What are the measures of Arc J K and ∠KIJ?
2 answers:
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Answer:
0.278
Step-by-step explanation:
Given that Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used.
Each box is equally likely to be selected.
In other words
If A is selected then probability of winning is using the correct key out of two keys i.e. 0.5
If B is selected then probability of winning is using the correct key out of three keys i.e. 0.333
If c is selected then probability of winning is using the correct key out of two keys i.e. 0.00
So the probability that Nuri can win the car
=
Benchmark are numbers that are used as standards to which the rest of the data is compared to. When counting numbers using a number line, the benchmark numbers are the intervals written on the axis. For benchmark numbers of 10, the number line on top of the attached picture is shown. Starting from 170, the tick marks are added by 10, such that the next numbers are 180, 190, 200, and so on and so forth. When you want to find 410, just find the benchmark number 410.
The same applies to benchmark numbers in intervals of 100. If you want to find 170, used the benchmark numbers 100 and 200. Then, you estimate at which point represents 170. For 410, you base on the benchmark numbers 400 and 500.
The same applies to benchmark numbers in intervals of 100. If you want to find 170, used the benchmark numbers 100 and 200. Then, you estimate at which point represents 170. For 410, you base on the benchmark numbers 400 and 500.
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have 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 mean 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.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01
X[bar] ±
174.5 ±
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!