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2 years ago

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A school replaced 20% of it computers with new ones . what the total number of computers in the school if 55 computers were repl

aced?

1 answer:

sladkih [1.3K]2 years ago

7 0

The total is 275 because 100% divided by 20% is 5 and if 20%= 55 then that means that 55 goes into 100% 5 times so you multiply 55 by 5 to get 275. Hope I helped! Please rate me as the brainliest!

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EAB and DCB are two right triangles. The figure has BED≅ BDE. Point B is the midpoint of segment AC. Prove: EAB≅ DCB

aleksklad [387]

See the attached picture.

<span>you are given that ABCE is an isosceles trapezoid. </span>

<span>you are given that AB is parallel to EC. </span>

<span>this means that AE is congruent to BC. </span>

<span>you are given that AE and AD are congruent. </span>

<span>triangle EAD is an isosceles triangle because AE and AD are congruent. </span>

<span>this means that angle 1 is equal to angle 3. </span>

<span>since angle 1 is equal to angle 2 and angle 3 is equal to angle 1, then angle 3 is also equal to angle 2. </span>

<span>this means that AD and BC are parellel because their corresponding angles (angles 3 and 2) are equal. </span>

<span>since AB is parallel to EC and DC is part of the same line, than AB is parallel to DC. </span>

<span>you have AB parallel to DC and AD parallel to BC. </span>

<span>if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. </span>

<span>that might be able to do it,depending on whether all these statements are acceptable without proof. </span>

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here's my diagram.

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5 0

1 year ago

A music competition on television had five elimination rounds. After each elimination, only half of the contestants were sent to

ikadub [295]

Its c, average rate, im sure of it

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1 year ago

Read 2 more answers

Just need to check these answers

amid [387]

Great Job! they are all correct. :)

Good luck in your next tests.

3 0

2 years ago

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution 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.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

2 years ago

The two-way table shows the preferred vacation destination for people in different age groups. A 5-column table has 4 rows. The

Mice21 [21]

Answer:

C. The probability that a randomly selected person who chose Mexico as the preferred destination is a child is $\frac{14}{113}$

Step-by-step explanation:

You are given a two-way table:

$\begin{array}{ccccc}&\text{Hawaii}&\text{Mexico}&\text{Florida}&\text{Total}\\\text{Child (less than13 years old)}&33&14&62&109\\\text{Teenager (13 - 17 years old)}&50&42&25&117\\\text{Adult (18 years old and up)}&64&57&8&129\\\text{Total}&147&113&95&355\end{array}$

Check all options:

A. The probability that a randomly selected adult chose Hawaii as the preferred destination.

Number of people who choose Hawaii = 147

Number of teenagers who choose Hawaii = 50

Probability is

$\dfrac{50}{147}$

False

B. The probability that a randomly selected child chose Florida as the preferred destination.

Number of childs = 109

Number of childs who prefer Florida = 62

The probability is

$\dfrac{62}{109}$

False

C. The probability that a randomly selected person who chose Mexico as the preferred destination is a child.

Number of people who choose Mexico = 113

Number of people who choose Mexico and is a child = 14

The probability is

$\dfrac{14}{113}$

True

5 0

1 year ago

Read 2 more answers