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Ulleksa
10 days ago
7

Find the average value of the function i = 15(1 - e1/2t) from t = 0 to t = 4. A. 7.5(1 + e-2) B. 7.5(2 - e-2) C. 7.5(2 + e-2) D.

7.5(3 - e-2)
Mathematics
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A students calculation was found to have a 15.6% error, and the actual value was determined to be 25.7 mL. What are the two poss
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The true value is 25.7 ml.
The calculated error is 15.6%.
Thus, the error amount equals 0.156 times 25.7, which calculates to 4.0092 ml.

The percentage error indicates that the student's measurement could either exceed or fall short of the true value by this error amount.

This leads to two potential readings:
one possibility is: 25.7 + 4.0092 = 29.7092 ml
the other possibility is: 25.7 - 4.0092 = 21.6908 ml

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An investment of $7,650 earns interest at the rate of 5% and is compounded quarterly. What is the accumulated value of the inves
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FV = P(1 + r/t)^nt, where P denotes the principal amount, r is the interest rate, t is the frequency of compounding per year, and n is the total number of years.

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What is the solution to the system of equations graphed below? Please Help (:
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C. (1,5) The solution represents where two lines intersect.
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Miguel is a golfer, and he plays on the same course each week. The following table shows the probability distribution for his sc
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1) 4.55

2) Short hit

Step-by-step explanation:

1)

The score and corresponding probabilities are shown in the following table:

Score 3 4 5 6 7

Probability 0.15 0.40 0.25 0.15 0.05

Let us define

X = Miguel's score on the Water Hole

The expected value of the variable X can be represented as:

E(X)=\sum x_i p_i

where

x_i represents the different possible outcomes of X

p_i denotes the associated probabilities

Accordingly, the expected value of Miguel's score is calculated as follows:

E(X)=3\cdot 0.15 + 4\cdot 0.40 + 5\cdot 0.25 + 6\cdot 0.15 + 7\cdot 0.05=4.55

2)

Here, we will again consider:

X = Miguel's score on the Water Hole

In this case:

- For a successful long hit, the expected value of X is

E(X)=4.2

- Conversely, if the long hit isn't successful, the expected value of X becomes

E(X)=5.4

We also know that the likelihood of succeeding with a long hit is

p(L)=0.4

Consequently, the chance of failure in a long hit is

p(L^c)=1-p(L)=1-0.4=0.6

Thus, the expected value of X when opting for the long hit strategy is:

E(X)=p(L)\cdot 4.2 + p(L^C)\cdot 5.4 = 0.4\cdot 4.2 + 0.6\cdot 5.4 =4.92

In the first part of this question, we calculated the expected value using the short hit strategy, yielding:

E(X)=4.55

Given that the expected value for X is smaller (which is better) when utilizing the short hit option, we can conclude that this approach is superior.

8 0
3 months ago
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