Answer:
<em>Rajan must use </em><em>0.75</em><em> liters of 5% hydrochloric acid solution and </em><em>0.25</em><em> liters of 45% hydrochloric acid solution.</em>
Step-by-step explanation:
Let us assume that, x liters of the 5% hydrochloric acid and y liters of the 45% hydrochloric acid solutions are combined.
As Rajan need total of 1 liter of solution, so
i.e 
--------------------1
As Rajan needs 5% hydrochloric acid and 45% hydrochloric acid to make a 1 liter batch of 15% hydrochloric acid, hence acid content of the mixture of two acids will be same as of the final one, so
i.e 
-------------2
Putting value of x from equation 1 in equation 2,
Putting the value of y in equation 1,
Therefore, Rajan must use 0.75 liters of 5% hydrochloric acid solution and 0.25 liters of 45% hydrochloric acid solution.
Answer:
$61,500
Step-by-step explanation:
For this problem, we will simply take her hourly wage and multiply that by the hours she worked and take her additional earnings for each loan processed and multiply that by the number of loans she processed.
Consider her hourly wage of $15 per hour. If she worked 2000 hours, then we can say she got paid ($15 * 2000) = $30000 for her work that year.
Consider her processing fee of $35 per loan processed. If she processed 900 loans, then we can say she got paid ($35 * 900) = $31500 for her processed loans that year.
Then, to find her yearly total earnings, we simply add those two values. $30000 + $31500 = $61500.
Thus the loan officer made $61500 that year.
Cheers.
IF BREADTH=b meters
THEN LENGTH=4 meter less than 3 times than the breadth of the hall.
LENGTH=3b-4 meters
the length of the rectangular hall is=3b-4 meters
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:
And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
