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katen-ka-za [31]
2 years ago
11

The corner section of a football stadium has 6 seats on the first row. Each row after that has an additional 3 seats. How many s

eats would be on the 20th row?
32


63


103


342
Mathematics
1 answer:
julia-pushkina [17]2 years ago
7 0

9514 1404 393

Answer:

  63

Step-by-step explanation:

The number of seats in a row will give an arithmetic sequence:

  6, 9, 12, 15, ...

The first term is 6; the common difference is 3. The general term is ...

  an = a1 +d(n -1) . . . . . . n-th term of sequence with first term a1, difference d

The 20th term of the sequence is ...

  a20 = 6 +3(20 -1) = 6 +57 = 63

There would be 63 seats on the 20th row.

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Evaluate the expression 6/5x when x=20
Sindrei [870]

Answer:

24

Step-by-step explanation:

6 divided by 5 times 20 =24

6/5*20= 24

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2 years ago
PL has endpoints P(4, −6) and L(−2, 1).
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I believe it'd be the one with the smiley emoji
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Bus stop Restaurant sell 3/4 as many cheeseburgers as they do with hamburgers on any given day. If there are a total of 42 burge
schepotkina [342]

Answer:

24cb 18 hb

Step-by-step explanation:

Let's call C the number of cheese burger sold and H the number of hamburgers.

As the cheeseburger sold are 3/4 of hamburgers, we can represent with the following equation:

C= (3/4)H (equation 1)

The total of burgers sold are the total of cheeseburger plus the total of hamburgers, and the sum 42. The equation representing it is:

C + H = 42

Replacing C by its value of eq. 1:

(3/4)H + H = 42

(3/4)H + (4/4)H = 42

(7/4) H = 42

Dividing both sides by (7/4)

H = 42/(7/4)

H= 42*(4/7)

H= (42/7)*4

H= 6*4

H=24

Then, C is:

C=(3/4)H

C=(3/4)*24

C= 3*24/4

C=72/4

C=18

We can verify it summing C+H

C+H=24+18=42 verified!

6 0
3 years ago
PLEASE HELP!! WILL REQARD A LOT
Soloha48 [4]

9514 1404 393

Answer:

  17,500

Step-by-step explanation:

The problem can be written as a proportion:

  hits/revenue = 1000/(£2) = x/(£35)

Multiplying by £35 gives ...

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3 0
2 years ago
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you currently have 24 credit hours and a 2.8 gpa you need a 3.0 gpa to get into the college. if you are taking a 16 credit hours
Juliette [100K]

Answer:

\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0

And we can solve for \sum_{i=1}^n w_f *X_f and solving we got:

3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f

And from the previous result we got:

3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f

And solving we got:

\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8

And then we can find the mean with this formula:

\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

Step-by-step explanation:

For this case we know that the currently mean is 2.8 and is given by:

\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8

Where w_i represent the number of credits and X_i the grade for each subject. From this case we can find the following sum:

\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0

And we can solve for \sum_{i=1}^n w_f *X_f and solving we got:

3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f

And from the previous result we got:

3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f

And solving we got:

\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8

And then we can find the mean with this formula:

\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

6 0
2 years ago
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