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3 years ago

BRAINLIEST PLUS 100 POINTS

Mathematics

2 answers:

scoundrel [369]3 years ago

3 0

The top and bottom of the rectangle need to be equal and the two sides also need to be equal.

Set the equations to equal each other and solve for the letter.

Top and bottom:

3x-5 = x +1

Subtract 1x from each side:

2x -5 = 1

Add 5 to both sides:

2x = 6

Divide both sides by 2:

x = 6/2

X = 3

SIdes:

Now you know what X is replace x in the left side equation and solve:

2x-4 = 2(3) -4 = 6-4 = 2

The left side = 2, so the right side also needs to equal 2:

y-3 = 2

Add 3 to both sides:

y = 5

Amiraneli [1.4K]3 years ago

3 0

Answer:

x = 3 and y = 5 and explanation below :)

Step-by-step explanation:

Because 2 side, up and down or right and left of the rectangle must be the same so we would have:

3x - 5 = x + 1

Subtract x from both side, we would get:

3x - x - 5 = x - x +1

2x - 5 = 1

Add 5 into both sides, we got:

2x - 5 + 5 = 1 + 5

2x = 6

Divide both sides by 2, we got:

2x / 2 = 6 / 2

x = 3

Becasue x = 3 and we know taht 2x - 4 = y - 3 so you can solve 2x - 4 first and then it would equal y -3

2x - 4

= 2 * 3 - 4

= 6 - 4

= 2

Because 2x - 4 = y -3

So y -3 = 2

Add 3 into both sides, we got:

y - 3 + 3 =2 + 3

y = 5

Hope this helped :3

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A teacher teaches two classes with 8 students each. Each student has a 95% chance of passing their class independent of the othe

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0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they pass, or they do not. The probability of an student passing is independent of other students, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Probability that all students pass in a class:

Class of 8 students, which means that $n = 8$

Each student has a 95% chance of passing their class independent of the other students, which means that $p = 0.95$

This probability is P(X = 8). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{8,8}.(0.95)^{8}.(0.05)^{0} = 0.6634$

Find the probability that, in exactly one of the two classes, all 8 students pass.

Two classes means that $n = 2$

0.6634 probability all students pass in a class, which means that $p = 0.6634$ .

This probability is P(X = 1). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,1}.(0.6634)^{1}.(0.3366)^{1} = 0.4466$

0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.

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