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yaroslaw [1]
3 years ago
5

(08.04 MC) Christina bought a yoyo from a company that claims that, with each retraction, the string rolls up by 70% of the orig

inal length. She sets up a tape measure and throws the yoyo 3 times. Her data are charted below. Throw Length of string (feet)
1 3
2 2.1
3 1.47

Answers:

9.99
9.72
3.24
0.12
Mathematics
2 answers:
egoroff_w [7]3 years ago
6 0
Here is the answer to the given problem. So, what Christina wants to find out is the sum of the length after 10 throws given that in each throw, it retracts 70% of the original length. So to continue the given data, here are the results:
4: 1.029
5: 0.7203
6: 0.50421
7: 0.352947
8: 0.2470629
9: 0.17294403
10: 0.12106082
Since she wants to find the sum of the length after 10 throws, we are just going to add the lengths above. So the answer would be this:  9.71752475. To round this off to the nearest hundredth, the answer would be 9.72. Hope this helps.
DanielleElmas [232]3 years ago
3 0

Answer: 9.72

Step-by-step explanation: We can use the equation, a1 - a1(r^n)/1 - r

a1 = first term of sequence (3)

r = common ratio (0.70)

n = position of term in sequence (10)

1. 3 - 3(0.70)^10/1 - 0.70

2. 3 - 3(0.0282475)/0.30

3. 3 - 0.0847426/0.30

4. 2.91526/0.30

5. 9.72 is the sum of the lengths.

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Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 ------ x<0

x²/25 ---- 0 ≤ x ≤ 5

1 ----- 5 ≤ x

Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(c) Calculate P(X > 4.5)

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

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Answer:

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Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

(c) Calculate P(X > 4.5).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

F(x) {-∝,μ} = ½ becomes

x²/25 {0,μ} = ½

μ² = ½ * 25

μ² = 12.5

μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

Then f(x) = d/dx (F(x))

where F(x) = x²/25 for 0 ≤ x ≤ 5

f(x) = d/dx(x²/25)

f(x) = 2x/25

When

F(x) = 0, f(x) = 2(0)/25 = 0

When

F(x) = 5, f(x) = 2(5)/25 = 2/5

f(x) = 2x/25 for 0≤x≤2/5

f. Calculating E(X).

E(x) = ∫xf(x) dx, 0,2/5

E(x) = ∫x * 2x/25 dx, 0,2/5

E(x) = 2∫x ²/25 dx, 0,2/5

E(x) = 2x³/75 , 0,2/5

E(x) = 2(2/5)³/75

E(x) = 16/9375

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