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Rudiy27
4 years ago
7

What operation would I do to find the average of something?

Mathematics
1 answer:
Lady bird [3.3K]4 years ago
5 0
Sum divided by count. Add up all the numbers and divide it by how many numbers it is.
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What’s an injective function?
Tanya [424]

Answer:

an Injective function is a function that maps distinct elements of its domain to distinct elements of its codomain. In other words, every element of the function's codomain is the image of at most one element of its domain.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Account A: $300 monthly deposits, interest rate is 4.2%. Account B: $250 monthly deposits, interest rate is 5.1%. Which account
Elan Coil [88]

Answer:

A

Step-by-step explanation:

We have to determine the future value of the annuity to determine which account has a greater value

Future value = Amount x annuity factor

annuity factor = Annuity factor = {[(1+r)^n] - 1} / r

Account A = 300 x[ (1.042)^15 - 1 ] / 0.042 = $6097.14

Account B = 250  x[ (1.051)^15 - 1 ] / 0.051 = $5435,42

Account A will be greater

4 0
3 years ago
Round to the nearest whole.<br> 782.36
ratelena [41]
The answer would be 782 because you would round down.
6 0
3 years ago
Read 2 more answers
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
The experimental probability of rain in a certain
devlian [24]

Answer:

18 days

Step-by-step explanation:

Find how many days it can be expected to rain by finding 60 percent of 30

30(0.6)

= 18

So, one can expect it to rain for a total of 18 days out of the 30 days

5 0
3 years ago
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