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3 years ago

The Perimeter of the rectangle below is 94 units find the length of side peek you write your answer without variables

Mathematics

1 answer:

ivolga24 [154]3 years ago

3 0

Answer:

DC = 21 ; BC = 26

Step-by-step explanation:

The perimeter of a rectangle is given by :

P = 2(length + width)

2x + 3

3x - 1

Perimeter, P = 94

Hence,

94 = 2(2x +3 + (3x - 1))

94 = 2(2x + 3 + 3x - 1)

94 = 2(5x + 2)

94 = 10x + 4

94 - 4 = 10x

90 = 10x

x = 90 /10

x = 9

DC = 2x + 3 = 2(9) + 3 = 18 + 3 = 21

BC = 3x - 1 = 3(9) - 1 = 27 - 1 = 26

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3 years ago

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HELP ME PLZZZ!!! Solve for x. Show each step of the solution. 4.5(4 − x) + 36 = 202 − 2.5(3 + 28)

Veronika [31]

4.5{4_x}+36=202-2.5{3+28}
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3 years ago

Use the figure to find the measures of the numbered angles,

Galina-37 [17]

Given:

A figure in which a transversal line t intersect the two parallel lines a and b.

To find:

The measure of the numbered angles.

Solution:

If a transversal line intersect the two parallel lines, then

1. Corresponding angles are equal.

2. Alternate interior angles are equal.

3. Alternate exterior angles are equal.

Now,

$m\angle 1+99^\circ=180^\circ$ (Linear pair)

$m\angle 1=180^\circ-99^\circ$

$m\angle 1=81^\circ$

$m\angle 2=99^\circ$ (Vertically opposite angles)

$m\angle 3=\angle 1=81^\circ$ (Vertically opposite angles)

$m\angle 4=99^\circ$ (Corresponding angles)

$m\angle 5=\angle 3=81^\circ$ (Alternate interior angles)

$m\angle 6=99^\circ$ (Alternate exterior angles)

$m\angle 7=\angle 3=81^\circ$ (Corresponding angles)

Therefore, $\angle 1=81^\circ, \angle 2=99^\circ,\angle 3=81^\circ, \angle 4=99^\circ,\angle 5=81^\circ, \angle 6=99^\circ,\angle 7=81^\circ$ .

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3 years ago

The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that

salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

$P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253$

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

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:

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

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

$P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253$

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0

3 years ago