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Karo-lina-s [1.5K]

1 year ago

15

a classroom of children has 13 boys and 17 girls in which five students are chosen to do presentations. in how many ways can the

five students be chosen so that more boys than girls are selected? a) 40,183 b) 1,287 c) 51,051 d) 52,338 e) 12,155 f) none of the above.

1 answer:

vazorg [7]1 year ago

4 0

Answer:

52,338

Step-by-step explanation:

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What is the difference in length between the longest insect and the shortest insect centimeters

Dmitry [639]

Can't answer this question coz there is no numbers or pictures

5 0

3 years ago

M∠A = 75 + 5x

Firlakuza [10]

Answer:

See below

Step-by-step explanation:

For a parallelogram m < A = m < C so:-

75 + 5x = 87 + 3x

2x = 12

x = 6

and m< A = 75 + 30 = 105

and m < C = 87 + 18 = 105

Now if the other 2 angles are = 75 degrees( that is the same side angles add up to 180 degrees) we have proved that it is a parallelogram.

m < B = m < D

111 - 6x = 39 + 6x

-12x = -72

x = 6

m < B = 111 - 36 = 75 and m < D = 39 + 36 = 75

So this has been proved.

6 0

2 years ago

Which statement best describes the equation y = 3 - 4x?

Ronch [10]

The correct answer is C. A linear equation is written in y=mx+b, which this equation can be. y=3-4x is the same as y=-4x+3, which is a linear function.

8 0

2 years ago

URGENT! if ABCD is dilated by a factor of 2, the coordinate of A' would be:

koban [17]

After dilating the quadrilateral ABCD by a factor of 2 and with respect to the origin, the point A(x, y) = (-3, -1) is transformed into the point A'(x, y) = (-9, -3).

<h3>How to determine the coordinates of an image after applying a rigid transformation</h3>

First of all, dilation is a type of <em>rigid</em> transformation. <em>Rigid</em> transformations are transformations applied on <em>geometric</em> loci such that the <em>Euclidean</em> distance at every point of the construction is conserved. Vectorially speaking, the dilation is expressed by the following formula:

A'(x, y) = A(x, y) + k · [A(x, y) - O(x, y)] (1)

Where:

A(x, y) - Original point
O(x, y) - Center of dilation
A'(x, y) - Resulting point
k - Dilation factor

If we know that A(x, y) = (-3, -1), k = 2 and O(x, y) = (0, 0), then the coordinates of A' are:

A'(x, y) = (-3, -1) + 2 · [(-3, -1) - (0, 0)]

A'(x, y) = (-3, -1) + (-6, -2)

A'(x, y) = (-9, -3)

After dilating the quadrilateral ABCD by a factor of 2 and with respect to the origin, the point A(x, y) = (-3, -1) is transformed into the point A'(x, y) = (-9, -3).

To learn more on dilations: brainly.com/question/13176891

#SPJ1

4 0

1 year ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

So

$P(X \geq 2) = 0.93803$

5 0

2 years ago