Answer:
(a) The value of P (None) is 0.062.
(b) The value of P(at least one) is 0.938.
(c) The value of P(at most one) is 0.253.
(d) The event is not unusual.
Step-by-step explanation:
Let X = number of households watching the show.
The probability of the random variable x is, P (X) = p = 0.18.
The sample selected for the survey is of size, n = 14
The random variable X follows a Binomial distribution with parameter n = 14 and p = 0.18.
The probability of a Binomial distribution is computed using the formula:
(a)
Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:
Thus, the value of P (None) is 0.062.
(b)
Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
Thus, the value of P(at least one) is 0.938.
(c)
Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)
Thus, the value of P(at most one) is 0.253.
(d)
An event that has a very low probability of occurrence is known as an unusual event.
The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.
This probability value is not low.
Hence, the event is not unusual.
Answer:
Step-by-step explanation:
The urn contains 3blue balls 5 red balls
a) probability of getting a red ball
P=no of favourable of outcomes /total no outcomes
P(red ball) = 5/8
b) Probability of blue ball
P(blue ball) = 3/8
c) Odds getting a red ball
odds in favour of any object = m/n
m : event to occur
n : event will not occur
Odds(red ball) = 5/3
d)
Odds(blue) = 3/5
Answer:
5 dots above zero
Step-by-step explanation:
Could the question be The dot plot shows the number of pets owned by the students in Jayod’s class.
The center, or the middle value, of the data set is _____.????
If it is that is the answer.
Hope this helps!! :)
Answer:
null hypothesis = µ1=µ2=µ3; µ1= popuation mean of inulin µ2 = population mean of fructicoligosaccharide =µ3=population mean of lactulose
alternative hypothesis =µ1≠µ2 ≠µ3
t1= µ-45/s1 / N-1
at the significance level 0.01
t at ᵅ/2 =0.005 and degree of freedom= 35-1=34 is 2.25
t1 = µ-32/s1/N-1
2.25= µ-45/s1/34
= s1/34= µ-45/2.25
=s1 =(µ-45/2.25)*34
t2= µ-32/s2/34
2.25 =µ-32/s2/34
s2/34= µ-32/2.25
s2=µ-32/2.25*34
T= 45-32/s1/sqrt34+s2/sqrt34
t at 0.005 and no of grees of freedom 68 =2.37
2.37=45-32/
or s1/sqrt34+s2/sqrt34 = 13/2.37
s1+s2 = 13/2.37 *5.83
s1+s2= 31.98 or 32
(µ1-45/2.25)*34+µ2-32/2.25*34=32
or µ1+µ2 = 2525
µ1=2525-µ2
µ1 and µ2 are not equal
thus null hypothesis is rejected
conclusion all the three components are not in equal amount in hydrogen production
Explanation : in this experiment we have to prove whether the means of insulin,fructicoligosaccharide and lactulose are equal. so even if we prove that two of them are not equal null hypothesis will be rejected. we use student-t test because we have to compare the means of two population.
Answer:
Step-by-step explanation:
a) The sum of opposite angles of an inscribed quadrilateral is 180°. This lets us use angles E and G to solve for x:
(x+15) + (2x) = 180
3x + 15 = 180 . . .simplify
x +5 = 60 . . . . . divide by 3
x = 55 . . . . . . . . subtract 5
Similarly, we can use angles F and H to solve for y:
(3y -60) + (y) = 180
4y -60 = 180 . . . . simplify
y -15 = 45 . . . . . . divide by 4
y = 60 . . . . . . . . . add 15
___
b) Then the measures of the angles are ...
G = 2x = 2·55 = 110
E = 180 -G = 70
H = y = 60
F = 180 -H = 120
The angle measures are ...
m∠E = 70°, m∠F = 120°, m∠G = 110°, m∠H = 60°
___
c) short arc HF is intercepted by inscribed angle E, so the arc will have twice the measure of the angle.
arc HF = 2·m∠E = 140°
_____
Comment on the problem
Throughout, the only relation being used is that the measure of an arc is twice the measure of the inscribed angle intercepting it. For opposite angles of the quadrilateral, the sum of the two intercepted arcs is 360° (the whole circle), so the sum of the two angles is 180°.