Astore allows customers to fill their own bags of candy. terri decides she only wants jelly beans and chocolate drops. jelly beans sell for $0.90 per pound, and chocolate drops sell for $0.70 per pound. terri’s bag weighs 5 pounds and it costs $4.10. a. 2 pounds of jellybeans; 3 pounds of chocolate drops b. 4 pounds of jellybeans; 1 pound of chocolate drops c. 1 pound of jellybeans; 4 pounds of chocolate drops d. 3 pounds of jellybeans; 2 pounds of chocolate drops

Option D is correct that she took 3 pounds of jellybeans and 2 pounds of chocolate drops

Step-by-step explanation:

Given :

Terri  wants jelly beans and chocolate drops.

Jelly beans sell for $0.90 per pound

Chocolate drops sell for $0.70 per pound.

Terri’s bag weighs 5 pounds

It costs $4.10.

To Find : How much she bought jellybeans and chocolate drops?

Solution :

Let she took x pounds of jellybeans.

Let she took y pounds of chocolate drops.

Since we are given that her bag weighs 5 pounds .

⇒  ---(A)

Now cost of 1 pound of jelly beans = $0.90

So, Cost of  x pounds of jellybeans. = $0.90 x

Now cost of 1 pound of chocolate drops.  =$0.70

So, Cost of  y pounds of chocolate drops.  = $0.70 y

Since we are given that total cost of her bag of 5 pounds is $4.10

⇒  ---(B)

Solving A and B by using substitution method

Substitute value of x from (A) in (B)

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Thus, she took y = 2 pounds of chocolate drops.

Now to calculate x substitute this value of y in (A)

⇒

⇒

⇒

Thus, she took x = 3 pounds of jellybeans

Hence , Option D is correct that she took 3 pounds of jellybeans and 2 pounds of chocolate drops

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: