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3 years ago

Y - y = m (x - x) represents

Mathematics

1 answer:

iragen [17]3 years ago

3 0

Answer:

point slope

Step-by-step explanation:

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The square root of 36y^3 =

Aleks04 [339]

The square roots of 36y^3 are plus and minus 6y^(3/2) .

6 0

3 years ago

There are 3 cards in Box A and 3 cards in Box B.

Anton [14]

Answer:

Step-by-step explanation:

<u>Possible number of pairs of given cards:</u>

3*3 = 9

<u>Number of sums with odd score, two odd numbers added to an even number from each box:</u>

2*1 + 1*2 = 4 (pairs of 3 and 5 with 2 or pairs of 9 and 3 with 4)

<u>Required probability:</u>

P(odd) = 4/9

8 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

If you divide any natural number n by 4, you get a remainder r.

blondinia [14]

Given:

If you divide any natural number n by 4, you get a remainder r.

To find:

The values of r if $n=13;34;43;100$ . Also find the domain and range.

Solution:

It is given that any natural number n by 4, you get a remainder r.

$\dfrac{n}{4}=q+\dfrac{r}{4}$

Where, n is a natural number, q is quotient, r is the remainder.

For $n=13$ ,

$\dfrac{13}{4}=3+\dfrac{1}{4}$

So, $r=1$ .

For $n=34$ ,

$\dfrac{34}{4}=8+\dfrac{2}{4}$

So, $r=2$ .

For $n=43$ ,

$\dfrac{43}{4}=10+\dfrac{3}{4}$

So, $r=3$ .

For $n=100$ ,

$\dfrac{100}{4}=25+\dfrac{0}{4}$

So, $r=0$ .

Therefore, the required value are $r=1,2,3,0$ if $n=13;34;43;100$ respectively.

The domain of the function is $\{13,34,43,100\}$ and the range of the function is $\{1,2,3,0\}$ .

8 0

3 years ago

3х – 4у = -20<br> 5х + 5 = 68

My name is Ann [436]

try using desmos calculator and it give you the answers and where they intersect

5 0

3 years ago