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cestrela7 [59]
3 years ago
15

Please help me!! A 20% solution of fertilizer is to be mixed with a 80% solution of fertilizer in order to get 150 gallons of a

60% solution. How many gallons of the 20% solution and 80% solution should be mixed?
Mathematics
2 answers:
dangina [55]3 years ago
7 0
Let n be the amount of 80% solution; and 150-n be the amount of 20% solution. Then:
.8n+.2(150-n)=.6(150)
.8n+30-.2n=90
.6n=60
n=100
150-n=50
☺☺☺☺
ahrayia [7]3 years ago
6 0
Hello,


Let x represent the number of gallons of 20% solution of fertilizer to be used be.


Let y represent the number of gallons of 80% solution of fertilizer to be used be.


You want to find out how many gallons of the 20% solution and 80% solution should be mixed. So, as the solution is being mixed, the final volume (gallons) adds up.


First to write an equation, change the percent to decimal form:


20%=.20
80%=.80

We now have:


x gallons of a 20% solution contains .20x gallons of fertilizer.
y gallons of an 80% solution contains .80y gallons of fertilizer.

To find how many gallons total will result after both solutions being mixed, write:


.20x+.80y = gallons of fertilizer 


Amount of fertilizer in 150 gallons of 60% solution of fertilizer:

.60*150=90 gallons of fertilizer.


Thus, .20+.80=90 is your second equation


Now, to find how many gallons of the 20 percent solution and 80 percent solution should be mixed in order to get 150 gals solve for your system of equations:


 x+y=150
.20x+.80y=90


Use the Substitution Method (if you want):

x=50

Substitute x=50 into the first equation:


50+y=150


Subtract 50 from both sides of the equation; 50 cancels out.

We now have y=100


Thus, when x=50 y=100


Therefore, 100 gallons of the 20% solution and 80% solution should be mixed in order to get 150 gallons of a 60% solution.


Faith xoxo






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Answer:

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Let X be the random variable that represents a row score, we find the values we are seeking in the following way

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you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number  x_{1} such that

P(X\leq x_{1}) = \int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9, this number is x_{1}=42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

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you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

P(41\leq X\leq 44)=\int\limits^{44}_{41} {f(x)} \, dx = 0.09

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

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