Olivia and Hudson each have an apple. Olivia cuts hers into halves and eats one piece. Hudson cuts his into quarters and eats tw
2 answers:
Answer:
No one ate more
Step-by-step explanation:
Olivia and Hudson each have an apple.
Olivia cuts her apple into halves and eats one piece.
1 ÷ 2 = +
Now she ate one piece =
Hudson cuts his apple into quarters and eats two pieces.
1 ÷ 4 = +
+
+
He ate 2 pieces = +
=
Therefore, No one ate more. They ate the same amount of apple.
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Answer:
Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.
To prove: Point A,D,P are collinear.
Proof:
→Case 1. When vertices A and P are opposite side of Base BC.
In Δ ABD and Δ ACD
AB= AC [Given]
AD is common.
BD=DC [median of a triangle divides the side in two equal parts]
Δ ABD ≅Δ ACD [SSS]
∠1=∠2 [CPCT].........................(1)
Similarly, Δ PBD ≅ Δ PCD [By SSS]
∠ 3 = ∠4 [CPCT].................(2)
But, ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]
2 ∠2 + 2∠ 4=360° [using (1) and (2)]
∠2 + ∠ 4=180°
But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.
So, point A, D, P lies on a line.
CASE 2.
When ΔABC and ΔPBC lie on same side of Base BC.
In ΔPBD and ΔPCD
PB=PC[given]
PD is common.
BD =DC [Median of a triangle divides the side in two equal parts]
ΔPBD ≅ ΔPCD [SSS]
∠PDB=∠PDC [CPCT]
Similarly, By proving ΔADB≅ΔADC we will get, ∠ADB=∠ADC[CPCT]
As PD and AD are medians to same base BC of ΔPBC and ΔABC.
∴ P,A,D lie on a line i.e they are collinear.
Answer:
C. At the 0.01 level of significance, the mean mass of the apples grown in the farmer's orchard is different from 100g.
Step-by-step explanation:
The null hypothesis is:
Because of the claim of the farmer.
The alternate hypothesis is:
The alternate hypothesis tests the farmer's claim at a significance level.
The results are calculated to be statistically significant at the 0.01 level.
This means that at the 0.01 level, the null hypothesis is rejected, that is, the mean mass of the farmer's orchard is different from 100. Since it is significant at the 0.01 level, it will be significant at the 0.05, 0.1, and increasing levels. So the correct answer is given by option C.