Answer:
There are 585 adults and children
Step-by-step explanation:
Let the number of adults be a, number of children be c and the number of seniors be a
Amount made per group;
adults; 52 * a = 52a
Children : 26 * c = 26c
Seniors = 20 * s = 20s
Adding all will give 20,490
52a + 20s + 26c = 20 490 ••••(i)
Now let us work with the ratios;
a : s = 6 : 1
a/s = 6/1
a = 6s •••••(ii)
Lastly;
a/c = 4/9
4c = 9a ••••(ii)
We want to get a + c
From the first equation , let’s substitute
52(6s) + 20s + 26c = 20,490
26c = 6.5 (4c)
but 4c = 9a; 6.5(9a)
But a = 6s
So we have; 6.5(9)(6s) = 351s
so we have;
312s + 351s + 20s = 20,490
683s = 20,490
s = 20490/683
s = 30
Recall;
a = 6s = 6 * 30 = 180
4c = 9a
4c = 9 * 180
c = (9 * 180)/4 = 405
So the total number of children and adult is a + c
405 + 180 = 585
Answer:
Below.
Step-by-step explanation:
To find the number of students in each grade you multiply the marks by the frequency density. The number of students is given by the area under the rectangles.
So to find the number of students with grade U (< 160) it is:
120 * 0.2 + 40*0.5
= 24 + 20
= 44.
The number for Grade E
= (200 - 160) * 0.5
= 40 * 0.5
= 20.
The others can be found in the same way.
(f∘g)(x) is equivalent to f(g(x)). We solve this problem just as we solve f(x). But since it asks us to find out f(g(x)), in f(x), each time we encounter x, we replace it with g(x).
In the above problem, f(x)=x+3.
Therefore, f(g(x))=g(x)+3.
⇒(f∘g)(x)=2x−7+3
⇒(f∘g)(x)=2x−4
Basically, write the g(x) equation where you see the x in the f(x) equation.
f∘g(x)=(g(x))+3 Replace g(x) with the equation
f∘g(x)=(2x−7)+3
f∘g(x)=2x−7+3 we just took away the parentheses
f∘g(x)=2x−4 Because the −7+3=4
This is it
g∘f(x) would be the other way around
g∘f(x)=2(x+3)−7
now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them.
g∘f(x)=2x+6−7
Next, +6−7=−1
g∘f(x)=2x−1
Its a lts easier than you think!
Hope this helped
Answer:
I am pretty sure it is dependent
Let me know if it is wrong.
