From the question, we know that the solutions of the system
![(s,c)](https://tex.z-dn.net/?f=%28s%2Cc%29)
is (14,6), which means the speed of the the boat in calm water,
![s](https://tex.z-dn.net/?f=s)
, is 14
![\frac{km}{h}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
, and the speed of the current,
![c](https://tex.z-dn.net/?f=c)
, is 6
![\frac{km}{h}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
. To summarize:
![s=14 \frac{km}{h}](https://tex.z-dn.net/?f=s%3D14%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
and
![c=6 \frac{km}{h}](https://tex.z-dn.net/?f=c%3D6%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
We also know that w<span>hen the boat travels downstream, the current increases the speed of the boat; therefore to find the speed of the boat traveling downstream, we just need to add the speed of the boat and the speed of the current:
</span>
![Speed_{downstream} =s+c](https://tex.z-dn.net/?f=Speed_%7Bdownstream%7D%20%3Ds%2Bc)
![Speed _{downstream} =14 \frac{km}{h} +6 \frac{km}{h}](https://tex.z-dn.net/?f=Speed%20_%7Bdownstream%7D%20%3D14%20%5Cfrac%7Bkm%7D%7Bh%7D%20%2B6%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
![Speed_{downstream} =20 \frac{km}{h}](https://tex.z-dn.net/?f=Speed_%7Bdownstream%7D%20%3D20%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
<span>
Similarly, to find the the speed of the boat traveling upstream, we just need to subtract the speed of the current from the speed of the boat:
</span>
![Speed_{upstream} =s-c](https://tex.z-dn.net/?f=Speed_%7Bupstream%7D%20%3Ds-c)
![Speed_{upstream} =14 \frac{km}{h} -6 \frac{km}{h}](https://tex.z-dn.net/?f=Speed_%7Bupstream%7D%20%3D14%20%5Cfrac%7Bkm%7D%7Bh%7D%20-6%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
![Speed_{upstream} =8 \frac{km}{h}](https://tex.z-dn.net/?f=Speed_%7Bupstream%7D%20%3D8%20%5Cfrac%7Bkm%7D%7Bh%7D%20)
<span>
We can conclude that the correct answer is </span><span>
C. The team traveled at 8 km per hour upstream and 20 km per hour downstream.</span>
No. of gallons required = csa of cuboid+area of rectangle/300
= 2(l+b)h + l*b /300
= 2(11+12)8 +11 *12 /300
=368+132/300
= 500/300
=5/3 =1.666.... =1.7
= 2 approx.
therefore 2 gallons should be used
Answer: A: there is not enough evidence to support a relationship between lunch preference and role at school
Answer: 14:3
Step-by-step explanation:
Given
Ratio between students(s) to female(f) adults is 7:6
while ratio between female(f) to male(m) is 4:1
So, we can write
![\Rightarrow \dfrac{s}{f}=\dfrac{7}{6}\quad \ldots(i)](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdfrac%7Bs%7D%7Bf%7D%3D%5Cdfrac%7B7%7D%7B6%7D%5Cquad%20%5Cldots%28i%29)
Also, we can write
![\Rightarrow \dfrac{f}{m}=\dfrac{4}{1}\quad \ldots(ii)](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdfrac%7Bf%7D%7Bm%7D%3D%5Cdfrac%7B4%7D%7B1%7D%5Cquad%20%5Cldots%28ii%29)
Multiply both equations
![\Rightarrow \dfrac{s}{f}\times \dfrac{f}{m}=\dfrac{7}{6}\times \dfrac{4}{1}\\\\ \Rightarrow \dfrac{s}{m}=\dfrac{14}{3}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdfrac%7Bs%7D%7Bf%7D%5Ctimes%20%5Cdfrac%7Bf%7D%7Bm%7D%3D%5Cdfrac%7B7%7D%7B6%7D%5Ctimes%20%5Cdfrac%7B4%7D%7B1%7D%5C%5C%5C%5C%20%5CRightarrow%20%5Cdfrac%7Bs%7D%7Bm%7D%3D%5Cdfrac%7B14%7D%7B3%7D)
So, the ratio of the students to male adults is 14:3 .
Begin by using simultaneous equations to find the values of x and y.
y=-0.38x+56.6
y=0.38x+43.4
-0.38x+56.6=0.38x+43.4
-0.38x-0.38x=-56.6+43.4
-0.76x=-13.3
x=-13.3/-0.76
x=17.5
y=0.38(17.5)+56.6
y=63.25
Knowing that the initial year was 1968, add 17.5 years to find to find out when the same percent of men and women received a bachelor's degree.
1968+17.5= 1985.5
So, in the year 1985 and half, both men and women got the same number of bachelor's degrees.
Hope I helped :)