All categories

3 years ago

How is 2.25 written as a fraction

Mathematics

2 answers:

Salsk061 [2.6K]3 years ago

6 0

2.25 as a fraction is 2 1/4

liraira [26]3 years ago

4 0

'There is not much that can be done to figure out how to write 2.25<span> as a </span>fraction<span>, except to literally use what the decimal portion of your number, the .25, means. Since there are 2 digits in 25, the very last digit is the "100th" decimal place. So we can just say that .25 is the same as 25/100.' -Google

So 25/100 is your answer.

</span>

You might be interested in

The perimeter of the sign is 28 meters. If the width of the sign is 4meters, what is the length in meters?

ValentinkaMS [17]

Answer:10

Step-by-step explanation:

28-4-4=20

20÷2=10

10+10+4+4=28

6 0

3 years ago

63% of 300 is what number

PolarNik [594]

Answer:

189

Step-by-step explanation:

You take 63% make it a decimal by dividing it by 100 and then you multiply it with 300. 300*.63=189.

6 0

3 years ago

Read 2 more answers

How can I learn to round better in math

love history [14]

I have always followed an easy rule: "five or more raises the score". What it means is if the number is 5 or more, it goes to the next whole. Make sense?

3 0

3 years ago

Read 2 more answers

Giving brainliest and max rating.

Yanka [14]

Given:

The cost of each carnival ticket is $5.

To find:

The equation, table of values and graph for the given problem.

Solution:

Let x be the number of tickets and y be the total money spent on tickets.

Cost of one ticket = $5

Cost of x tickets = $5x

So, total cost is

$y=5x$

The required equation is $y=5x$ .

At x=1,

$y=5(1)$

$y=5$

At x=2,

$y=5(2)$

$y=10$

At x=3,

$y=5(3)$

$y=15$

The required table of values is

x y

1 5

2 10

3 15

So, the required table of values is table A.

From the above table, it is clear that the graph passes through the point (1,5), (2,10) and (3,15). The graph B passes through these points.

So, the required graph is graph B.

Since the required answers are $y=5x$ , table A, graph B, therefore the correct option is B.

8 0

3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago