1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Marina CMI [18]
3 years ago
6

John is making shells for his wrong he was 3 shelves for each shells requires a 8 2/3 be of lumber how many feet of lumber does

he need
Mathematics
1 answer:
Reil [10]3 years ago
7 0

Answer: 26 feet.

Step-by-step explanation:he

The exercise is not written clearly. Assuming that it asks to find the amount of lumber (in feet) John needs for 3 shelves, let "x" represents that amount of lumber in feet.

Based on the data provided, you know that one shelf requires 8\frac{2}{3} feet of lumber.

You can convert the Mixed number to Decimal number applying these steps:

1. You must divide the numerator of the fraction by the denominator.

2. You must add the result obtained to the Whole number part.

Then you get:

8\frac{2}{3}=8+0.6667=8.6667

Knowing this, you can set up the following proportion:

\frac{8.6667}{1}=\frac{x}{3}

Finally, you must solve for "x" in order to find its value. This is:

(3)(8.6667)=x\\\\x=26.0

You might be interested in
What are measures of dispersion?
Alex_Xolod [135]
It is the extent to which a distribution is stretched
4 0
3 years ago
I am confused help plz
Wewaii [24]

Answer:


Step-by-step explanation:

a 6(m+3)

b expression

equation

5 0
3 years ago
What is the percent error in using 3.14 as an approximation of pi
inessss [21]
Pi is an irrational number meaning it has no end, 3.14 is just an estimate which means that it is not accurate.
5 0
3 years ago
single die is rolled twice. Find the probability of rolling an oddodd number the first time and a number greater than 33 the sec
givi [52]

Answer:

\frac{1}{4}

Step-by-step explanation:

A single die is rolled twice, we have to find the probability of rolling an odd number in first throw and a number greater than 3 in the second throw.

a) Rolling an Odd number in first throw

A die has total 6 possible outcomes, out of which 3 are odd numbers i.e. 1,3 and 5

So, total number of possible outcomes = 6

Total Favorable outcomes (Odd numbers) = 3

Probability is defined as the ratio of favorable outcomes to total number of outcomes. So,

The probability of rolling an odd number would be = \frac{3}{6} = \frac{1}{2}

b) Rolling a number greater than 3 in second throw

Here again total possible outcomes = 6

Favorable outcomes (Numbers greater than 3 are 4, 5 and 6) = 3

So,

The probability of rolling a number greater than 3 =  \frac{3}{6} = \frac{1}{2}

These two events(rolls) are independent of each other, so the overall probability of both events occurring would be the product of individual probabilities.

So,

Probability of rolling an odd number the first time and a number greater than 3 the second time = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

6 0
3 years ago
Which shows how the distributive property can be used to evaluate 7×8 4/5
Greeley [361]
First you find 4/5 of 8 and then 4/5of 7 and multiply them
5 0
3 years ago
Read 2 more answers
Other questions:
  • What is the inverse of the function f(x) = x +3?
    10·1 answer
  • Help please?<br> 8 + (n - 1)6<br> n=10..
    14·2 answers
  • A/2 - 1/2 and please explain!!
    14·1 answer
  • Name two pairs of adjacent angels and two pairs of vertical angels in figure
    7·1 answer
  • What tunnel leads you to the wormhole in Run 3?
    13·2 answers
  • 1/4x +7=10 What is X?
    7·2 answers
  • A shop repairs both audio and video components. Let A denote the event that the next component brought in for repair is an audio
    8·1 answer
  • I need help with solving inequalitiy but in fraction form:/
    5·1 answer
  • Evaluate the expression 5* – 31 for = 2.<br> Help pls!
    9·1 answer
  • What do rhombuses always have
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!