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Vesnalui [34]
2 years ago
9

the table below shows the total cost of attendance to a basket ball game, y, for x students. which explains how the table can be

used to predict the cost of any number of students attending the game?​

Mathematics
1 answer:
maxonik [38]2 years ago
3 0

Answer:

Hi there it's simple we can use direct proportion for predicting like that

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Help help help help
deff fn [24]

Answer:

b = 12

Step-by-step explanation:

b/3 + 4 = 8

subtract 4 on both sides

b/3 = 4

multiply 3 with 4

b = 12

4 0
2 years ago
If you plant 48 tulips in 12 minutes how many tulips did you plant per minute
8090 [49]
The answer is 8 because in a minute there is 60 seconds so you divide 48 by 60 and it gives you 8
8 0
3 years ago
A cafe sells puffs. 5/9 are beef and 1/3 are chicken. The remaining 18 are tuna. Calculate 1. What fraction of the puffs are tun
Ksivusya [100]
Let n be the total number of puffs. Now we can write:
n=\frac{5}{9}n+\frac{1}{3}n+18
Adding the fractional value of n, we get:
n=\frac{5+3}{9}n+18
Simplifying and rearranging gives us:
n-\frac{8}{9}n=18
Therefore we can simplify to get:
\frac{1}{9}n=18
and finally
n=18\times9=?
The fraction of the puff that are tuna is found from:
\frac{18}{18\times9}=\frac{1}{9}
7 0
3 years ago
Read 2 more answers
Please answer <br> ( will give brainlst)
serg [7]

1. cluster

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3. association

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5 0
2 years ago
Read 2 more answers
6. If the net investment function is given by
Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated  capital for the period.

  • (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

  • (b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.

Reasons:

(a) The given investment function is presented as follows;

I(t) = 100 \cdot e^{0.1 \cdot t}

(a) The capital formation is given as follows;

\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot  e^{0.1 \cdot t}} + C

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore,  for the three years; Year 3, year 4, and year 5, we have;

\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

\displaystyle  \mathbf{\left[1000 \cdot  e^{0.1 \cdot t}} + C \right]^t_0} = 100,000

Which gives;

\displaystyle 1000 \cdot  e^{0.1 \cdot t}} - 1000 = 100,000

\displaystyle \mathbf{1000 \cdot  e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000

\displaystyle e^{0.1 \cdot t}} = 101

\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

6 0
2 years ago
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