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3 years ago

Write the equation....

Mathematics

1 answer:

BaLLatris [955]3 years ago

5 0

1-0/-1-0 = -1
Y- 0 = -(x-0)
Y=-x
Y - 1 = -(x+1)
Y-1= -x-1
Y = -x

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A marketing consultant observed 40 consecutive shoppers to estimate the average money spent by shoppers in a supermarket store.

KonstantinChe [14]

Answer:

0.998 is the probability that the average money spent by a sample of 40 shoppers is within $10 of the actual population mean.

Step-by-step explanation:

We are given the following information in the question:

Standard Deviation, σ = $21.51

We are given that the distribution of average money spend is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

We have to find:

P( average money spent is within $10 of the actual population mean.)

$= P( z \leq \displaystyle\frac{10\times \sqrt{40}}{21.51}}) = P(z \leq 2.94)$

Calculation the value from standard normal z table, we have,

$P(z \leq 2.94) = 0.998$

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3 years ago

On each round, Ann and Bob each simultaneously toss a fair coin. Let Xn be the number of heads tossed in the 2n flips which occu

CaHeK987 [17]

Answer:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

Step-by-step explanation:

In a coin toss the probability of tossing a head is 0.5 (50% head/50% tails)

If n is the number of rounds and 2n the number of coins tossed (one for each player), the probability of having m heads tossed is:

$R=\frac{2n!}{m!*(2n-m)!}$

R is the number of cases (combination of coins tossed) that gives a m number of heads. Each case has a probability of $P_{case}=0.5^{2n}$ so:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

<u>For example, to toss 4 heads in 5 rounds: </u>

n=5
2n=10
m=4

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$P=\frac{10*9*8*7*6!}{4!*6!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}=0.205$

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Answer:

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Step-by-step explanation:

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