The school band is ordering health bars to sell for a fundraiser. The company that sells the bars charges $0.40 per bar plus $20
2 answers:
Answer: They can order from 0 to 449 bars, but no more.
Step-by-step explanation:
Given: The school band is ordering health bars to sell for a fundraiser. The company that sells the bars charges $0.40 per bar plus $20.00 for shipping regardless of the size of the order. The band must spend less than $200.00 on bars for the fundraiser.
The inequality below relates x, the number of bars that could be ordered, with the shipping costs and their spending requirements.
To solve for x, subtract 20 on both sides, we get
Divide 0.40 on both sides, we get
......... → the number of bars that could be ordered, with the shipping costs and their spending requirements must less than 450.
Thus, they can order from 0 to 449 bars, but no more.
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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!