ON THE LINE! Sierra has to pay 7% taxes on her purchase at IKEA. If the tax is $38.36, what
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$31.36

Step-by-step explanation:

If she has $38.36 and you subtract 7 you would get the answer $31.36. I hope I helpped (◡ ω ◡)

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3 years ago
32​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
Tems11 [23]

Answer:

a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

c) P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.32)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

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One-thousand five-hundred and sixty divided by twenty four is sixty five

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