Three classes of school children are selling tickets to the school play. the number of tickets sold by these classes, and the nu
2 answers:
We know that by employing the formula for the average here, we can find the average number of tickets sold per school child as the ratio of the total number of tickets sold to the total number of children in the school.
Let A be the average number of tickets sold per school child.
Let T be the total number of tickets sold.
Let C be the total number of children in the school.
Then from the information provided in the question,
T=tickets1+tickets2+tickets3 and C=class1+class2+class3
Thus, A can be expressed as:
The above is the expression for the average number of tickets sold per school child.
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Answer:
Step-by-step explanation:
a)
Test statistic:
here test statistic lie in rejection region,that why null hypothesis fails
so Yes, its significant.
b)
Test statistic:
c)
sample variability increases, therefore likelihood of rejecting the null hypothesis decreases.
Answer:
a). x = 11
b). m∠DMC = 39°
c). m∠MAD = 66°
d). m∠ADM = 36°
e). m∠ADC = 18°
Step-by-step explanation:
a). In the figure attached,
m∠AMC = 3x + 6
and m∠DMC = 6x - 49
Since "in-center" of a triangle is a points where the bisectors of internal angles meet.
Therefore, MC is the angle bisector of angle AMD.
and m∠AMC ≅ m∠DMC
3x + 6 = 8x - 49
8x - 3x = 49 + 6
5x = 55
x = 11
b). m∠DMC = 8x - 49
= (8 × 11) - 49
= 88 - 49
= 39°
c). m∠MAD = 2(m∠DAC)
= 2(30)°
= 60°
d). Since, m∠AMD + m∠ADM + m∠MAD = 180°
2(39)° + m∠ADM + 66° = 180°
78° + m∠ADM + 66° = 180°
m∠ADM = 180° - 144°
= 36°
e). m∠ADC =
=
= 18°