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Olin [163]
2 years ago
12

In the equation above k is a constant if y = 2 when x= 5 what is the value of x when y= 5

Mathematics
1 answer:
Tresset [83]2 years ago
6 0

The value f x when y= 5 is 2

<h3>Variations </h3>

Let the given equation be y = k/x

where k is the constant

if y = 2 when x= 5, then;

k = xy

k= 2(5)

k = 10

In order to determine the value of x when y = 5

x = k/y

x = 10/5

x = 2

Hence the value f x when y= 5 is 2

Learn more on variation here: brainly.com/question/6499629

#SPJ1

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45% of what number is 300?
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Answer:

666  2/3

Step-by-step explanation:

45 %  *  x  = 300

.45 x = 300

x = 300 / .45 = 666  2/3

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3 years ago
What is <br> 35g−13=−103<br> g=
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g= 13/138

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8 0
3 years ago
A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience
bulgar [2K]

Answer:

a) P(X \geq 1) = 1-P(X

P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}

And replacing we got:

P(X \geq 1) =1 -1x10^{-18} \approx 1

b) P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355

P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And adding we got:

P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992

c) P(X \leq 8) =1 -P(X>8) = 1-P(X=9)

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And replacing we got:

P(X \leq 8)= 1-0.913517=0.086483

Step-by-step explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

X \sim Binom(n=9, p=0.99)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

We want to find this probability:

P(X \geq 1) = 1-P(X

P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}

And replacing we got:

P(X \geq 1) =1 -1x10^{-18} \approx 1

Part b

P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)

P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355

P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And adding we got:

P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992

Part c

P(X \leq 8) =1 -P(X>8) = 1-P(X=9)

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And replacing we got:

P(X \leq 8)= 1-0.913517=0.086483

4 0
3 years ago
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