REALLY NEED HELP ON THIS PROBLEM
image attached
Answers
Answer:
average rate of change = - 0.5
Step-by-step explanation:
the average rate of change of f(x) in the closed interval [ a, b ] is
here the closed interval is [ 1, 3 ] , then
f(b) = f(3) = 4 ← point (3, 4 ) on graph
f(a) = f(1) = 5 ← point (1, 5 ) on graph
Then
average rate of change = =
= - 0.5
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1. Julia's goal is to run faster than two of her friends in an upcoming6.2-mile race. The table shows the results of the last race that eachrunner finished. Assume they each run the race at the same ratethey ran their last race. Complete the table. Who will finish firstamong the three friends, and by how much time will she beat thesecond-place finisher?RunnerJuliaHazelMekenaLast RaceDistance(mi)856Last RaceTime(min: sec)62:0039:1046:00Rate(min:secper mi)Time for6.2 Miles(min: sec)
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PLEASE HELP The probability distribution for arandom variable x is given in the table.
Answers
Answer:
a) According to probability theory, the following are possible:
(1) The probability that a customer will buy 5 items in a single purchase is 30%. This represents a 0.30 chance of the customer buying 5 items.
(3) The probability that a customer will buy 3 items in a single purchase is 99% This means that there a 0.99 chance that the customer will buy 3 items.
b) According to probability theory, the following are not possible:
2) The probability that a customer will buy 3 items in a single purchase is 101%. Probability is always equal to 1 or 100%. It cannot exceed 100%.
(4) The probability that a customer will buy 8 item. This part lacks the necessary information to solve the problem.
Step-by-step explanation:
Probability describes the likelihood or chance of an event happening out of the many possible events that can take place. The sum of all probabilities for a particular event is always equal to 1. This shows that probability does not exceed 100%.
Answers
Answers
Answer
Relative risk= 0.742
Odds ratio= 0.745
Detailed calculation shown in diagram:
Answers
Final answer:
The exponential function representing the bacteria population after t hours is f(t) = 2000 * e^(ln(0.5)/3 * t).
Explanation:
To find the exponential function that represents the size of the bacteria population after t hours, we can use the formula N = N0 * e^(kt), where N0 is the initial population, e is Euler's number (approximately 2.71828), k is the growth/decay constant, and t is the time in hours.
In this case, the initial population N0 is 2,000 and the population after 3 hours is 1,000. Plugging these values into the formula, we get:
N = 2000 * e^(3k) = 1000
Solving for k, we find k = ln(0.5)/3. Therefore, the exponential function representing the bacteria population after t hours is f(t) = 2000 * e^(ln(0.5)/3 * t).
Learn more about exponential function here:
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Final answer:
The exponential decay function representing the bacteria population after t hours is f(t) = 2000 × 0.5^(t/3), where t is the number of hours passed.
Explanation:
The student has observed a population of bacteria decreasing from 2,000 to 1,000 over three hours and seeks an exponential function to model the decay of the population over time, expressed as f(t). Since the population is halving every three hours, we can represent this with the function f(t) = 2000 × 0.5^(t/3), where 2000 is the initial population, 0.5 represents the halving, and t is the time in hours. The exponent (t/3) is used because the halving occurs every three hours.
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Answers
Answer:
-40
Step-by-step explanation:
You start at -5 then you add 20 to get 15 . Then you subtract 25 from 15 to get -10. Then you add 40 to -10 to get 30. Then you have to subtract 70 from 30 to get -40 F.
Sample Mean 25 23
Sample Variance 27 7.56
Sample Size 45 36
As the statistical advisor to Ajax, would you recommend purchasing Allied's machine? Explain.
Answers
Answer:
z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine
Step-by-step explanation:
We must evaluate the differences of the means of the two machines, to do so, we will assume a CI of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).
New machine
Sample mean x₁ = 25
Sample variance s₁ = 27
Sample size n₁ = 45
Old machine
Sample mean x₂ = 23
Sample variance s₂ = 7,56
Sample size n₂ = 36
Test Hypothesis:
Null hypothesis H₀ x₂ - x₁ = d = 0
Alternative hypothesis Hₐ x₂ - x₁ < 0
CI = 90 % ⇒ α = 10 % α = 0,1 z(c) = - 1,28
To calculate z(s)
z(s) = ( x₂ - x₁ ) / √s₁² / n₁ + s₂² / n₂
s₁ = 27 ⇒ s₁² = 729
n₁ = 45 ⇒ s₁² / n₁ = 16,2
s₂ = 7,56 ⇒ s₂² = 57,15
n₂ = 36 ⇒ s₂² / n₂ = 1,5876
√s₁² / n₁ + s₂² / n₂ = √ 16,2 + 1.5876 = 4,2175
z(s) = (23 - 25 )/4,2175
z(s) = - 0,4742
Comparing z(s) and z(c)
|z(s)| < | z(c)|
z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine
The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean