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Alenkasestr [34]
3 years ago
7

Evaluate -x + 2.7 for x =0.9

Mathematics
2 answers:
HACTEHA [7]3 years ago
6 0
Hey!

x = 0.9
-x = -0.9

-0.9 + 2.7 ⇒ 2.7 - 0.9 

2.7 - 0.9 = 1.8

Answer ⇒ 1.8

Hope this helps! :)
DiKsa [7]3 years ago
3 0
Plug in 0.9 for x

-(x) + 2.7

- (0.9) + 2.7

simplify

2.7 - 0.9 = 1.8

1.8 is your answer

hope this helps
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Re write the statement 19+7 divided by 2-5 by including one pair of brackets to make the total equal to 8
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The solution is given below:

Step-by-step explanation:

The computation is shown below:

= 19 +  7 divided by 2 - 5

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The figure below shows a quadrilateral ABCD. Sides AB and DC are congruent and parallel:
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SAS postulate

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What is the value of x when 9x=45
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Read 2 more answers
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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

6 0
3 years ago
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