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3 years ago

Please help, seriously confused

Mathematics

1 answer:

Rufina [12.5K]3 years ago

6 0

Original height = original width = x mm

1. She made an enlarged copy...
height = 2x
width = 2x

2. She cut off a rectangle...
height = 2x
width = $\frac{2}{3} *2x= \frac{4}{3}x$

3.She doubled the width...
height = 2x
width = $2*\frac{4}{3}x = \frac{8}{3}x$

height * width = area

$2x * \frac{8}{3}x=139 \ 968 \\ \\ \frac{16}{3}x^2= 139 \ 968 \\\\ x^2=139 \ 968: \frac{16}{3}=139 \ 968 * \frac{3}{16}= 8 \ 748*3 = 26 \ 244 \\ \\ x= \sqrt{26 \ 244} = 162 \ mm$

Original height was 162 mm

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Answer:

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Part B:

Value of p=0.8

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Part C:

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Part D:

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Step-by-step explanation:

Part A:

Binomial Distribution is used because the number of tracks and the probability to find the intruding object is constant for all tracks

It means:

$P(X=x)=n_{C_{x}}*p^x*(1-p)^{n-x}$

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Part B:

Value of p=0.8

Value of n=20

Part C:

x=15

$P(X=15)=20_{C_{15}}*0.8^{15}*(1-0.8)^{20-15}\\$

By solving above Expression:

P(X=15)=0.17456

Part D:

P(X≥15)=P(X=15)+P(X=17)+P(X=16)+P(X=18)+P(X=19)+P(X=20)

$P(X=15)=20_{C_{15}}*0.8^{15}*(1-0.8)^{20-15}\\$

P(X=15)=0.17456

$P(X=16)=20_{C_{16}}*0.8^{16}*(1-0.8)^{20-16}\\P(X=16)=0.21819$

$P(X=17)=20_{C_{17}}*0.8^{17}*(1-0.8)^{20-17}\\P(X=17)=0.20536$

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$P(X=15)+P(X=17)+P(X=16)+P(X=18)+P(X=19)+P(X=20)\\=0.17456+0.21819+0.20536+0.13691+0.05765+0.01153\\$

P(X≥15)=0.80417

Part E:

Mean Distribution:

E(x)=n*p

E(x)=20*0.8

E(x)=16

E(x) tells us that out of 20 tracks the SBIRS will detect from 16 tracks.

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