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2 years ago

5

Could you help me solve?

1 answer:

Ilya [14]2 years ago

8 0

Answer:

1.14200738982×10^26

Step-by-step explanation:

Substitution can make this integral much easier to evaluate.

<h3>Substitution</h3>

Let u = 7x² -x. Then du = (14x -1)dx. The limits on x become different limits for u:

for x = 1: u = 7(1²) -1 = 6

for x = 3: u = 7(3²) -3 = 60

<h3>Integral</h3>

$\displaystyle\int_1^3{(14x-1)e^{(7x^2-x)}}\,dx=\int_6^{60}{e^u}\,du=\left.e^u\right|^{60}_6\\\\=e^{60}-e^6\approx e^{60}\approx\boxed{1.14200738982\times10^{26}}$

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4 years ago

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Alex17521 [72]

Here is the full question:

$The \ shell \ of \ the \ land \ snail \ Limocolaria \ martensiana \ has \ two \ possible \ color$ $forms: \ streaked \ and \ pallid. \ In \ a \ certain \ population \ of \ these \ snails, \ 60\% \ o f \ the$ $individuals \ have \ streaked \ s hells.\ 16 \ Suppose \ that \ a \ random \ sample \ of \ 10 \ snails$ $is \ to \ be \ chosen \ from \ this \ population. \ F ind \ the \ probabilit y \ that \ the \ percentage \ of$

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Answer:

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(b) 0.2510

(c) 0.2150

Step-by-step explanation:

Given that:

The sample size = 10

Sample proportion= 60% 0.6

Let X represents the no of streaked-shell snails.

$X \sim Binom (n =1 0, p = 0.60)$

The Probability mass function (X) is:

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= 6

Standard deviation σ = $\sqrt{np(1-p)}$

= $\sqrt{10*0.6*(1-0.6)}$

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The probability that the percentage of the streaked-shelled snails in the sample will be:

a)

$P(X = 0.5) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_5 * (0.6)^5(1-0.6)^{10-5}$

$= \dfrac{10!}{5!(10-5)!} * (0.6)^5(1-0.6)^{10-5}$

= 0.2007

b)

$P(X = 0.6) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_6 * (0.6)^6(1-0.6)^{10-6}$

$= \dfrac{10!}{6!(10-6)!} * (0.6)^6(1-0.6)^{10-6}$

= 0.2510

c)

$P(X = 0.7) = ^nC_x p^x (1 -p) ^{(n-x)}$

$= ^{10}^C_7 * (0.6)^7(1-0.6)^{10-7}$

$= \dfrac{10!}{7!(10-7)!} * (0.6)^7(1-0.6)^{10-7}$

= 0.2150

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