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Nadusha1986 [10]
3 years ago
6

Mya is a performance athlete. She wants to use a function to represent the number of calories that she burns while training, whe

re y
represents the total calories burned and a represents her training time in minutes. Mya estimates that the exponential function
y = 45.30.0548.2 is the best fit for the data. Given the following table, what is the actual number of calories that Mya burned when she
trained for 30 minutes? Round your answer to the nearest whole number.
I Need help plz

Mathematics
1 answer:
Lerok [7]3 years ago
4 0

Answer:

The actual number of calories that Mya burned when she

trained for 30 minutes is 240.

Step-by-step explanation:

We have an exponential function that is the best fit for the number of calories Mya burns, in function of time (minutes).

The function is:

y=45.3e^{0.0548x}

Where y: calories burned, and x: training time.

Whenever we use regression models, we have a real value and a predicted value. The difference between them, the measure we want to minimize when we adjust with this typo of models, is called residual.

In the table is shown as one of the columns.

For example, for x=10 minutes, we have a real value of 30 and a predicted value of 78.36, so the residual becomes e=30-78.36=-48.36.

e=y-\hat y=30-78.36=-48.36

For x=30 minutes, we have only the residual, that has a value of 5.53. That means that the predicted value, which we can calculate, is 5.53 below the real value.

The predicted value for x=30 is:

y(30)=45.3e^{0.0548*30}=45.3*5.18=234.47

Using the equation for the residual e, we can calcualte the actual number of calories that Mya burned in 30 minutes:

e=y-\hat y=y-234.47=5.53\\\\y=234.47+5.53\\\\y=240

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Write the equation for the parabola that has x− intercepts (1.2,0) and (4,0) and y− intercept (0,12).
zzz [600]
Equation of a parabola is written in the form of f(x)=ax²+bx+c.
The equation passes through points (4,0), (1.2,0) and (0,12), therefore; 
 replacing the points in the equation y = ax² +bx+c 
we get  0 = a(4)²+b(4) +c   for (4,0)
             0 = a (1.2)²+ b(1.2) +c for (1.2,0)
             12 = a(0)² +b(0) +c   for (0,12)
simplifying the equations we get
16a + 4b + c = 0
1.44a +1.2b + c = 0
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thus the first two equations will be
16a + 4b = -12
1.44 a + 1.2b = -12 solving simultaneously 
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3 0
3 years ago
Read 2 more answers
A festival charges $3 for child admission and $5 for adult admision. At the end of the festival they have sold 779 tickets for a
lutik1710 [3]

Answer:

562 child tickets were sold

217 adult tickets were sold

Step-by-step explanation:

A festival charges $3 for children admission and $5 for adult admission

At the end of the festival they have sold a total number of 779 tickets for $2771

Let x represent the child ticket

Let y represent the adult ticket

x + y= 779..............equation 1

3x + 5y= 2771..........equation 2

From equation 1

x + y = 779

x= 779 -y

Substitute 779-y for x in equation 2

3x + 5y= 2771

3(779-y) + 5y= 2771

2337 - 3y + 5y= 2771

2337 +2y= 2771

2y= 2771 -2337

2y = 434

y = 434/2

y = 217

Substitute 217 for y in equation 1

x + y= 779

x + 217= 779

x = 779-217

x= 562

Hence 562 child tickets were sold and 217 adult tickets were sold

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3 years ago
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lozanna [386]

Answer:

<h2>A = 52.5 m²</h2>

Step-by-step explanation:

The formula of an area of a kite:

A=\dfrac{e\cdot f}{2}

<em>e, f</em><em> - diagonals</em>

<em />

We have <em>e = 15m, f = 7m.</em>

Substitute:

A=\dfraC{(15)(7)}{2}=\dfrac{105}{2}=52.5\ m^2

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Given ∠ DBC = 90° and ∠ DBE + ∠ EBC = ∠ DBC, then

∠ DBE = ∠ DBC - ∠ EBC = 90° - 64.5° = 25.5°

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If a kite is 40 feet off the ground and the string holding the kite is 42 feet long, what is the angle of elevation to the kite
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