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Verizon [17]
4 years ago
9

During the baseball season, marvins team won 5 games and lost 14 games

Mathematics
1 answer:
777dan777 [17]4 years ago
7 0
The answer is 9 games
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FICA stands for
Alja [10]

Answer:

A.Federal insurance Contributions Act

Step-by-step explanation:

Please mark me as brainleist

3 0
3 years ago
A like has two-fold Symmetry
Vikki [24]

Answer:

yes

Step-by-step explanation:


5 0
3 years ago
Harvard University accepts 6 students for every 100 applicants. How many students will be accepted if 850 applicants apply for a
patriot [66]

Answer:

51 accepted

Step-by-step explanation:

Well, if 6 are accepted out of 100, that's 6% or 0.06. To find how many are accepted out of 850, multiply 850 by 0.06.

4 0
3 years ago
Read 2 more answers
If gas prices are now $3.75 and they were up 10% from 6 months ago, how much were gas prices 6 months ago?
Hatshy [7]

Given

gas prices are now $3.75 and it up 10% from 6 months ago.

Find out how much were gas prices 6 months ago.

To proof

gas prices now = $3.75

let the gas price 6 month ago = x

gas price now is rise up 10% from 6 months ago

First write 10% in simpler form

= \frac{10}{100}

= 0.1

Now the equation becomes

0.1x + x = 3.75

1.1x = 3.75

x =\frac{3.75}{1.1}

x = 3.41 (approx)

Thus gas prices 6 months ago be $ 3.41 (approx )

Hence proved

7 0
3 years ago
What is the sum of the first 70 consecutive odd numbers? Explain.
expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N

So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

5 0
3 years ago
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