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Artist 52 [7]
4 years ago
9

What is -7(x+9)=9(x-5)-14x equal to in math

Mathematics
2 answers:
tester [92]4 years ago
8 0

Answer:

x = -9

Step-by-step explanation:

Distribute the -7 to (x + 9) and distribute 9 to (x - 5), creating the equation, -7x - 63 = 9x - 45 -14x. Then you add the like terms, 9x - 14x, which equals -5x. So the equation is now -7x - 63 = -5x -45. Add the -5 to both sides which makes -2x -63 = -45. Then add the -63 to both sides, which equals 18, then divide that by -2 and you get -9 as your answer.

Alik [6]4 years ago
3 0

____________________________________________________

Answer:

x = -9

____________________________________________________

Step-by-step explanation:

Solve for x in the equation: -7(x+9)=9(x-5)-14x

-7(x+9)=9(x-5)-14x\\\\\text{Distribute the -7}\\\\-7x-63=9(x-5)-14x\\\\\text{Distribute the 9}\\\\-7x-63=9(x-5)-14x\\\\-7x-63=9x-45-14x\\\\-7x-63=-5x-45\\\\\text{Add 63 to both sides}\\\\-7x=-5x+18\\\\\text{Add 5x to both sides}\\\\-2x=18\\\\\text{Divide by -2}\\\\x = -9

____________________________________________________

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10 snails is to be chosen from this population. Find the probability that the percentage of streaked-shelled snails in the sampl
Alex17521 [72]

Here is the full question:

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

streaked-shelled  \ snails   \ in  \ t he  \ sample \  will  \ be

(a) 50\%. \  (b) 60\%.  \ (c) 70\%.

Answer:

(a) 0.2007

(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:

P(X =x)= (^n_x) p^x(1-p)^{n-x}; x = 0,1,2,3...

The Binomial probability with mean μ = np

= 10 * 0.6

= 6

Standard deviation σ = \sqrt{np(1-p)}

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

= 1.55

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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