Thank you, I will reward the braniest

A beekeeper has 65,000 female bees and 700 male bees in one colony. A disease kills 25% of the female bees and 14% of the male bees. After the disease hits the colony, the beekeeper decides to sell LaTeX: \frac{1}{2} 1 /2 of the remaining female bees to a friend.

How many total bees are left in the beekeeper’s colony after the disease and sale?

Answer:

24977 bees

Step-by-step explanation:

From the above question:

Number of female bees = 65,000

Number of male bees = 700

A disease kills 25% of the female bees and 14% of the male bees.

Number of female bees killed = 25% × 65,000

= 16,250 bees

Amount of female bees left = 65,000 - 16,250

= 48750 bees

Number of male bees killed = 14% × 700

= 98

Amount of male bees left = 700 - 98

= 602 bees

After the disease hits the colony, the beekeeper decides to sell LaTeX: \frac{1}{2} 1 /2 of the remaining female bees to a friend.

Hence:

1/2 × 48750 bees

= 24375 bees

Total bees are left in the beekeeper’s colony after the disease and sale is calculated as:

24375 female bees + 602 male bees

= 24977 bees

4 0

3 years ago

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

Please help will give brianliest

neonofarm [45]

All of these transformations are equivalent to reflection across the line y=x:

a reflection across the y-axis followed by a clockwise rotation 90° about the origin
a clockwise rotation 90° about the origin followed by a reflection across the x-axis
a counter-clockwise rotation 90° about the origin followed by a reflection across the y-axis
a reflection across the x-axis followed by a counter-clockwise rotation 90° about the origin

_____

These are all but the second choice.

... reflection across y: (x, y) ⇒ (-x, y); 90°CW: (-x, y) ⇒ (y, x)

... 90°CW: (x, y) ⇒ (y, -x); reflection across x: (y, -x) ⇒ (y, x)

... 90°CCW: (x, y) ⇒ (-y, x); reflection across y: (-y, x) ⇒ (y, x)

... reflection across x: (x, y) ⇒ (x, -y); 90°CCW: (x, -y) ⇒ (y, x)

Note that the second choice gives a different result:

... reflection across y: (x, y) ⇒ (-x, y); 90°CCW: (-x, y) ⇒ (-y, -x)

5 0

3 years ago

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Thank you, I will reward the braniest​

Thank you, I will reward the braniest