Ask question

Ask question

All categories

4 years ago

15

A circular track is 1/4 mile long. You run 13 1/3 laps around the track. How many miles did you run?

1 answer:

rusak2 [61]4 years ago

3 0

So all you have to do is times $\frac{1}{4}$ x 13 and 1/3 which is the same as 1/4 x 40/3 which is equal to 40/12 = 3 1/3mile

⭐ Answered by Foxzy0⭐

⭐ Brainliest would be appreciated, I'm trying to reach genius! ⭐

⭐ If you have questions, leave a comment, I'm happy to help! ⭐

You might be interested in

10² = ?<br><br> What is 10² = to?<br><br> Thanks in advance. :P

romanna [79]

To answer this exponent we will multiply the base by the power. Lets do it:-

10 = base
2 = power

10² = 10 × 10 = 100
10² = 100

CHECK OUR WORK:-

√100 = 10

We were RIGHT!!!

So, 10² = 100.

Hope i helped ya!! :)

8 0

3 years ago

Read 2 more answers

Please help, the first part is x squared, if people need clarification

Ainat [17]

Use the quadratic formula .
x^2-2x+11=0
a=1
b=-2
c=11

x=-2+- (-2)^2-4*1*11 divided by 2
x=2+- the square root of -40 divided by 2
There is no real solutions

6 0

3 years ago

Given a circle with measures of (C, d, and r) and a circle with measures of (C', d', and r'), what is C' if

Anestetic [448]

Answer:

The answer can be calculated by doing the following steps;

Step-by-step explanation:

5 0

3 years ago

In a large school, it was found that 77% of students are taking a math class, 74% of student are taking an English class, and 70

Iteru [2.4K]

Answer:

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

Step-by-step explanation:

We solve this question working with the probabilities as Venn sets.

I am going to say that:

Event A: Taking a math class.

Event B: Taking an English class.

77% of students are taking a math class

This means that $P(A) = 0.77$

74% of student are taking an English class

This means that $P(B) = 0.74$

70% of students are taking both

This means that $P(A \cap B) = 0.7$

Find the probability that a randomly selected student is taking a math class or an English class.

This is $P(A \cup B)$ , which is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

So

$P(A \cup B) = 0.77 + 0.74 - 0.7 = 0.81$

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

Find the probability that a randomly selected student is taking neither a math class nor an English class.

This is

$1 - P(A \cup B) = 1 - 0.81 = 0.19$

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

6 0

3 years ago

Translate the variable expression into a word phrase " 5c "

Marina CMI [18]

5 lots of the value of c added together

Explanation:

By example:

<span>2×6=12=6+6=12</span>
<span>3×6=18=6+6+6=18</span>

<span>5×2=10=2+2+2+2+2=10</span>

<span>5×c=5c=c+c+c+c+c=5<span>c</span></span>

6 0

3 years ago