Select the correct answers.

A jar contains sugar. The jar and the sugar have a total weight of 850 g. Anna uses 2/3 of the sugar. The jar and the sugar now

maria [59]

Answer:

$J = 370g$

Step-by-step explanation:

Given

Represent the weight of the Jar with J and the sugar with S

Initially, we have:

$J + S = 850g$

After 2/3 of sugar is removed we have:

$J + S -\frac{2}{3}S= 530g$

Required

Determine the weight of the jar

$J + S = 850g$ --- (1)

$J + S -\frac{2}{3}S= 530g$ --- (2)

Simplify (2)

$J + S -\frac{2S}{3}= 530g$

Take LCM

$J + \frac{3S - 2S}{3}= 530g$

$J + \frac{S}{3}= 530g$

Make S the subject in (1)

$J + S = 850g$

$S = 850g - J$

Substitute 850g - J for S in $J + \frac{S}{3}= 530g$

$J + \frac{850g - J}{3}= 530g$

Multiply through by 3

$3 * J + 3*\frac{850g - J}{3}= 530g * 3$

$3J + 850g - J= 1590g$

Collect Like Terms

$3J - J= 1590g-850g$

$2J= 740g$

Make J the subject

$J = \frac{1}{2} * 740g$

$J = 370g$

The jar weighs 370g

5 0

1 year ago

Read 2 more answers

The data set below shows the number of books checked out from a library during the first two weeks of the month: 10, 89, 80, 95,

kompoz [17]

I am pretty sure your answer is going to be B, There are two outliers, indicating very few books were rented out on those two days. I hope this helps!! :)

5 0

1 year ago

Read 2 more answers

Thomas invested $8,500 for one year. Part of the money was invested at6% and the rest at 9%. The total interest earned was $667.

oee [108]

Okay so say that x=amount invested with 6% and y=amount invested with 9%
x+y=8,500 so > x=8500-y
6%=0.06 9%=0.09
0.06x +0.09y=667.5 (Substitute in x=8500-y so only numbers and y)
0.06(8500-y)+0.09y=667.5 (expand brackets)
510-0.06y+0.09y=667.5 (-510)
0.03y=117.5 (/0.03)
$3916.67=Y ->9%
X=8500-Y
x=$4583.33 ->6%

4 0

2 years ago

Half of j minus 5 is the sum of k and 13

barxatty [35]

.5j - 5 = k + 13. Is that what you want?

4 0

2 years ago

A photocopy machine can reduce copies to 80% of their original size. By copying an already reduced copy, further reductions can

alukav5142 [94]

Part a) if a page is reduced to 80%, what percent enlargement is needed to return it to its original size?

Let

x---------> the percent enlargement

we know that

the original size is the 100%

so

x*80%=100%

x=(100%/80%)

x=1.25--------> 1.25=(125/100)=125%

therefore

the answer Part a) is

the percent enlargement is 125%

Part b) Estimate the number of times in succession that a page must be copied to make the final copy less than 15% of the size of the original

we know that

A photocopy machine can reduce copies to 80% of their original size

so

Copy N 1

0.80*100%=80%

Copy N 2

0.80*80%=64%

Copy N 3

0.80*64%=51.2%

Copy N 4

0.80*51.2%=40.96%

Copy N 5

0.80*40.96%=32.77%

Copy N 6

0.80*32.77%=26.21%

Copy N 7

0.80*26.21%=20.97%

Copy N 8

0.80*20.97%=16.78%

Copy N 9

0.80*16.78%=13.42%-------------> 13.42% < 15%

therefore

the answer Part b) is

the number of times in succession is 9

6 0

2 years ago