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2 years ago

Help solving this math

Mathematics

1 answer:

VMariaS [17]2 years ago

3 0

Answer:

(1) 9n = 56

n = 6

(2) 10 + 4y = 38

y = 7

(3) 3x + 8x = 55

x = 5

(4) 6m + 20 = 8

m= -2

(5) 17 - 3a = 23

a = -2

(6) 10x + 4 = 6x + 16

x = 3

(7) 4(x - 2) + x = 17

x = 5

(8) 2(x + 3) + 3x = x + 22

x = 4

(9) D. 10

(10) -2(3x - 2.4) = 3(3x - 2.4)

im not sure about this equation

Hope this helps

-Unknown

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Answer: The required answers are

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(b) the probability that X greater or equal than 5

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(a) The probability of getting 5 heads is given by

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(b) The probability of getting 5 or more than 5 heads is

$P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.$

(c) Proceeding as in parts (a) and (b), we see that

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