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3 years ago

10

Emmett went to the chocolate store and bought 5 pounds of chocolate. If 2⁄3 of the chocolate was white chocolate, how many pound

s of white chocolate did he purchase?

2 answers:

AleksandrR [38]3 years ago

7 0

10/3 pounds white chocolate he bought

mr_godi [17]3 years ago

7 0

he is right 10/3 but as a mixed number it is 3 1/3

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G 3.6 Dice rolls. If you roll a pair of fair dice, what is the probability of (a) getting a sum of 1

Paladinen [302]

Answer:

0

Step-by-step explanation:

In the roll of a pair of fair dice, The sample space is as follows:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

If x,y represent the possible outcome of rolling the two die

Then the total number of possibilities for the sample space = (x,y)

= (6 × 6) = 36

Now, the probability of getting a sum of 1 does not exist in the roll of a pair of fair dice.

Therefore, Probability(1) = 0

8 0

3 years ago

21q - 7p and (3q - p)(7)

S_A_V [24]

Answer:

1

Step-by-step explanation:

4 0

2 years ago

Can u answer all three pls

faust18 [17]

Answer:

1. 28ft

2. 26ft

3. 20ft

Step-by-step explanation:

here ya go

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2 years ago

Read 2 more answers

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

3 years ago

Which situation describes a car that can travel $40 miles per gallon of gasoline?

Salsk061 [2.6K]

Answer:

A car can travel 80 miles on 2 gallons of gasoline.

Step-by-step explanation:

8 0

2 years ago