The table shows the precipitation in December, in inches, for Washington state from 2005 to 2013. What was the mean amount of pr
ecipitation during this time? Round to the nearest hundredth.
A) 1.97 Inches
B) 2.61 Inches
C) 5.33 Inches
D) 5.98 Inches
A) 1.97 Inches
B) 2.61 Inches
C) 5.33 Inches
D) 5.98 Inches
2 answers:
Answer : C) 5.33 Inches
The table shows the precipitation in December, in inches, for Washington state from 2005 to 2013.
We are given with 9 values.
number of values =9
To find the mean amount of precipitation during this time we add all the precipitation then divide by number of values
=
The mean amount of precipitation = 5.33 inches
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Answer:
See below.
Step-by-step explanation:
Well first, we need to find the weight of the table. We know that 8 boxes weighs a total of 240kg (since each box weights 30kg). Thus, we can conclude that the table weighs 70kg by doing 310-240=70.
Now, we can write our function. Let equal the amount of boxes.
The table is a set weight, so that would be our constant.
Thus, we will have:
30x represents the weight each box of book adds to the total. One box equals 30kg, 2 boxes equal 60kg, etc.
The 70 represents the unchanging weight of the table.
In terms of W(x), it will be:
Answer: 96 teaspoons
Step-by-step explanation:
If 1 cup contains 16 tablespoons the 2 cups will contain;
= 16 * 2
= 32 tablespoons
If 1 tablespoon contains 3 teaspoons then 32 tablespoons will contain;
= 32 * 3
= 96 teaspoons
The given dimensions of the base of the box are 2 x 3 units.
Let h be the height of the box.
So, the volume of the box = 2 x 3 x h ...(i)
The shape of baseball is spherical and one baseball fits in a cubical box having the minimum dimensions equal to the diameter of the spherical ball.
Assuming the diameter of the baseball is d units.
So, the volume occupied by one baseball= volume of the cubical box having the side units.
cubic units
So, the volume required to pack 48 baseballs
Assuming that all the baseballs fit in the box, i.e peripheral balls in every horizontal and vertical layer touched the box as shown in the figure.
So, the length, 3 units and the width, 2 units, must be the integral multiple of the diameter, units, of the baseball. i.e
The length,
The width,
The height,
where and
belong to the natural number.
This required volume must be equals to the volume of the box.
From equations (i) and (ii),
[from (iii) and (iv)]
On comparing with the equation (iv)
Now, if the diameter of the baseball is unity,i.e, d=1 unit
( integer) and (integer) [from (iii) and (iv)]
(also an integer) [from (v)]
Now, the height of the box,
unit.
Hence, the other dimension of the box, is 8 units corrosponding to the unity diameter of the baseball.
Note the height of the box depends on the diameter of the baseball, so it will have infinite possible value for different value of the diameter.
For example, the value of height for d=1 unit has been shown.
and another example may be if units, then (integer) and (integer)
(also an integer) [from (v)]
Now, the height of the box,
unit. [from (iv)]
Here, the height of the box is 2 units corrosponding to 0.5 unit diameter of the baseball.
The possivle values of d can be taken for which and
must be integers.
