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sergiy2304 [10]
3 years ago
10

What is 1 2/3 times 2 3/5

Mathematics
2 answers:
Orlov [11]3 years ago
8 0

Answer:

4 1/3

Step-by-step explanation:

Hope this helps :)

kobusy [5.1K]3 years ago
4 0
1 2/3 x 2 3/5 = 13/3 = 4 1/3
You might be interested in
{20, 68, 70, 70, 71, 72, 75, 76}
guapka [62]

Answer:

Mean: 65.25

Median: 70.5

Mode: 70

Step-by-step explanation:

Mean is the sum of the set of numbers divided by the number of numbers in the set.

20+68+70+70+71+72+75+76=522

522/8=65.25

The median is the number that is in the middle of the set when put in numerical order.

Since there is an even amount of numbers, the two middle numbers are 70 and 71, so you would take the mean of these two numbers to find the median.

70+71=141

141/2=70.5

The mode is the number that occurs most frequently in the set.

70 appears more times than any other number, therefore it is the mode.

5 0
4 years ago
Read 2 more answers
Which method would be the most efficient method to solve x^2-11=0
telo118 [61]
All three methods would be efficient

4 0
3 years ago
I'll give brainliest
Ray Of Light [21]
The answer is 60. The formula is v=bxhxh/b or in this case 6x4x15/6. After you multiply the numerator you will get 360 and then you divide by 6 to get your answer: 60.
7 0
3 years ago
Math help please its urgent. I'll mark brainliest!!!
OleMash [197]
3n + 15 >= 3n + 8     distribution.
3n - 3n >= 8 - 15       arrange the like terms.
0 >= -7                      combine the like terms.
3 0
3 years ago
What is the midpoint of the line segment with endpoints (–1, 7) and (3, –3)?
Kipish [7]
The coordinates consist of x coordinate and y coordinate
(x₁,y₁) = (-1,7)
(x₂,y₂) = (3,-3)

To find the midpoint of x coordinate, use this following formula
x midpoint = (x₁ + x₂)/2
x midpoint = (-1 + 3) / 2
x midpoint = 2/2
x midpoint = 1

To find the midpoint of y coordinate, use this following formula
y midpoint = (y₁ + y₂)/2
y midpoint = (7 + (-3))/2
y midpoint = (7 - 3)/2
y midpoint = 4/2
y midpoint = 2

ANSWER
The midpoint is (1,2)
6 0
4 years ago
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