Percent is calculation of quantity per 100 quantity of other thing. The percentage of bikes out of of total surveyed vehicles was 13%
Suppose the value of which a thing is expressed in percentage is "a'
Suppose the percent that considered thing is of "a" is b%
Then since percent shows per 100 (since cent means 100), thus we will first divide the whole part in 100 parts and then we multiply it with b so that we collect b items per 100 items(that is exactly what b per cent means).
Thus, that thing in number is
Assume that total T number of vehicles were surveyed.
Then, as it is given that 70% of the vehicles were cars and 17% were buses.
So, number of cars =
Similarly, the number of buses =
Thus, number of bikes = Total vehicles - number of cars and buses
Number of bikes = T - 0.7T - 0.17T = T(1 - 0.87) = 0.13T
Taking its percentage with total vehicles, let it be x% of total vehicles, then
(we could have used short cut that total there is 100%, since number of cars = 70% of total, number of buses = 17% of total, thus
total vehicle's percent being bike = all vehicle (100%) - percent of cars - percent of buses) = (100 - 70 -17)% = 13% )
Thus, the percentage of bikes out of of total surveyed vehicles was 13%
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Answer:
13%
Step-by-step explanation:
There are three groups: bikes, buses, and cars. The percentages of each must add up to 100%.
bikes + buses + cars = 100%
bikes + 17% + 70% = 100%
bikes + 87% = 100%
bikes = 13%
Answer:
2.8
Step-by-step explanation:
Note that 1 rotation = 360°
Divide 1008 by 360 to find number of rotations
rotations = = 2.8
Answer:
Number of cubes that fills the prism=24
Step-by-step explanation:
The number of cube that completely fill the prism=Volume of Prism/Volume of cube with side 1/2 cm=0.5 cm
Volume of the prism=
Volume of Prism =
Volume of the Cube:
Number of cubes that fills the prism:
So, the number of cubes that can fill the prism is '24'
Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]
⇒ a + (9-1)d = 0
⇒ a + 8d = 0 ⇒ a = -8d ...(i)
Now, its 19th term , T19 = a + (19-1)d
= - 8d + 18d [from Eq.(i)]
= 10d ...(ii)
and its 29th term, T29 = a+(29-1)d
= -8d + 28d [from Eq.(i)]
= 20d = 2 × T19
Hence, its 29th term is twice its 19th term
Answer:
Proved below.
Step-by-step explanation:
a9 = a1 + 8d = 0 where a1 = first term and d = common difference.
we need to prove that
a1 + 28d = 2(a1 + 18d
simplifying:-
a1 + 36d - 28d = 0
a1 + 8d = 0 which is what we are given.
Therefore the proposition is true.
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