Answer:
AB = √18 , BC=√18 and CA =4
AB²+BC² = CA² and AB=BC
ΔABC isosceles right angled triangle.
Step-by-step explanation:
Given vectors are 7j+ 10k,-i + 6j+6k and - 4i + +9j + 6k
A( 0,7,10), B( -1,6,6) C(-4,9,6)
AB⁻ = OB-OA = -I+6j+6k-(7j+10k) = -I-j-4k
AB =
BC = OC-OB = -4i+9j+6k-(-I+6j+6k) = -3i+3j
BC=
CA = OA-OC = 7j+10k - (- 4i + +9j + 6k ) = 4i-2j+4k
CA =
Since AB²+BC² = CA²
And AB=BC
Therefore it follows that ΔABC is a right angled isosceles triangle
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!