Marissa has twice as much money as Frank. Christina has $20 more than Marissa. If Christina has $100, how much money does Frank
have? Let f represent the amount of money Frank has in dollars and m represent the amount of money Marissa has in dollars.
a. Draw a tape diagram to represent the amount of money Frank has.
b. Draw a tape diagram to represent the amount of money Marissa has.
c. Draw a tape diagram to represent the amount of money Christina has.
d. Which tape diagram provides enough information to determine the value of the variable m?
e. Write and solve the equation.
a. Draw a tape diagram to represent the amount of money Frank has.
b. Draw a tape diagram to represent the amount of money Marissa has.
c. Draw a tape diagram to represent the amount of money Christina has.
d. Which tape diagram provides enough information to determine the value of the variable m?
e. Write and solve the equation.
1 answer:
Answer:
a. Take a grid that shows represents the amount of money Frank has,
Say F,
( shown below ),
b. Since, Marissa has twice as much money as Frank,
So, the amount Marissa has = 2F
If M represents the amount Marissa has,
M = 2F
c. Now, Christina has $20 more than Marissa,
Add 20 to the grid M
d. The below diagram provides enough information to determine the value of the variable m
e. The amount Christina has = M + 20 = 2F + 20
According to the question,
2F + 20 = 100
2F = 80
F = 40
Hence, the amount frank has = $ 40,
Amount Marissa has = 2(40) = $ 80,
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Answer:
(a) 0.6579
(b) 0.2961
(c) 0.3108
(d) 240
Step-by-step explanation:
The random variable <em>X</em> can be defined as the number of particles in a suspension.
The concentration of particles in a suspension is 50 per ml.
Then in 5 mL volume of the suspension the number of particles will be,
5 × 50 = 250.
The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.
The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.
The mean of the approximated distribution of X is:
μ = λ = 250
The standard deviation of the approximated distribution of X is:
σ = √λ = √250 = 15.8114
Thus,
(a)
Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:
Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.
(b)
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
Thus, the value of .
(c)
A 10 mL sample is withdrawn.
Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:
Thus, the value of .
(d)
Let the sample size be <em>n</em>.
The value of <em>z</em> for this probability is,
<em>z</em> = 1.96
Compute the value of <em>n</em> as follows:
Thus, the sample selected must be of size 240.