It is given in the question that
BD is the median. So it divides the opposite sides in two equal parts .
Therefore in triangles BAD and BCD,
AB and AC are congruent because of isosceles triangle.
AD and CD are congruent because of the median BD.
And BD and BD are congruent .
So the two triangles are congruent by SSS and the correct option is the first option .
Answer:
(3, 5)
Step-by-step explanation:
Point (3, -5) is 5 units below the x-axis.
After it is reflected, it will be 5 units above the x-axis.
Answer: (3, 5)
Answer:
im 70% sure that its D
Step-by-step explanation:
I hope i helped sorry if im wrong :D
Answer:
1899
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 3234
Standard deviation = 871
Percentage of newborns who weighed between 1492 grams and 4976 grams:
1492 = 3234 - 2*871
So 1492 is two standard deviations below the mean.
4976 = 3234 + 2*871
So 4976 is two standard deviations above the mean.
By the Empirical Rule, 95% of newborns weighed between 1492 grams and 4976 grams.
Out of 1999:
0.95*1999 = 1899
So the answer is 1899
Answer:
140
Step-by-step explanation:
To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.
First, let's count the number of subsets that contain the element 3.
Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is
.
Now, let's count the number of subsets that contain the element 4.
4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in
ways.
We conclude that there are 70+70=140 required subsets of S.