Large balloons are sold in packages of 12. Select the expressions that can represent the total number of balloons in p packages
1 answer:
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Answer: On the 29th day
Step-by-step explanation:
According to this problem, no lilypad dies and the lilypads always reproduce, so we can apply the following reasoning.
On the first day there is only 1 lilypad in the pond. On the second day, the lilypad from the first reproduces, so there are 2 lilypads. On day 3, the 2 lilypads from the second day reproduce, so there are 2×2=4 lilypads. Similarly, on day 4 there are 8 lilypads. Following this pattern, on day 30 there are 2×N lilypads, where N is the number of lilypads on day 29.
The pond is full on the 30th day, when there are 2×N lilypads, so it is half-full when it has N lilypads, that is, on the 29th day. Actually, there are lilypads on the 30th, and
lilypads on the 29th. This can be deduced multiplying succesively by 2.
Answer:
The conditions necessary for the analysis are met.
The probability the newspaper’s sample will lead them to predict defeat is 0.7881
Step-by-step explanation:
We are given;
population proportion; μ = 52% = 0.52
Sample size;n = 400
The conditions are;
10% conditon: sample size is less than 10% of the population size
Success or failure condition; np = 400 x 0.52 = 208 and n(1 - p) = 400(1 - 0.52) = 192.
Both values are greater than 10
Randomization condition; we assume that the voters were randomly selected.
So the conditions are met.
Now, the standard deviation is gotten from;
σ = √((p(1 - p)/n)
where;
p is the population proportion
n is the sample size
σ is standard deviation
Thus;
σ = √((0.52(1 - 0.52)/400)
σ = √((0.52(0.48)/400)
σ = 0.025
Now to find the z-value, we'll use;
P(p^ > 0.5) = P(z > (x - μ)/σ)
Thus;
P(p^ > 0.5) = P(z > (0.5 - 0.52)/0.025)
This gives;
P(p^ > 0.5) = P(z > - 0.8)
This gives;
P(p^ > 0.5) = 1 - P(z < -0.8)
From the table attached we have a z value of 0.21186
Thus;
P(p^ > 0.5) = 1 - 0.21186 = 0.7881
Thus, the probability the newspaper’s sample will lead them to predict defeat is 0.7871
