Answer:
It is a quadrilateral.
Step-by-step explanation:
Quadrilateral is a 4-sided shape where the total angles is 360°.
So when you add up all the angles, you will get 360°.
Answer:
ASA
ΔFGH ≅ ΔIHG ⇒ answer B
Step-by-step explanation:
* Lets revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and
including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ
≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles
and one side in the 2ndΔ
- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse
leg of the 2nd right angle Δ
* Lets prove the two triangles FGH and IHG are congruent by on of
the cases above
∵ FG // HI and GH is transversal
∴ m∠FGH = m∠IHG ⇒ alternate angles
- In the two triangles FGH and IHG
∵ m∠FHG = m∠IGH ⇒ given
∵ m∠FGH = m∠IHG ⇒ proved
∵ GH = HG ⇒ common side
∴ ΔFGH ≅ ΔIHG ⇒ ASA
* ASA
ΔFGH ≅ ΔIHG
Answer:
- <u>The correct statement is the first one: </u><u><em>The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the right of the mean</em></u><em> </em>
<em />
Explanation:
To calculate how many<em> standard deviations</em> a particular value in a group is from the mean, you can use the z-score:

Where:
is the number of standard deviations the value of x is from the mean
is the mean
is the standard deviation
Substitute in the formula:

Which means that <em>the number of blue-eyed students in Mr. Garcia's class is 2 standard deviations</em> above the mean.
Above the mean is the same that to the right of the mean, because the in the normal standard probability graph the central value is Z = 0 (the z-score of the mean value is 0), the positive values are to the right of the central value, and the negative values are to the left of the central value.
Therefore, the correct statement is the first one: <em>The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the right of the mean, </em>