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nika2105 [10]
3 years ago
8

Yuma is solving the inequality 3x−23≥−5.

Mathematics
2 answers:
Mademuasel [1]3 years ago
6 0

3x \geqslant 23 - 5 \\ 3x \geqslant 18 \\ x \geqslant 6 \\

ololo11 [35]3 years ago
5 0

Answer:      x≥ 6

Step-by-step explanation:

To solve this inequality

3x - 23 ≥ -5

First, what you need to do is to add 23 to both-side of the inequality, Hence;

3x  -23  +  23  ≥ -5 + 23

(-23 + 23  = 0,   also  -5 +  23  = 18)

3x   ≥   18

To get the value of x, we need to divide both-side of the inequality by 3

3x /  3       ≥   18/3

(On the left hand side of the equation, the three at the numerator will cancel out that of the denominator, then on the right hand side of the equation, 3 will divide 18 which will give 6)

x ≥    6

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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
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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

6 0
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antoniya [11.8K]

I hope this helps :)

Your welcome

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2 years ago
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