7.
lateral surface area of traingular prism=perimeter of traingle ×length=[5+3+4]×6=72cm²
curved surface area=2[area oftraingle]=2×1/2×4×3
=12cm²
now
total surface area=72cm²+12cm²=84cm²
8.
volume of traingular prism=area of traingle ×length
=1/2×12×3.4×4=81.6in³
7.
Find the lateral surface area of the figure below."
qfi
87-
10 ft
4 ft7.
Find the lateral surface area of the figure below."
qfi
87-
10 ft
4 ft
Answer: minimum = 0.5
maximum = 9.0
first quartile = 3
median = 5.5
third quartile = 7.5
Step-by-step explanation:
The histogram is shown below and used to obtain the various information
Time (hour) (X) No of People (F) FX
0.5 19 9.5 -- minimum = 0.5
1 20 20
1.5 21 31.5
2 22 44
2.5 23 57.5
3 24 72 ----first quartile = 3
3.5 26 91
4 27 108
4.5 28 126
5 29 145
5.5 30 165 --median/ inter quartile =5.5
6 31 186
6.5 32 208
7 33 231
7.5 34 255-- third quartile = 7.5
8 35 280
8.5 36 306
9 37 333 --- maximum = 9
T=85.5 T = 507 T= 2666.82 T = TOTAL
mean = sumFX/sumF = 2666.82/507 = 5.26
Median = middle number = inter quartile =5.5 (507/2 = 253.5 = 254 and add the frequency from top = 19+20+21+ 22+ 23+ 24+26+ 27+ 28+ 29 =269
and 254 close to 269 that is the frequency is within the range and corresponds to 5.5
first quartile = 1/4 *507 = 126.75 = 127 ( add the frequency from top= 19+20+21+ 22+ 23+ 24= 129 and 127 is close to 129)
third quartile = 3/4 * 507 = 380.25 ( add the frequency from top = 19+20+21+ 22+ 23+ 24+26+ 27+ 28+ 29+ 30+ 31+32+ 33+ 34 = 399 and 380 is close to 399 so the third quartile is 7.5
minimum is the one with lowest frequency = 0.5 at 19
maximum or mode is the one with highest frequency = 9 at 37
Answer:
The 0 would most likely effect the mean the most. This is because the mean accounts for value of the numbers. But the median, however, just depends on how much of the numbers there are. Although both will decrease, the mean will increase the most. The median will just go down slightly. While the mean will take tremendous damage.
Apologies if it's wrong and have a good day!~