You would convert one of the units to the other, then find the ratio between them when they are the same.
Example: 2 inches and 10 centimeters
1 inch = 2.54 cm
2 x 2.54 = 5.08
10 / 5.08 = 1.969
10 cm is 1.969 times longer than 2 inches.
Step-by-step explanation:
15) 50 ÷ 2 = 25
17) Mean = 301, Mode = 40-50
(10+20) ÷ 2 = 15, (20+30) ÷ 2 = 25, (30+40) ÷ 2 = 35
(40+50) ÷ 2 = 45, (50+60) ÷ 2 = 55, (60+70) ÷ 2 = 65
(70+80) ÷ 2 = 75
• 15×4 = 60, 25×8 = 200, 35×10 = 350, 45×12 = 540
55×10 = 550, 65×4 = 260, 75×2 = 150
Mean = (60+200+350+540+550+260+150) ÷ 7
= 2110 ÷ 7
= 301.4285....
= 301
Mode : the highest frequency
Answer:
242 m³ of space is there in the fort.
Step-by-step explanation:
Given that,
The dimensions of first box is 4 meters long, 8 meters wide, and 8 meters high.
The dimensions of the second box is 2 meters long, 7 meters wide, and 1 meter high.
Space left = Volume of first box - volume of second box
= (4)(8)(8) - 2(7)(1)
= 242 m³
So, 242 m³ of space is there in the fort.
We have that the workdone in stretching natural length to is mathematically given as
W=9/2lbft
<h3>Workdone in stretching natural length</h3>
Question Parameters:
- A force of 128 lb is required to hold a spring stretched 2 ft <em>beyond </em>its natural length.
- Stretching it from its natural length to 9 inches beyond its natural length
Generally the Hookes equation for the Force is mathematically given as
F=Kx
Where
3.2lb=2ft*k
Thereofore
k=16
Force becomes
f=kx
f=16x
Hence
![W=8[x^2]^{3/4}_{0}](https://tex.z-dn.net/?f=W%3D8%5Bx%5E2%5D%5E%7B3%2F4%7D_%7B0%7D)
W=9/2lbft
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