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jeka94
3 years ago
15

If point C is not located between points A and B then AC+CB__ AB

Mathematics
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer:

\neq

Step-by-step explanation:

If Point C <em>was </em>located between A & B, then:

AC+CB=AB

However, we are told Point C is <em>not</em>; therefore:

AC+AB\neq AB

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Let x ∼ bin(9, 0.4). find
Kobotan [32]
A.
\mathbb P(X>6)=\mathbb P(X=7)+\mathbb P(X=8)+\mathbb P(X=9)
=\dbinom97(0.4)^7(1-0.4)^{9-7}+\dbinom98(0.4)^8(1-0.4)^{9-8}+\dbinom99(0.4)^9(1-0.4)^{9-9}
\approx0.025

b.
\mathbb P(X\ge2)=1-\mathbb P(X
=1-\dbinom90(0.4)^0(1-0.4)^{9-0}-\dbinom91(0.4)^1(1-0.4)^{9-1}
\approx0.929

c.
\mathbb P(2\le X
=\dbinom92(0.4)^2(1-0.4)^{9-2}+\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}
\approx0.663

d.
\mathbb P(2
=\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}+\dbinom95(0.4)^5(1-0.4)^{9-5}
\approx0.669

e.
\mathbb P(X=0)=\dbinom90(0.4)^0(1-0.4)^{9-0}\approx0.01

f.
\mathbb P(X=7)=\dbinom97(0.4)^7(1-0.4)^{9-7}\approx0.021

g, h.
For X\sim\mathcal B(n,p), recall that \mathbb &#10;E[X]=\mu_X=np and \mathbb V[X]={\sigma_X}^2=np(1-p). So

\mu_X=9(0.4)=3.6
{\sigma_X}^2=9(0.4)(0.6)=2.16
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Step-by-step explanation:

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Its equivalent percent is less than 100% and its equivalent decimal is less than 0.99 or less.

For example, 1/2 is a fraction between 0 and 1. Its percentage is 50%, which is less than 100%. Its decimal is 0.50, which is less than 1 or 0.99
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