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deff fn [24]
3 years ago
6

An archery range calculated the average speed of wooden arrows shot during a competition. Which is the best representation of th

e average speed?
Mathematics
1 answer:
GalinKa [24]3 years ago
8 0
To find an average, add all the speeds of each arrow together and divide that number by the total number of arrows shot.

example:
Arrow 1:
20mph
Arrow 2:
23 mph
Arrow 3:
23 mph
= 20+23+23 = 66

66÷ 3= 22 mph on average
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16=5/4(6x+20) <br>what's it's simplest form<br> ​
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-6/5

Step-by-step explanation:

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A school has 200 students and spends $40 on supplies for each student. The principal expects the number of students to increase
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Answer:

\mathbf{S(t)=200(\frac{105}{100})^{x}}

\mathbf{A(t)=40(\frac{98}{100})^{x}}

\mathbf{E(t)=S(t) \cdot A(t)=200(\frac{105}{100})^{x} \cdot 40(\frac{98}{100})^{x}=8000(\frac{10290}{10000})^{x}}

Step-by-step explanation:

<h3>The predicted number of students over time, S(t) </h3>

Rate of increment is 5% per year.  

A function 'S(t)' which gives the number of students in school after 't' years.  

S(0) means the initial year when the number of students is 200.

S(0) = 200  

S(1) means the number of students in school after one year when the number increased by 5% than previous year which is 200.  

S(1) = 200 + 5% of 200 = 200+\frac{5}{100}\time200 = 200(1+\frac{5}{100}) = 200(\frac{105}{100})  

S(2) means the number of students in school after two year when the number increased by 5% than previous year which is S(1)  

S(2) = S(1) + 5% of S(1) = \textrm{S}(1)(\frac{105}{100}) = 200(\frac{105}{100})(\frac{105}{100}) = 200(\frac{105}{100})^{2}  

.  

.  

.  

.  

.  

Similarly \mathbf{S(x)=200(\frac{105}{100})^{x}}  

<h3>The predicted amount spent per student over time, A(t) </h3>

Rate of decrements is 2% per year.  

A function 'A(t)' which gives the amount spend on each student in school after 't' years.  

A(0) means the initial year when the number of students is 40.  

A(0) = 40  

A(1) means the amount spend on each student in school after one year when the amount decreased by 2% than previous year which is 40.  

A(1) = 40 + 2% of 40 = 40-\frac{2}{100}\time40 = 40(1-\frac{2}{100}) = 40(\frac{98}{100})  

A(2) means the amount spend on each student in school after two year when the amount decreased by 2% than previous year which is A(1)  

A(2) = A(1) + 2% of A(1) = \textrm{A}(1)(\frac{98}{100}) = 40(\frac{98}{100})(\frac{98}{100}) = 40(\frac{98}{100})^{2}  

.  

.  

.  

.  

.  

Similarly \mathbf{A(x)=40(\frac{98}{100})^{x}}  

<h3>The predicted total expense for supplies each year over time, E(t)</h3>

Total expense = (number of students) ×  (amount spend on each student)

E(t) = S(t) × A(t)

\mathbf{E(t)=S(t) \cdot A(t)=200(\frac{105}{100})^{x} \cdot 40(\frac{98}{100})^{x}=8000(\frac{10290}{10000})^{x}}

\mathbf{E(t)=8000(\frac{10290}{10000})^{x}}

(NOTE : The value of x in all the above equation is between zero(0) to ten(10).)

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