An airline operates in only 10 destinations in 2015, but plan on adding more destinations each year. In 2016, the number of dest
inations will be twice the initial number of destinations. Then, in 2017, the number of destinations will be twice the number of destinations the previous year. If this pattern continues, which of the following graphs represents the number of destinations in which the airline will operate over time?
Graph W
Graph X
Graph Y
Graph Z
Graph W
Graph X
Graph Y
Graph Z
2 answers:
Answer:
The correct option is 1.
Step-by-step explanation:
Let x-axis represents the number of years since 2015 and y-axis represents the number of destination.
It is given that the n airline operates in only 10 destinations in 2015, but plan on adding more destinations each year. It means the graph of passes through the point (0,10).
In 2016, the number of destinations will be twice the initial number of destinations. So, the number of destinations in 2016 is
It means the graph of passes through the point (1,20).
In 2017, the number of destinations will be twice the number of destinations the previous year. So, the number of destinations in 2017 is
It means the graph of passes through the point (2,40).
From the given graphs, only graph W passes through the points (0,10), (1,20) and (2,40).
Therefore the option is 1.
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Julius’ original number was 3 & Laila cheated as accordingly number was 20 ! but she said 108 .
<u>Step-by-step explanation:</u>
Let Julius pick number x , So we have following instructions as :
Add three to your number and multiply the sum by five
⇒
"Now take that number and subtract seven from it and tell me the new number.” “Twenty-three,”
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⇒
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So , Julius’ original number was 3 .
In the next round Leila is supposed to pick a number between 1 and 10 and follow the same instructions. She gives her final result as 108. i.e.
⇒
⇒
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Therefore , Julius’ original number was 3 & Laila cheated as accordingly number was 20 ! but she said 108 .
Answer:
Step-by-step explanation:
Hello!
To test the accuracy of a machine said to package exactly 1.20 pounds od cereal a random sample of 36 cereal boxes was taken.
The study variable is:
X: Weight of a cereal box.
This variable has no known distribution but it's population standard deviation is known σ= 0.66 pounds.
The sample weight is X[bar]= 1.48 pounds.
a.
The claim is that the machine packages exactly 1.20 pounds per box and the parameter of interest is the population mean, so you have to test is the population mean weight packaged by the machine is 1.20 pounds or not, symbolically:
H₀: μ = 1.20
H₁: μ ≠ 1.20
b.
As I said before there is no information about the distribution of the study variable so I'll apply the Central Limit Theorem to approximate the distribution of the sample proportion to normal.
Remember: Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample means tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
Then we can use the standard normal as a statistic for the test:
≈N(0;1)
c.
The level of significance of this test is α: 0.10
The critical is two-tailed, wich means you will have two critical values:
Critical value 1:
Critical value 1:
This means that you will reject the null hypothesis if ≤ -1.648 or if
≥ 1.648.
You will not reject the null hypothesis if -1.648 ≤ ≤ 1.648.
d.
is greater than the upper critical value, so the decision is to reject the null hypothesis.
Using 10% confidence level there is enough statistical evidence to conclude that the population average weight of the cereal boxes is different from 1.2 pounds.
I hope it helps!