the expression below gives the cost of green beans(x) and cucumbers (y). what is the cause of 6 bunches of green beans and 4 cuc
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Answer:
Hope this helps
Answer:
Step-by-step explanation:
Given that X is a normal random variable with parameters µ = 10 and σ 2 = 36,
X is N(10, 6)
Or z =
is N(0,1)
a) P(X > 5),
=
(b) P(4 < X < 16),
=
(c) P(X < 8),
=
(d) P(X < 20),
=
(e) P(X > 16).
=P(Z>-0.6667)
= 0.2524
Answer:
Step-by-step explanation:
Hello!
To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:
<u>Group 1</u>: Ate healthy breakfast
X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.
n₁= 26
<u>Group 2: </u>Did not eat healthy breakfast
X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.
n₂= 20
After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:
p'₁= 0.50
p'₂= 0.40
The parameters of study are the population proportions, if the claim is true then p₁ > p₂
And you can determine the hypotheses as
H₀: p₁ ≤ p₂
H₁: p₁ > p₂
α: 0.05
≈N(0;1)
pooled sample proportion:
p-value: 0.2514
The decision rule is:
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.
At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.
So having a healthy breakfast doesn't seem to improve the grades of students.
I hope this helps!