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Ameeta finished 12 math problems. David finished 3 times as many. How many more math problems did David finish than Ameeta?

Volgvan

Answer:
david finished 24 problems more than ameetha
Step-by-step explanation:
problems finished by ameetha = 12 problems
problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36
problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36the problems finished by David more than ameetha =
problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36the problems finished by David more than ameetha =36-12= 24
problems finished by ameetha = 12 problemsproblems finished by David = 3 times more than ameethas problem = 3*12 = 36the problems finished by David more than ameetha =36-12= 24 so David finished 24 problems more than ameetha

8 0

3 years ago

A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

3 years ago

What is the midpoint of the line segment with endpoints (-1, 3) and (-5,5)?

Debora [2.8K]

Step-by-step explanation:

it is the ans to the midpoint.

7 0

3 years ago

From a candy machine a random handful of colored chocolate candies are dispensed with 2 quarters. In one turn the machine dispen

Gemiola [76]

Answer:

the answer should be 1/11

Step-by-step explanation:

p(blue first) = 3/12

p(yellow second) = 4/11

so p = 3/12 x 4/11

1/11 is your probability

8 0

2 years ago

The price of a computer component is decreasing at a rate of 15% per year.

Alex_Xolod [135]

Answer:

exponential
$30.71

Step-by-step explanation:

Usually, this wording means the decrease is proportional to the current price, characteristic of an exponential function.

__

The decay factor is 1-(decay rate), so is 1-0.15 = 0.85. After 3 years, the value has been multiplied by this factor 3 times:

$50×0.85³ ≈ $30.71 . . . . cost in 3 years

_____

Comment on percent per year

Less often, the indicated decrease is intended to be that percentage of the original price. The result is that the price decreases by a constant amount each year, a linear decrease. This condition most often arises in conjunction with figuring depreciation in value for tax or accounting purposes.

The upshot is that you always need to be careful to understand what the base of a percentage is intended to be.

5 0

3 years ago