Answer:
<em>5.5</em>
Step-by-step explanation:
Given the set of data
5, 4, 2, 1, 1, 2, 10, 2, 3, 5.
The average of the least and the greatest value is known as the midrange
The formula for calculating the midrange is expressed as shown:
Midrange = (Greatest value + Least value)/2
Given
Greatest value = 10
Least value = 1
Midrange = 10+1/2
Midrange = 11/2
Midrange = 5.5
<em>Hence the midrange of the data is 5.5</em>
Answer:
16 pieces of candy per bag
Step-by-step explanation:
The answer is... A hope this helps
<em><u>The inequality can be used to find the interval of time taken by the object to reach the height greater than 300 feet above the ground is:</u></em>

<em><u>Solution:</u></em>
<em><u>The object falls, its distance, d, above the ground after t seconds, is given by the formula:</u></em>

To find the time interval in which the object is at a height greater than 300 ft
Frame a inequality,

Solve the inequality
Subtract 1000 from both sides


Time cannot be negative
Therefore,
t < 6.61
And the inequality used is: 