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3 years ago

A proportion compares two what

Mathematics

2 answers:

blsea [12.9K]3 years ago

6 0

Answer:

it compares rates to cats

kupik [55]3 years ago

3 0

Answer:

A proportion compares two numbers by division

(by the way can you also help me with one of my questions part 2 would be nice on my profile it shouldbe the first one :) )\

Step-by-step explanation:

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PLEASE HELP IM BEGGING YOU ITS DUE IN 15 MIN ILL GIVE BRAINEST AND PTS

lys-0071 [83]

answer:

you just have to shrink it down to 4cm then to 2cm

i wish i could help out more, but i would need to have access to the form

hopefully, this helps anyhow :)

good luck

have a great day !!

4 0

3 years ago

How to prove all pairs of corresponding sides and angles are congruent

Lera25 [3.4K]

The lines on the triangles

5 0

3 years ago

David is going to install tiles on the floor of two rooms. One of the floors is 18 feet long by 12 feet wide. The other floor is

crimeas [40]

Dimension of one of the floors of one room that David wants to install tiles is 18feet long by 12 feet wide
Then
Area of the above room = 18 * 12 square feet
   = 216 square feet
Dimension of the floor of the other room that David wants to install tiles is 24 feet long and 16 feet wide
Then
Area of the other room = 24 * 16 square feet
   = 384 square feet
Then
The total square feet of the
rooms that David wants to install tiles = 216 + 384
   = 600 square feet
Cost of the tile that covers 1 square feet = $5
Cost of the 4 tiles that cover 4 square feet = $17
Then
Area that can be covered with 4 square feet of tiles = 600/4 square feet
      = 150 square feet
Minimum cost of covering
the two rooms that David wants to install tiles = 150 * 17 dollars
      = 2550 dollars
So the minimum cost of installing the tiles on the two floors of David's two rooms is $2550. I hope the procedure is simple enough for you to understand.

8 0

3 years ago

Exercise 10.4.1: Bayes' Theorem - detecting a biased coin. About (a) Sally has two coins. The first coin is a fair coin and the

8_murik_8 [283]

Answer:

0.026

Step-by-step explanation:

Given the result of 10 coin flips :

T,T,H,T,H,T,T,T,H,T

Number of Heads, H = 3

Number of tails, T = 7

Let :

B = biased coin

B' = non-biased coin

E = event

Probability that it is the biased coin:

P(E Given biased coin) / P(E Given biased coin) * P(E Given non-biased coin) * P(non-biased coin)

P(E|B)P(B) / (P(E|B)*P(B) + P(E|B')P(B')

([(0.75^3) * (0.25^7)] * 0.5) /([(0.75^3) * (0.25^7)] * 0.5) + (0.5^10) * 0.5

0.0000128746 / 0.00050115585

= 0.0263671875

7 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

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

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

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

$X \sim Binom(n=20, p=0.2)$

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:

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

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

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

3 0

3 years ago