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xxTIMURxx [149]
2 years ago
5

Forty-four students go on a field trip. The students are divided into groups of 7 students. How many groups of 7 will there be?.

Mathematics
1 answer:
strojnjashka [21]2 years ago
4 0

6

To solve this problem, you need to divide the total number of students going on the field trips with the number of students in each group.

44/7

= 6 R 2

Since, I assume there can <em>only</em> be groups of 7, the answer is 6.

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Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10 Supersc
algol13

Answer:

You need to sum at least 5000 terms

Step-by-step explanation:

Remember what the estimation theorem for alternating series says.

Given a series

s = \sum\limits_{n=0}^{\infty} (-1)^n a_n

An upper bound for the error of the series R_n  is a_{n+1} so

R_n = |s-s_n| \leq a_{n+1}

For this case you want

\frac{1}{2n+1}  \leq \frac{1}{10^4}\\10^4 \leq 2n+1\\(10^4-1 )/2 \leq n

(10^4-1 )/2 is approximately 5000. So you must sum at least 5000 terms.

5 0
3 years ago
Need help with stats!
Brut [27]

Answer:

a) 1,440 ways

b) 59,280 or 64,000

Step-by-step explanation:

a) Aircraft boarding.

8 people, 2 in first class, boarding first, then 8 economy class.

The 2 people in first class board first, but they can board as AB or BA... so 2 ways here.

For the 6 economy class passengers, we have a permutation of 6 out of 6, so 720, as follows:

P(6,6) = \frac{6!}{(6 - 6)!} = 6! = 720

Since the two are independent, we multiply them to have a global number of ways: 2 * 720 = 1,440 different ways for the 8 passengers to board that plane.

b) combination lock.

Here we do have a little problem... the question doesn't specify if the 3 numbers are different numbers of not.  So, we'll calculate both:

Numbers go from 1 to 40 inclusively... so 40 possibilities.

Normally, in a combination lock, the numbers are different, so let's start with that one:

First number: 40 options available

Second number: 39 options available (cannot take the first one again)

Third number: 38 different options (can't take First or Second number again)

Overall, we then have 40 * 39 * 38 = 59,280 different lock combinations.

If we can pick pick the same number twice:

First number: 40 options available

Second number: 40 options available

Third number: 40 options available

Overall 40 * 40 * 40 = 64,000 different lock combinations

8 0
3 years ago
Please show work. Thanks​
3241004551 [841]

Answer:

625

Step-by-step explanation:

So it says its off by 25%, so we have to do 25% of 2500, we get 625 :D

4 0
3 years ago
I do not understand this math problem I have to do. It is 32-6x=20 and I need to find the variable but it's too confusing for me
sineoko [7]

Answer:

The answer to your question x=2.

Step-by-step explanation: So, first you simplify both sides of the equation which, would look like this after, -6x+32=20. After that, you subtract 32 from both sides which will look like this after -6x=-12. Then you would divide each side by -6 which will get x=2. Hope this helps!


4 0
3 years ago
Solve the equation. Determine whether the equation has one solution, no solution, or infinitely many solutions.
Montano1993 [528]
The equation has infinitely many solutions
4 0
3 years ago
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