Answer:
0.30
Step-by-step explanation:
Probability of stopping at first signal = 0.36 ;
P(stop 1) = P(x) = 0.36
Probability of stopping at second signal = 0.54;
P(stop 2) = P(y) = 0.54
Probability of stopping at atleast one of the two signals:
P(x U y) = 0.6
Stopping at both signals :
P(xny) = p(x) + p(y) - p(xUy)
P(xny) = 0.36 + 0.54 - 0.6
P(xny) = 0.3
Stopping at x but not y
P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06
Stopping at y but not x
P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24
Probability of stopping at exactly 1 signal :
P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30
Answer:
a) 15
b) 2
Step-by-step explanation:
a) The sum of the enrollments in chemistry (60), physics (45), and biology (30) counts those triply enrolled 3 times and those doubly-enrolled twice. This sum will exceed the total number of students by 1 times those double-enrolled and twice those triply-enrolled.
We know that there are 10 students triply-enrolled, so the difference ...
(60 +45 +30) -2(10) = 15
is the number of doubly-enrolled students.
There are 15 students enrolled in exactly 2 science classes.
__
b) There are 9+4 = 13 students doubly-enrolled in physics and something else. Using the result from part A, there will be 15 -13 = 2 students doubly-enrolled in chemistry and biology, but not physics.
Answer:
1.) It is biased
2.) No
3.) Yes
4.) No
Step-by-step explanation:
1.) It is biased as it is predisposition to one particular outcome over another.
2.) No, because biased in research leads to unrepresentative outcomes as the estimates is predisposed to the left or to the right of the true values.
3.) Yes. Since the selection of shift has an equal probability of being chosen.
4.) No. The probability that the sample accurately reflects the efficiency of the workers, the standard should be 95%.
Expanded form: 9.00+0.70+0.06
Word form: Nine and seventy-six hundredths
Round it to 10: 10
Rounded to the nearest tenth: 9.80
First one
remember that all triangles internal angles (the 3 at the vertices) add to 180
we want BCD
we know BAD already
and ABC is a right angle
the measure of BDA is just there to throw us off
so
all adds to 180
BAD+ABC+BCD=180
59+90+BCD=180
149+BCD=180
minus 149 both sides
BCD=31
BCD=31 degrees
2nd problem
all angles add to 180 degrees
so
CAB+ABC+BCA=180
6x+147+2x+78+126+x=180
6x+2x+x+147+78+126=180
9x+351=180
minus 251 both sides
9x=-171
divide both sides by 9
x=-19
x=-19 degrees
