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Shtirlitz [24]
3 years ago
9

In a normal distribution, the mean is ____ it's mean and mode

Mathematics
1 answer:
Blizzard [7]3 years ago
6 0
In a normal distribution, the median (I think you meant to say) is equal to the mean and mode.
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Answer:

\frac{13}{3} ÷ (-\frac{5}{6})

Step-by-step explanation:

4\frac{1}{3}/(-\frac{5}{6})=\\\\\frac{13}{3}/(-\frac{5}{6})

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Area of a triangle is 1/2 x base x height:

Area = 1/2 x 8 x 20.5 = 82 square cm.

It is hone-half the area of a rectangle with dimensions 20.5 by 8 cm.

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Factorise x cubed - x
kirill [66]

Answer:

= x(x+1)(x-1)

Step-by-step explanation:

Given the expression x³ - x

On factorizing

x³ - x

= x(x²-1)

Using difference of two square

a² - b² = (a+b) (a-b)

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= (x+1)(x-1)

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7 0
3 years ago
Compute the standard deviation for the set of data.
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Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoti
Archy [21]

This question is incomplete, the complete question is;

Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen.

a) What is E[G] or range of G

b) Give the distribution over the random variable G

Answer:

a) E[G] is [ 0, 1 , 2 ]

b)

the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

Step-by-step explanation:

Given the data in the question;

Number to be chosen is 2

from 7 girls and 3 boys

a) range of G

number of girls can be; [ 0, 1 , 2 ]

Therefore, E[G] is [ 0, 1 , 2 ]

b) he distribution over the random variable G.

Distribution of the random variable G is Hypergeometric;

so

with P( G=g ) = P( getting g girls from 7 and 2-g boys from 3)

= ((^7C_g) × (^3C_{2-g)) / ^{10}C_2

now since, the range of G is [ 0, 1 , 2 ]

P( G=0 ) = ((^7C_0) × (^3C_{2)) / ^{10}C_2

= [(7!/(0!(7-0)!)) × (3!/(2!(3-2)!))] / (10!/(2!(10-2)!))

= [1 × 3] / 45 = 3/45 = 0.0667

P( G=1 ) = ((^7C_1) × (^3C_{1)) / ^{10}C_2

= [(7!/(1!(7-1)!)) × (3!/(1!(3-1)!))] / (10!/(2!(10-2)!))

= [ 7 × 3 ] / 45 = 21/45 = 0.4667

P( G=2 ) = ((^7C_2) × (^3C_{0)) / ^{10}C_2

= [(7!/(2!(7-2)!)) × (3!/(0!(3-0)!))] / (10!/(2!(10-2)!))

= [ 21 × 1 ] / 45 = 21/45 = 0.4667

Therefore, the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

3 0
3 years ago
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