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Arlecino [84]
4 years ago
12

Candy selling at $6.00 per pound will be mixed with candy selling at $9.00 per pound. How many pounds of the more expensive cand

y are needed to produce a 15-pound mixture that sells for $7.00 per pound?
Mathematics
2 answers:
Gennadij [26K]4 years ago
8 0
Let us call X the number of pounds of $6 candies and Y the number of pounds of $9 candies.
We know that:
X+Y=15 (we are picking a total of 15 pounds)
(6X+9Y)/15=7 (the average of each pound out of those 15 is "valued" $7)

So you can deduce that Y=15-X
Then replace Y in the other equation by 15-X which is:
(6X+9*(15-X))/15=7
(6X+135-9X)/15=7
(135-3X)/15=7
135-3X=7*15
135-3X=105
-3X=-30
X=10

So you can deduce that Y=15-10=5

As a matter of fact you can verify that 10 pounds of the first candies are valued $60 and 5 pounds of the seconds are valued $45 which is a total of 105.
105/15=7 which means that the overall 15 pounds do have an average per-pound-value of $7.
shutvik [7]4 years ago
8 0

Answer:

10

Step-by-step explanation:

look at the answer above for explanation

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Answer: The equation is 16g+6 ,lmk if you need the answer :)

Step-by-step explanation:

4 0
3 years ago
Plz Help, and solve. Show your work. I will give Brainliest. A - 7 = -13 solve and show your work 10X - 8 = 9X + 8 Can a right t
likoan [24]

Answer:

A - 7 = -13

Add 7

A = -6

10X - 8 = 9X + 8

Add 8

10X = 9X + 16

Subtract 9X

X = 16

In a right triangle, where a and b are the shorter sides, and c is the longer side a^2+b^2=c^2

Thus, plug in the values.

12^2+16^2=20^2

144+256=400

400=400.

Because the equation is true, 12, 16, and 20 can be a right triangle

<em>Hope it helps <3</em>

3 0
3 years ago
Read 2 more answers
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Valentin [98]

Answer:

o. k this is a required answere

6 0
3 years ago
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
Writing an equation of a probably given the vertex and focus
White raven [17]

The equation of the vertical parabola in vertex form is written as

y=\frac{1}{4p}(x-h)^2+k

Where (h, k) are the coordinates of the vertex and p is the focal distance.

The directrix of a parabola is a line which every point of the parabola is equally distant to this line and the focus of the parabola. The vertex is located between the focus and the directrix, therefore, the distance between the y-coordinate of the vertex and the directrix represents the focal distance.

p=1-6=-5

Using this value for p and (3, 1) as the vertex, we have our equation

y=-\frac{1}{20}(x-3)^2+1

6 0
1 year ago
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