Pablo's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Pablo $5.50 per pound, and type B coffee costs $4.40 per pound. This month, Pablo made 174 pounds of the blend, for a total cost of $869.00. How many pounds of type B coffee did he use?
Number of pounds of type B coffee:

We know blend A costs 5.50 and blend B costs 4.30. We know that Pablo made 153 pounds of coffee this month. Let's let A=amount of blend A used and B=amount of blend B used. Since he made 153lbs coffee, we know that

A+B=153.

But he spent 756.30. The total amount he spent on a blend of coffee is the price * amount used. So he spent 5.50*A on blend A and 4.30*B on B, so in total, he spent 5.50*A+4.30*B in coffee. But this must add to 756.30! So

5.5A+4.3B = 756.30

Now we have two equations:

A+B = 153

5.5A + 4.3B = 756.30

We need to solve for B. I hate decimals so let's get rid of them in the second equation by multiplying both sides of the equation by 100 (which simply moves the decimals on everything two to the right). So now we have

A+B = 153

550A + 430 B = 75630

Now let's multiply the first equation by 550 (so we can eventually get rid of pesky A). So multiply the left and right side of the first equation by 550, so we have

550A + 550B = 84150

550A + 430B = 75630

Now subtract the second equation from the first, notice that since both have 550A, they cancel, so we obtain

120B = 8520

Now divide by 120 into both sides to solve for B, we get

B= 8520/120 = 71

So Pablo used 71lbs of blend B.

Step-by-step explanation:

Hope this helps you :D

24/56=hx^7

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;