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3 years ago

If someone drank 3 fourthes of a 16 ounce bottle . how many ounces are left

2 answers:

4 0

How many fourths are there in 16 wholes?

$\bf 16\implies \cfrac{16}{1}\implies \cfrac{16\cdot 4}{1\cdot 4}\implies \stackrel{\textit{fourths in 16 wholes}}{\cfrac{64}{4}}
\\\\\\
\textit{now someone takes away }\frac{3}{4}\textit{ from that}\qquad \cfrac{64}{4}-\cfrac{3}{4}
\\\\\\
\cfrac{64-3}{4}\implies \stackrel{leftover}{\cfrac{61}{4}}$

7nadin3 [17]3 years ago

3 0

Look at the demoninator, 1/4 of 16 is 4 ,you have 3 fourths so multiply 4 by 3 and you get 12 16-12 and the answer is 4

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Solve for x<br><br> 2(x-3)+5=4-7(2x-1)<br><br> A. -1/10<br> B. 3/4<br> C. 1/2<br> D. 1/16

Stolb23 [73]

Hello
<span>2(x-3)+5=4-7(2x-1)
2x-6+5 = 4 -14x+7
2x+14x =6-5+4+7
16x=12
x=12/16
x=(4×3)/(4×4)
x=3/4 (simplified by 4)
answer B</span>

6 0

3 years ago

Read 2 more answers

(4 × 6) ÷ (2 + 4) ÷ (8 ÷ 4) =

rodikova [14]

74.839 is the answer

4 0

3 years ago

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Mr. Martin is giving a math test next period. The test, which is worth 100 points, has 35 problems. Each problem is worth either

fenix001 [56]

Answer:

x + y = 100

5x + 2y = 35

Step-by-step explanation:

3 0

2 years ago

A bag contains an unknown number of marbles. You know that P(red) = 1 and P(green) = 1. What can you conclude about how many mar

Nataly [62]

The number of marbles in the bag illustrates probability

The conclusion about the number of marbles in the bag is that the bag can only contain one type of marble

<h3>The number of marbles in the bag?</h3>

The probabilities are given as:

P(Red) = 1

P(Green) = 1

For a distribution, the sum of all probabilities must equal to 1.

This means that:

P(Red) + P(Green) = 1

So, we have:

1 + 1 = 1

This gives

2 = 1

The above equation is false because 2 does not equal to 1.

So, the conclusion about the number of marbles in the bag is that the bag can only contain one type of marble i.e. either red marbles are in the bag or blue marbles are in the bag

Read more about probability at:

brainly.com/question/251701

5 0

2 years ago

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is r

svp [43]

Answer:

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.

This means that $a = 50, b = 52$

If one such class is randomly selected, find the probability that the class length is between 51.5 and 51.7 min.

$P(51.5 \leq X \leq 51.7) = \frac{51.7 - 51.5}{52 - 50} = \frac{0.2}{2} = 0.1$

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

3 0

2 years ago