<h3>
The monthly cost of cell phone is A = 40 + 0.005 m</h3>
Step-by-step explanation:
The monthly fixed charges by mobile company = $40
The cost of sending each text message = 0.05 cents
Now, 100 cents = 1 dollar
⇒ 1 cent =$ 
⇒ 0.05 cent s =$ 
Also, the number of text messages send = m
So, the cost of sending m text message = m x ( cost of 1 text )
= m x ( $ 0.0005) = 0.0005 m
Now, Total bill = Fixed rate + cost of m texts
⇒ A =$40 + $ 0.0005 m
Hence, the monthly cost of cell phone is A = 40 + 0.005 m
Answer: The ratio 16:12 describes the ratio of the number of students who play the viola to the students who play the clarinet.
Step-by-step explanation:
Given : Number of students who play the clarinet= 12
Number of students who play the viola=16
We know that ratio of A to B is represented by A:B.
Since, the number '12' is representing " students who play the clarinet" and number '16' is representing " students who play the viola"
So , the ratio 16:12 describes the ratio of the number of students who play the viola to the students who play the clarinet.
A sinusoid is cyclical and has a period or frequency.
y = sin x is a sinusoid, with a period of 2π.
y = √x is not a sinusoid.
y = cos x is a sinusoid with a period of 2π.
Answer: y = √x is not a sinusoid.
Answer:
probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3
Step-by-step explanation:
First of all;
Let B1 be the event that the card with two red sides is selected
Let B2 be the event that the
card with two black sides is selected
Let B3 be the event that the card with one red side and one black side is
selected
Let A be the event that the upper side of the selected card (when put down on the ground)
is red.
Now, from the question;
P(B3) = ⅓
P(A|B3) = ½
P(B1) = ⅓
P(A|B1) = 1
P(B2) = ⅓
P(A|B2)) = 0
(P(B3) = ⅓
P(A|B3) = ½
Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;
P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]
Thus;
P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]
P(B3|A) = (1/6)/(⅓ + 0 + 1/6)
P(B3|A) = (1/6)/(1/2)
P(B3|A) = 1/3