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zhenek [66]
2 years ago
15

heather is training for a long-distance run. her data points listed below represent the days of practice, x, and the number of m

iles run, y.(1, 2.5), (2, 4.2), (4, 5.6), (6, 7), (8, 8.1), (10, 11)use the equation to interpolate the value and estimate the distance that she could have run on day 3. round to the nearest tenth of a mile.
Mathematics
2 answers:
MAXImum [283]2 years ago
3 0
Interpolation equation
Poit 1: (a,b)
Point 2: (c,d)
Intermediate point (x,y)

(d - b)/(c - a) = (y - b) / (x - a)

(a,b) = (2, 4.2)
(c,d) = (4, 5.6)

x = 3

(5.6 - 4.2) / (4 - 2)  =  (y - 4.2) / (3 -2)

y =  [1.4/2]*[1] + 4.2

y = 0.7 + 4.2 = 4.9

Answer 4.9 miles

 




Airida [17]2 years ago
3 0

Answer:

4.6

Step-by-step explanation:

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Lionel is planning a one-day outing.
pychu [463]

Answer:

Option B - y=5x+40 and  y=3x+60

Step-by-step explanation:

Given : The Thrill amusement park charges an entry fee of $40 and an additional $5 per ride, x. The Splash water park charges an entry fee of $60 and an additional $3 per ride, x.

To find : Which system of equations could be used to determine the solution where the cost per ride of the two amusement parks, y, is the same?

Solution :

Let x be the number of rides and

y be the cost per ride.

According to question,

The Thrill amusement park charges an entry fee of $40 and an additional $5 per ride.

The equation form is y=40+5x

The Splash water park charges an entry fee of $60 and an additional $3 per ride.

The equation form is y=60+3x

Therefore, The required system of equations form are

y=5x+40 and  y=3x+60

So,Option B is correct.

6 0
2 years ago
Read 2 more answers
"The Highest Common Factor (HCF) of my two numbers is 3 The Lowest Common Multiple (LCM) of my two numbers is 45" (b) Write down
Mila [183]

Answer:

Step-by-step explanation:

15 and 3

6 0
2 years ago
A boat travels at constant speed. After 20 minutes the boat had traveled 2.5 miles. How long does it take the boat to reach the
schepotkina [342]

120 minutes

Step by step:

divide 20 by 2.5 to find how many minutes it takes to drive 1 mile, then multiply that by 15 miles

7 0
2 years ago
A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3
butalik [34]

Answer:

The answer is below

Step-by-step explanation:

A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3 miles.

a) Start by drawing a diagram of this situation. Your diagram might include a coordinate plane, the cell tower, and a circle representing the cell tower's coverage boundary.

a) A house is located 5 miles north of the center of the town and is to the east of the cell tower. If the house lies on the boundary of the cell tower's coverage, how far east of the center of the town is the house?

Answer:

Let the center of the town represent the origin (0,0). Also 1 unit = 1 mile. Since the cell tower is located 3 miles east and 4 miles north of the center of a small town, it is represented by A(3, 4)

The cell tower has a coverage of radius 3 miles. This can be represented by a circle with equation:

(x - a)² + (y - b)² = r². where (a,b) is the center of the circle and r is the radius. Hence:

(x - 3)² + (y - 4)² = 3²

(x - 3)² + (y - 4)² = 9

The diagram is drawn using geogebra.

b) The house is 5 miles north. It can be represented by y = 5 line.

To find the distance east of the house we have to substitute y = 5 and solve for x, hence:

(x - 3)² + (y - 4)² = 9

(x - 3)² + (5 - 4)² = 9

(x - 3)² + 1 = 9

(x - 3)² = 8

(x - 3) = √8

x - 3 = ±2.83

x = 3 ± 2.83

x = 5.83 or 1.83

Since it is to the east to the cell tower, hence x = 5.83.

Therefore the house is located 5.83 miles to the east of the cell tower

8 0
2 years ago
The rabbit population on a small island is observed to be given by the function P(t) = 130t − 0.3t^4 + 1100 where t is the time
Montano1993 [528]
The maximum occurs when the derivative of the function is equal to zero.
P(t)=-0.3t^{4}+130t+1100 \\ P'(t)=-1.2t^{3}+130 \\ 0=-1.2t^{3}+130 \\ 1.2t^{3}=130 \\ t^{3}= \frac{325}{3}  \\ t=4.76702
Then evaluate the function for that time to find the maximum population.
P(t)=-0.3t^{4}+130t \\ P(4.76702)=-0.3*4.76702^{4}+130*4.76702+1100 \\ P(4.76702)=1564.79201
Depending on the teacher, the "correct" answer will either be the exact decimal answer or the greatest integer of that value since you cannot have part of a rabbit.
7 0
2 years ago
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