The difference between two numbers is 9 and the product of the numbers is 162. Find the two numbers

1 answer:

7 0

Let the no be X and Y

acc to ques....

x-y=9 .........1

xy=162 ..........2

substituting value from 1 in 2 we get;

x=9+y

[9+y][y] = 162

y^2+9y = 162

y^2 + 9y - 162=0

y^2 + 18y - 9y - 162=0

y[y+18] + 9[y+18]=0

[y+9][y+18}

y= -9.................................3

y= -18......................................4

case 1 :

y= -9

x = 9-9=0

case 2:

y= -18

x= 9-18 = -9

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Which set of population data is the least dispersed from its mean? 2, 3, 2, 9 4, 0, 4, 0 6, 2, 2, 2 9, 3, 5, 3.

exis [7]

The set of data 6, 2, 2, 2 will have the least dispersion from its mean.

From four sets of data, we take the mean of 6,2,2,2

$\rm mean =\dfrac{ Sum \ of\ data}{Number \ of\ data}$

So the mean will be

$\rm mean =\dfrac{ 6+2+2+2}{4}=\dfrac{12}{4} =3$

So the mean of the data (6,2,2,2) is 3 which has the least dispersion from its every data as compared to the other data

Thus the set of data 6, 2, 2, 2 will have the least dispersion from their mean.

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2 years ago

Suppose that 10% of all homeowners in an earthquake-prone area of California are insured against earthquake damage. Four homeown

guapka [62]

Remainder of question:

Find the probability distribution of x

Answer:

The random variable x is defined as: X = {0, 1, 2, 3, 4}

The probability distribution of X:

P(X = 0) = 0.656

P(X = 1) = 0.2916

P(X= 2) = 0.0486

P(X=3) = 0.0036

P(X = 4) = 0.0001

Step-by-step explanation:

Sample size, n = 4

Random variable, X = {0, 1, 2, 3, 4}

10% (0.1) of the homeowners are insured against earthquake, p = 0.1

Proportion of homeowners who are not insured against earthquake, q = 1 - 0.1

q = 0.9

Probability distribution of x,

$P(X = r) = ^nC_r *p^r q^{n-r} \\\\P(X= 0) =(^4C_0 *p^1 q^4 )\\P(X=0) = (^4C_0 *0.1^0 0.9^4 ) = 0.656\\P(X= 1)= (^4C_1 *p^1 q^3 )\\P(X=1) = (^4C_1 *0.1^1 0.9^3 ) = 0.2916\\P(X= 2)=( ^4C_2 *p^2 q^2) \\P(X=2) = (^4C_2 *0.1^2 0.9^2 ) = 0.0486\\P(X= 3) = (^4C_3 *p^3 q^3) \\ P(X=3) = (^4C_3 *0.1^3 0.9^1 ) = 0.0036\\P(X= 4) = (^4C_4 *p^4 q^0 )\\ P(X=4) =(^4C_4 *0.1^4 0.9^0 ) = 0.0001$