Ask question

Ask question

All categories

3 years ago

7

daisy mixes 50ml of orange juice with 200ml of water. Write down the ratio of Orange Juice to Water. Give your answer in simples

t form

1 answer:

Vadim26 [7]3 years ago

4 0

Answer:

1:4

Step-by-step explanation:

Given

Volume of orange juice = 50ml

Volume of water = 200mL

Required

Ratio of volume of orange juice to water

Ratio = volume of orange juice/volume of water

Ratio =50mL/200mL

Ratio =5/20

Ratio =1/4 or 1:4

Hence the required ratio is 1:4

You might be interested in

Can someone tell me the answerz

anygoal [31]

Answer:

A is the right anser. Anything passt 0 is a negative.

Step-by-step explanation:

6 0

3 years ago

Help is needed folks plz will give brainiest!

irina1246 [14]

Answer:

I belive the answer is F

Step-by-step explanation:

In question/ option F it tells us that she started with 12 bannanas before getting 4 more bunchs, the question then asks how many bannanas were in each bunch, which then added plus 12 would of gotten you 32 bannans.

6 0

3 years ago

Read 2 more answers

One minus five sixths equals

Gnesinka [82]

Answer: 0.16666666666

Step-by-step explanation:lol

5 0

3 years ago

Read 2 more answers

What is the slope? (simplify your answer as much as possible) THERE IS THE PICTURE! Please help me

nasty-shy [4]

Answer:

-5x

Step-by-step explanation:

4 0

3 years ago

Brown Law Firm collected data on the transportation choices of its employees for their morning commute. The table shows the perc

yanalaym [24]

The two events out of the listed events which are independent events are given by: Option A: A and C

<h3>What is chain rule in probability?</h3>

For two events A and B, by chain rule, we have:

$P(A \cap B) = P(B)P(A|B) = P(A)P(A|B)$

<h3>What is law of total probability?</h3>

Suppose that the sample space is divided in n mutual exclusive and exhaustive events tagged as

$B_i \: ; i \in \{1,2,3.., n\}$

Then, suppose there is event A in sample space.

Then probability of A's occurrence can be given as

$P(A) = \sum_{i=1}^n P(A \cap B_i)$

Using the chain rule, we get

$P(A) = \sum_{i=1}^n P(A \cap B_i) = \sum_{i=1}^n P(A)P(B_i|A) = \sum_{i=1}^nP(B_i)P(A|B_i)$

<h3>How to form two-way table?</h3>

Suppose two dimensions are there, viz X and Y. Some values of X are there as $X_1, X_2, ... , X_n$ values of Y are there as $Y_1, Y_2, ..., Y_k$ rows and left to the columns. There will be $n \times k$ values will be formed(excluding titles and totals), such that:

Value( $i^{th}$ row, $j^{th}$ column) = Frequency for intersection of $X_i$ and $Y_j$ values are going in rows, and Y values are listed in columns).

Then totals for rows, columns, and whole table are written on bottom and right margin of the final table.

For n = 2, and k = 2, the table would look like:

$\begin{array}{cccc}&Y_1&Y_2&\rm Total\\X_1&n(X_1 \cap Y_1)&n(X_1\cap Y_2)&n(X_1)\\X_2&n(X_2 \cap Y_1)&n(X_2 \cap Y_2)&n(X_2)\\\rm Total & n(Y_1) & n(Y_2) & S \end{array}$

where S denotes total of totals, also called total frequency.

n is showing the frequency of the bracketed quantity, and intersection sign in between is showing occurrence of both the categories together.

<h3>How to calculate the probability of an event?</h3>

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

<h3>How to find if two events are independent?</h3>

Suppose that two events are denoted by A and B.

They are said to be independent event if and only if:

$P(A \cap B) = P(A)P(B)$

The given frequency table is:

$\begin{array}{ccccc} &\text{Public}&\text{Own}&\text{Others}&\text{Total}\\\text{Male}&12&20&4&36\\\text{Female}&8&10&6&24\\\text{Total}&20&30&10&60\end{array}$

The probability table for the same labels would be:

$\begin{array}{ccccc} &\text{Public}&\text{Own}&\text{Others}&\text{Total}\\\text{Male}&12/60&20/60&4/60&36/60\\\text{Female}&8/60&10/60&6/60&24/60\\\text{Total}&20/60&30/60&10/60&1\end{array}$

The events A, B,C,D and E are given as:

A: The employee is male.
B: The employee is female.
C: The employee takes public transportation.
D: The employee takes his/her own transportation.
E: The employee takes some other method of transportation.

Checking all the listed options one by one, for them being independent:

Case 1: A and C

P(A) = P(The employee is male. ) = 36/60

P(C) = P(The employee takes public transportation.) = 20/60 $P(A \cap C) = 12/60 \\\\ P(A)P(C) = \dfrac{36 \times 20}{60^2} = 12/60$

$P(A \cap C) = P(A)P(C)$

Thus, A and C are independent events.

Case 2: A and D

P(A) = P(The employee is male. ) = 36/60

P(D) = P(The employee takes his/her own transportation.) = 30/60

$P(A\cap D) = 20/60\\\\P(A)P(D) = \dfrac{30 \times 36}{60^2} = 12/60 \neq P(A \cap D)$

Thus, A and D are not independent events.

Case 3: B and D

P(B) = P(The employee is female. ) = 24/60

P(D) = P(The employee takes his/her own transportation.) = 30/60

$P(B \cap D) = 10/60 \neq P(B)P(D)=\dfrac{24 \times 30}{60^2} = 12/60$

Thus, B and D are not independent events.

Case 4: B and E

P(B) = P(The employee is female. ) = 24/60

P(E) = P(The employee takes some other method of transportation.) = 10/60

$P(B \cap E) = 6/60 \neq P(B)P(E)= \dfrac{24 \times 10}{60^2} = 4/60$

Thus, B and E are not independent events.

Thus, the two events out of the listed events which are independent events are given by: Option A: A and C

Learn more about independent events here:

brainly.com/question/3898488

5 0

2 years ago