Beans A= x
Beans B=(110-x)
0.08x+0.32(110-x) = 0.27(110)
Solve for x
0.08x+35.20-0.32x = 29.70
0.08x-0.32x=29.7-35.2
−0.24x=−5.5
X=5.5÷0.24
X=22.9 of Beans A
110−22.9=87.1 of Beans B
This is reasonable if one can weigh beans to the nearest tenth of a pound.
Hope it helps:-)
I found the answer to the same question you have.
Answer:
Mass= 1.479 KG
Step-by-step explanation:
Density of the cuboid ![= 9.86 cm^3](https://tex.z-dn.net/?f=%3D%209.86%20cm%5E3)
Length= 10 cm
Width= 5 cm
Height=3 cm
Volume of the Cuboid is:
Length*Width*Height
Volume:
![= 10*5*3\\\\ = 150 cm^3](https://tex.z-dn.net/?f=%3D%2010%2A5%2A3%5C%5C%5C%5C%20%3D%20150%20cm%5E3)
![Density = Mass(gm)/Volume(cm^3)](https://tex.z-dn.net/?f=Density%20%3D%20Mass%28gm%29%2FVolume%28cm%5E3%29)
![9.86= Mass/150\\\\Mass(gm)=9.86*150](https://tex.z-dn.net/?f=9.86%3D%20Mass%2F150%5C%5C%5C%5CMass%28gm%29%3D9.86%2A150)
Mass=1479 grams
Or, Mass=1479/1000= 1.479 KG
1/8 is least,3/4second,3/16 great
Answer:
![r=3.8\%](https://tex.z-dn.net/?f=r%3D3.8%5C%25)
Step-by-step explanation:
we know that
The simple interest formula is equal to
![I=P(rt)](https://tex.z-dn.net/?f=I%3DP%28rt%29)
where
I is the interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have
![t=(91/365)\ years\\ P=\$2,400\\ I=\$22.74\\r=?](https://tex.z-dn.net/?f=t%3D%2891%2F365%29%5C%20years%5C%5C%20P%3D%5C%242%2C400%5C%5C%20I%3D%5C%2422.74%5C%5Cr%3D%3F)
substitute in the formula above
![22.74=2,400(91/365)r](https://tex.z-dn.net/?f=22.74%3D2%2C400%2891%2F365%29r)
![22.74*365=2,400(91)r](https://tex.z-dn.net/?f=22.74%2A365%3D2%2C400%2891%29r)
![r=22.74*365/[2,400(91)]](https://tex.z-dn.net/?f=r%3D22.74%2A365%2F%5B2%2C400%2891%29%5D)
![r=0.038](https://tex.z-dn.net/?f=r%3D0.038)
![r=3.8\%](https://tex.z-dn.net/?f=r%3D3.8%5C%25)