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2 years ago

Five lemons cost $1.80. What is the cost per lemon?

Mathematics

1 answer:

kap26 [50]2 years ago

6 0

Answer:

0.36 cents

Step-by-step explanation:

1.80/5=0.36 <----answer

To check you can do 0.36x5 and get 1.80

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Please help me thanks please

s344n2d4d5 [400]

Answer:

um all you do is multiply 21 as the length by 14 was the width (LxW)

Step-by-step explanation:

=294 is your answer I think ...

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2 years ago

Jason has a piece of wire that is 63.9 inches long. He cuts the wire into 3 equal places. Estimate the length of one piece of wi

Shtirlitz [24]

Answer: 21.3

Step-by-step explanation:

63.9 divided by 3 equal pieces is 21.3!

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2 years ago

Please help solve equation

gregori [183]

<h2><u>Solution</u> :-</h2>

Height of Eric = 6 + 1/6 feet

= 37/6 feet

Height of his brother = 5 + 3/4 feet

= 23/4 feet

Difference between their height :

= 37/6 - 23/4

= (74 - 69)/12

= 5/12

<u>Hence</u>,

Eric is 5/12 feet taller than his brother.

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2 years ago

Hey can someone help me with this

ivanzaharov [21]

Step-by-step explanation:

The median is the middle number is an ordered set of data. In a frequency table, the observations are already arranged in an ascending order. We can obtain the median by looking for the value in the middle position. If there is an odd number of observations, the median is the middle number

7 0

2 years ago

An NFL coach sometimes uses a defense that utilizes 3 defensive linemen, 4 linebackers, and 4 defensive backs. His roster (the p

Alexandra [31]

Answer:

68,600

Step-by-step explanation:

The order of the players is not important. For example, a defensive line of Shaq Lawson, Ed Oliver and Jerry Hughes is the same as a defensive line of Ed Oliver, Shaq Lawson and Jerry Hughes. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Defensive Lineman:

3 from a set of 8. So

$C_{8,3} = \frac{8!}{3!5!} = 56$

56 combinations of defensive lineman

Linebackers:

4 from a set of 7. So

$C_{7,4} = \frac{7!}{4!3!} = 35$

35 combinations of linebackers

Defensive backs:

4 from a set of 7. So

$C_{7,4} = \frac{7!}{4!3!} = 35$

35 combinations of defensive backs

How many different ways can the coach pick the 11 players to implement this particular defense?

56*35*35 = 68,600

68,600 different ways can the coach pick the 11 players to implement this particular defense

7 0

2 years ago