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Katyanochek1 [597]
2 years ago
3

Bruce had an EKG to measure his heartbeat rate. After conversion, the function produced could be modeled by a cosine function, a

nd the wave produced a maximum of 4, minimum of −2, and period of pi over 2. Which of the following functions could represent Bruce's EKG read-out?
f(x) = 4 cos pi over 2x − 2
f(x) = 3 cos 4x + 1
f(x) = 3 cos pi over 2x + 1
f(x) = 4 cos 4x − 2
Mathematics
2 answers:
zimovet [89]2 years ago
6 0

Answer:

Option B.

Step-by-step explanation:

Bruce had an EKG to measure his heartbeat rate. After conversion, the function produced was modeled by a cosine function.

Now we will form this function.

Function will be in the form of f(x) = a cos(Bx) + d

Amplitude a=\frac{Maximum-minimum}{2}

a=\frac{4+2}{2}=3

Period = π/2

And Period=\frac{2\pi }{B}

⇒\frac{\pi }{2}=\frac{2\pi }{B}

⇒ B = 4

Since minimum is (-2) and maximum is (4), means cosine graph was shifted upwards.

Mid line of the graph is x=\frac{4+2}{2}=3 which shows graph is shifted by one unit above the x-axis.

Now the function we get is f(x) = 3 cos4x + 1

Therefore option B is the answer.

Vesna [10]2 years ago
4 0
Y= a.cos (bx) + midline

a = Amplitude = |4+2}/2 = |3|

Period = 2π/b = (2π) / (π/2) = 4

midline = (4-2)/2 = 1

Then the equation is:

y=3.cos(4x) + 1
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Suppose that the exam scores for students in a large university course are normally distributed with an unknown mean and standar
HACTEHA [7]

Answer:

Critical value t-score=2.701.

Step-by-step explanation:

When we calculate a confidence interval with an unknown population standard deviation, we estimate it from the sample standard deviation and use the t-score instead of the z-score.

The critical value for t depends on the level of confidence and the degrees of freedom.

If the sample size is 42, the degrees of freedom are:

df=n-1=42-1=41

For a confidence level of 99% and 41 degrees of freedom, the critical value of t is t=2.701.

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2 years ago
Miss Watson runs a distance of 200 meters in 25 seconds. Workout the average speed of Mrs Watson in meters per second.
Readme [11.4K]
Hey there! 

Divide 200 by 25. 

200/25= 8

She is running an average of 8 m/s. (that's how we represent meters per second)

I hope this helps!
~kaikers
8 0
2 years ago
Read 2 more answers
Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total dist
GaryK [48]

Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile =  1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

5 0
2 years ago
A new car is purchased for 15300 dollars. The value of the car depreciates at 14.25% per year. What will the value of the car be
Andrei [34K]

Answer:

\$6,082.70  

Step-by-step explanation:

we know that

The  formula to calculate the depreciated value  is equal to  

V=P(1-r)^{x}  

where  

V is the depreciated value  

P is the original value  

r is the rate of depreciation  in decimal  

x  is Number of Time Periods  

in this problem we have  

P=\$15,300\\r=14.25\%=0.1425\\x=6\ years

substitute in the formula

V=15,300(1-0.1425)^{6}  

V=15,300(0.8575)^{6}  

V=\$6,082.70  

3 0
3 years ago
Which sequence of transformations produces R'S'T' from RST? On a coordinate plane, triangle R S T has points (0, 0), (negative 2
arsen [322]

Answer:

Step-by-step explanation:

Given is a triangle RST and another triangle R'S'T' tranformed from RST

Vertices of RST are (0, 0), (negative 2, 3), (negative 3, 1).

Vertices of R'S'T' are  (2, 0), (0, negative 3), (negative 1, negative 1).

Comparing the corresponding vertices we find that x coordinate increased by 2 while y coordinate got the different sign.

This indicates that there is both reflection and transformation horizontally to the right by 2 units

So first shifted right by 2 units so that vertices became

(2,0) (0,3) (-1,1)

Now reflected on the line y=0 i.e. x axis

New vertices are

(2,0) (0,-3) (-1,-1)

8 0
2 years ago
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