Help me please. 20+(-16)4

1 answer:

8 0

P.E.M.D.A.S
First , you multiply.

200+(-16)4
200+(-64)

Since they have different signs, you subtract and keep the sign of the greater absolute value.

200+(-64)=136

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<img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B12x-7%3D0%20%5C%5C" id="TexFormula1" title="2x^{2} +12x-7=0 \\" alt="2x^{2}

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-2. 46

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2x²+12x-7=0
2x²+12x+18= 25
2*(x²+6x+9)=25
2*(x+6)²= 5²
x+6= 5√2/2 ⇒ x= 5√2/2 -6 ≈ -2. 46
x+6= -5√2/2 ⇒ x= -5√2/2 -6 ≈ - 9.54

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For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

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The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

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