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Finger [1]
3 years ago
10

What is the average rate of change of the function on the interval from x = 0 to x = 5?

Mathematics
2 answers:
bixtya [17]3 years ago
8 0
The average rate is 3^x/2
damaskus [11]3 years ago
6 0

Answer:

Average rate of change = 24.2

Step-by-step explanation:

f(x)=\frac{1}{2}3^x

Average rate of change formula is on interval a to b is

\frac{f(b)-f(a)}{b-a}

Given interval is x=0 to x=5

a=0 and b=5

Average rate = \frac{f(5)-f(0)}{5-0}

f(0)=\frac{1}{2}3^0=\fracc{1}{2}=0.5

f(5)=\frac{1}{2}3^5=\frac{3^5}(2}=121.5

Average rate = \frac{f(5)-f(0)}{5-0}

Average rate = \frac{121.5-0.5}{5}=24.2

Average rate of change = 24.2

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Trucks in a delivery fleet travel a mean of 120 miles per day with a standard deviation of 18 miles per day. The mileage per day
kvv77 [185]

The probability that a truck drives between 150 and 156 miles in a day is 0.0247. Using the standard normal distribution table, the required probability is calculated.

<h3>How to calculate the probability distribution?</h3>

The formula for calculating the probability distribution for a random variable X, Z-score is calculated. I.e.,

Z = (X - μ)/σ

Where X - random variable; μ - mean; σ - standard deviation;

Then the probability is calculated by P(Z < x), using the values from the distribution table.

<h3>Calculation:</h3>

The given data has the mean μ = 120 and the standard deviation σ = 18

Z- score for X =150:

Z = (150 - 120)/18

   = 1.67

Z - score for X = 156:

Z = (156 - 120)/18

  = 2

So, the probability distribution over these scores is

P(150 < X < 156) = P(1.67 < Z < 2)

⇒ P(Z < 2) - P(Z < 1.67)

From the standard distribution table,

P(Z < 2) = 0.97725 and P(Z < 1.67) = 0.95254

On substituting,

P(150 < X < 156) = 0.97725 - 0.95254 = 0.02471

Rounding off to four decimal places,

P(150 < X < 156) = 0.0247

Thus, the required probability is 0.0247.

Learn more about standard normal distribution here:

brainly.com/question/26822684

#SPJ1

5 0
1 year ago
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