All categories

3 years ago

Sam ate 1/3 of the Halloween candy. Tim ate 2/5 of the Halloween candy. How much candy us left

Mathematics

2 answers:

rusak2 [61]3 years ago

8 0

Answer:

4/15

Step-by-step explanation:

convert the 2 fractions to have the same denominator

5/15 and 6/15

together they are 11/15

you have 4/15 left

Georgia [21]3 years ago

5 0

Answer:

4/15

Step-by-step explanation:

1/3=5/15

2/5=6/15

15/15-11/15=4/15

You might be interested in

A mirror frame in the shape of an oval is shown below. The ends of the frame form semicircles: (5 points)

guapka [62]

Answer:

696.265 inches

Step-by-step explanation:

Radius = 27/2 = 13.5

2 semicircles + 2 lengths

(3.14 × 13.5²) + 2(62)

696.265 inches

4 0

3 years ago

devi and her brother had the same amount of money. after devi spent 2/5 of her money and her brother spent 3/10 of his money, th

krok68 [10]

To answer this question you have to create a system of equations. The first equation will be that Devi's money (x) equals her brother's money (y), or x = y. The next equation would be that (3/5)x + (7/10)y = 78. You than can substitute x in for y because x = y. The equation would know be (3/5)x + (7/10)x = 78. You then combine the like terms to create an equation of (13/10)x = 78. Then, multiply both sides by 10 / 13 in order to isolate x. This will create the equation x = 60. This means that Devi and her brother each had 60 dollars. You then find out how much they spent and add it together. You can do so with the equation (2/5)x + (3/10)y = z, with z being total money spent. You substitute 60 in for x and for y then solve. When you solve you see that 24 + 18 = z, or that z equals 42. In other words, they spent 42 dollars altogether.

6 0

3 years ago

Mrs. Bailey gives a test, and her students’ scores range from 30 to 70. She decides to curve the scores, so that they range from

Harman [31]

Answer:

f(x) = x*3/4 + 42.5

Step-by-step explanation:

The original difference between the pair is 70 - 30 = 40

The new difference between the pair is 95 - 65 = 30

Since the differences are not the same, Mrs Bailey must first perform a (slope) multiplication by a factor of 30/40 or 3/4

Then 30 * 3/4 = 22.5

Then she can shift the scores up by 65 - 22.5 = 42.5 in order to get the range from 65 to 95

Therefore, f(x) = x*3/4 + 42.5. We can test that

f(30) = 30*3/4 + 42.5 = 65

f(70) = 70*3/4 + 42.5 = 95

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 ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$