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Bond [772]
3 years ago
13

....... ....... ....... =5

Mathematics
1 answer:
Sauron [17]3 years ago
6 0
What? Ig that equals 5 lol
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What is the median for 76,84,93,67,82,and 76
Lapatulllka [165]
First arrange the numbers from least to greatest:

67, 76, 76, 82, 84, 93

The median is the middle number. But since we have two numbers that are in the middle we have to find the average of them.

76 + 82 = 158
158/2 = 79

So your answer is 79
3 0
4 years ago
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3
julsineya [31]

Answer:

In order to ride on the ride an individual must be at least 4 feet tall.

This means that the individual must be equal to 4 ¾ feet tall.

h = 4 ⅓

and the individual could be greater than 4¾ feet tall.

h > 4 ¾

So the individual could be greater than and equal to 4¾ feet tall.

h ≥ 4¾

4 0
3 years ago
What is x minus 4.7 divided by 3 =-8
AleksandrR [38]
X= -193 over 30
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Convert 12 and a half of 30
tamaranim1 [39]

Answer:

375

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12½[30] → 25⁄2[30]

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I am joyous to assist you anytime.

7 0
3 years ago
A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is
tatiyna

Answer:

Hence to get same number of students in each classroom,the sufficient condition is that assign 13n students to each classroom.

Step-by-step explanation:

Given:

There are m classrooms and n be the students

3<m<13<n.

To Find:

Whether it is possible to assign each of n students to one of m classrooms with same no.of students.

Solution:

This problem is related to p/q form  has to be integer in order to get same no of students assigned to the classroom.

As similar as ,n/m ratio

So 1st condition is that,

If it is possible to assign the n/m must be integer and n should be multiple of m,

when we assign 3n students to m classrooms ,we cannot say that 3n/m= integer so that  n is greater than 13 i.e n=14 and m=6

hence they are not multiple of each other so they will not make same students in each classrooms.

Otherwise,n=14 and m=7 they will give same number but this condition is not sufficient condition to assign the student.

So 2nd condition is that ,

When we assign 13n students to m classrooms, as 13 is prime number and

3<m<13 which implies the 13n/m to be integer so n and m must be multiple of each other.

Suppose n=20 and m=5 classrooms

then 13*20=260 ,

260/5=52 students in each classroom,

4 0
3 years ago
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