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givi [52]
3 years ago
11

Which intervals are most appropriate for a histogram

Mathematics
1 answer:
Lorico [155]3 years ago
5 0

Answer:

C. 10–15, 15–20, 20–25, 25–30, 30–35

Step-by-step explanation:

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4/10 means 4 tenths (4 times 1/10)
4/10
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Simplify the expression 13. 8+(-11. 5). Use pencil and paper. How are 13. 8 + (-11. 5) and (-13. 8) +11. 5
VashaNatasha [74]

On simplifying the expression 13.8+ (-11.5) , we get 2.3 . And additions

13.8 + (-11.5) and (-13.8) +  11.5 are numbers will same value and opposite sign.

<h3>What are rules for adding a positive and a negative number?</h3>

For adding a negative and a positive number, use the sign of the larger number and subtract.

For example: (–8) + 2 = -6

Given that we have to solve 13.8 + (-11.5) = 13.8-11.5 (we subtract smaller no from larger and will use the sign of lager number that is + here)

13.8+(-11.5) = 13.8-11.5 =2.3

13.8+(-11.5) simply means that 13.8 -11.5 which is subtraction of 11.5 from 13.8

but (-13.8) +  11.5 = -13.8 +11.5  =  -( 13.8 -11.5) [taking -1  common from the expression]

Therefore , -13.8 + 11.5 has same value as 13.8- 11.5 but opposite  in sign.

Learn, More about addition of numbers from here:'

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7 0
1 year ago
At time, t=0, Billy puts 625 into an account paying 6% simple interest. At the end of year 2, George puts 400 into an account pa
svetoff [14.1K]

Answer:

26

Step-by-step explanation:

Given that:

At time, t=0, Billy puts 625 into an account paying 6% simple interest

At the end of year 2, George puts 400 into an account paying interest at a force of interest, 1/(6+t), for all t ≥ 2.

If both accounts continue to earn interest indefinitely at the levels given above, the amounts in both accounts will be equal at the end of year n. Calculate n.

In order to calculate n;

Let K constant to be the value of time for both accounts

At  time, t=0, the value of time K when Billy puts 625 into an account paying 6% simple interest is:

K = 625 \times (1+ 0.06 K)

K = 625 +37.5 K

At year end 2; George  amount of 400 will grow at a force interest, then the value of  K = 400 \times e^{\int\limits^2_k {\dfrac{1}{6+t}} \, dx }

K =400 \times \dfrac{6+K}{6+2}

K =400 \times \dfrac{6+K}{8}

K =50 \times ({6+K})

K =300+50K

Therefore:

If K = K

Then:

625 + 37.5 = 300 +50 K

625-300 = 50 K - 37.5 K

325 = 12.5K

K = 325/12.5

K = 26

the amounts in both accounts  at the end of year n = K = 26

7 0
3 years ago
PLSS HELP IF YOU TURLY KNOW THISS
victus00 [196]

Answer:

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Step-by-step explanation:

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The new fraction would be 48/46 or 1 2/46 or 1 1/23
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