Don claims that the trianlges are congruent by SAS triangle congruence criterion because a sequence of rigid motions carries \tr
iangle ABC△ABC onto \triangle DEF△DEF. Select all that can be used to support Don's claim.
A -a rotation of 30^o30 clockwise about vertex A, follwed by a translation that carries AB to DE, and a reflection over AC.
B- a rotation of 30^o30 clockwise about vertex A, follwed by a translation that carries AB to DE, and a reflection over AB.
C- a rotation of 90^o90 counterclockwise about vertex A, follwed by a translation that carries AC to DF, and a reflection over AC.
D- a rotation of 90^o90 counterclockwise about vertex A, follwed by a translation that carries AC to DF, and a reflection over AB.
A -a rotation of 30^o30 clockwise about vertex A, follwed by a translation that carries AB to DE, and a reflection over AC.
B- a rotation of 30^o30 clockwise about vertex A, follwed by a translation that carries AB to DE, and a reflection over AB.
C- a rotation of 90^o90 counterclockwise about vertex A, follwed by a translation that carries AC to DF, and a reflection over AC.
D- a rotation of 90^o90 counterclockwise about vertex A, follwed by a translation that carries AC to DF, and a reflection over AB.
2 answers:
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Answer:
a) 14.46%
b) 0.00%
c) 1.54%
Step-by-step explanation:
According to the survey, 10% of Americans are afraid to fly.
This means and
.
If 1100 Americans are sampled, then the sample size is .
The mean of the distribution is .
This means
The standard deviation is
We substitute the values to get:
a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.
We first apply the continuity correction factor to get:
We now convert to Z-scores to get:
From the standard normal distribution table P(z>1.06)=0.1446
As a percentage, the probability is 14.46%
b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.
We apply the CCF to get:
We convert to z-scores:
From the normal distribution, P(z>164.5)=0
c) First, 8% of 1100 is 88.
We want to find the probability that 88 or less Americans in the survey are afraid to fly.
We apply the CCF to get:
We convert to z-scores:
From the normal distribution, P(z>\:-2.16)=0.0154
As a percentage, we get 1.54%