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3 years ago

How do you put this in an equation (The number of checkers is 24 times the number of checkerboards.)

Mathematics

1 answer:

anyanavicka [17]3 years ago

4 0

Let checkers be x and checker boards be y

x = 24 * y or

y = x/24

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What is 6 divided by 1/2?

Lemur [1.5K]

Answer:

Step-by-step explanation:

Make 6 a reciprocal which would be 1/6 then change the division sign to multiplication so 1/6 x 1/2 will be 12

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3 years ago

Find all angles, 0º < 0 < 360°, that solve the following equation.<br> cos 0 = 1

Annette [7]

Tbh not completely sure but I think its 180 .

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3 years ago

A pizza with radius 7 in. is cut into 12 equal-sized pieces. What is the area of each piece? Round to the nearest hundredth of a

dimaraw [331]

Answer:

Area of one piece of pizza ≈ 12.83 inches

Step-by-step explanation:

First of all let's calculate the area of the pizza.

Pizza is circular in shape, thus,

Area; A = πr²

We are given the radius;r = 7 inches

So, A = π(7²) = 153.94 in²

Now, since the area is divided into 12 equal sized pieces, thus we will divide the total area we got by 12 to get the area of one piece.

Area of one piece of pizza = 153.94/12 ≈ 12.83 inches

5 0

4 years ago

50 POINTS + BRAINLIEST

Vedmedyk [2.9K]

X + 2y = 6
2y = -x + 6
y = -1/2 x + 3 so slope = -1/2

parallel lines has same slope = -1/2
perpendicular lines slope is opposite and reciprocal so slope = 2
if slope not equal either -1/2 or 2 then neither.

y = -1/2 x - 5 ....m= - 1/2.... parallel

-2x + y = -4
y = 2x - 4...m = 2 ....perpendicular

-x + 2y = 2
2y = x + 2
y = 1/2 x + 1 ...m = 1/2 ...neither

x + 2y = -2
2y = -x - 2
y = -1/2 x - 1 ...m = -1/2 .....parallel

hope it helps

3 0

3 years ago

Read 2 more answers

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:

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