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sergeinik [125]
2 years ago
11

I need help again please 5/6-1/6

Mathematics
1 answer:
Murljashka [212]2 years ago
6 0
The answer is 2/3 in lowest terms.
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A box of chocolates includes 32 chocolates. If 1/4 contain nuts, how many chocolates do not contain nuts?
const2013 [10]

Answer: There should be 24 chocolates without nuts.

Step-by-step explanation:

32 divided by 4 is 8. 8 Chocolates will have nuts, and that leaves 24 without.

7 0
3 years ago
Johns family has $12 to spend after buying tickets to the Aquarium. Parking costs $3 an hour.
TEA [102]

Answer:

4 hours

Step-by-step explanation:

4 x 3 =12 or 12÷3 = 4

7 0
2 years ago
Jose now has $500. How much would he have after 6 years if he leaves it invested at 5.5% with annual compounding?
arsen [322]
345.78 is the answer
6 0
3 years ago
b. Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box.
Ainat [17]

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

<u>Step-by-step explanation:</u>

Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:

We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.

⇒ 2x

The third box contained two more pounds than the first box i.e.

⇒ x+2

The last box contained one-fourth the amount in the second box i.e.

⇒ (\frac{1}{4})2x = \frac{x}{4}

Therefore , Sum of pounds of candy are :

⇒ \frac{x}{2} +x+2+2x+x=65

⇒ \frac{x}{2} +4x=63

⇒ \frac{9x}{2}=63

⇒ x=63(\frac{2}{9} )

⇒ x=14

Therefore , Candy in each box is :

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

8 0
3 years ago
Rite the first ten terms of a sequence whose first term is -10 and whose common difference is -2.
Lelechka [254]
-10, -10-2·1= -12, -10-2·2=-14, -10-2·3=-16, -10-2·4= -18, -10-2·5= -20,
 -10-2·6= -22, -10-2·7 =-24, -10-2·8=-26, -10-2·9=-28

a=-10
d=-2
xn=a+d(n-1)

the first ten terms are
-10, -12, -14, -16, -18, -20, -22, -24, -26, -28

8 0
3 years ago
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