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Temka [501]
3 years ago
15

Given the graph of the function f(x) below what happens to f(x) when x is a number between 0 and 1

Mathematics
2 answers:
GenaCL600 [577]3 years ago
5 0

Answer:

The answer would be D. Or F(x) is a negative number with a large absolute value.

Step-by-step explanation:

Apex

Andrews [41]3 years ago
4 0

Answer: correct answer is D.


Step-by-step explanation:


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Option D

Step-by-step explanation:

Option A represents the circle

Option B represents the parabola

Option C represents the ellipse

Option D represents the line

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What property says that -12+12+0
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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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(x +4) * (x -1)
****************************************************


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