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liraira [26]
3 years ago
8

1. Find the value of x, rounded to the nearest tenth A. 3.7 B. 4.4 C. 8.9 D. 4.2

Mathematics
1 answer:
Yuki888 [10]3 years ago
8 0

Answer:

C 8.9

Step-by-step explanation:

all these lines (from the same point of origin, cutting into and through a circle, ending at the far end of the circle they are cutting into) have the same relation of their lengths :

a×(a+b) is the same for all these lines.

with a being the first segment from the point of origin to touching the outside of the circle, and b being the length of the line inside the circle.

so,

4×(4+16) = 8×(8+2) = 80

this is correct for these two lines.

now, the line with the length x is a special case, as it does not really go through a circle, it only touches them.

but the length relation applies also to this line. it is just that the length inside the circle is 0.

so,

x×(x+0) = 80 (the same value as for the other lines)

must be true here too.

=>

x² = 80

x = sqrt(80) ≈ 8.9

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What the answer? 8x-(2x-3)=12
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Answer:

x= 3/2 or 1.5

Step-by-step explanation:

First of all, you can take out the parenthesis because 8x is subtracting 2x-3.

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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

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