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3 years ago

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Mikayla is a waitress who makes a guaranteed $50 per day in addition to tips of 20% of all her weekly customer receipts, t. She

works 6 days per week. Which of the following functions best represents the amount of money that Mikayla makes in one week?f(t) = 50 + 20tf(t) = 300 + 20tf(t) = 50 + 0.2tf(t) = 300 + 0.2t

2 answers:

motikmotik3 years ago

7 0

$50 per day x 6 days = 300

20% = 0.2, times t = 0.2t

The equation would be: f(t) = 300 + 0.2t

Minchanka [31]3 years ago

3 0

Answer:

Option D: $f(t)=300+0.2 t$

Step-by-step explanation:

We are given that Mikayla is a waitress

She makes a guaranteed per day =$50

Her tips per day =20 % of weekly tips t

We have to find a function that represents the amount of money makes by Mikayla in one week

$20% of t=\frac{20}{100}t=0.2t$

She works per week=6 days

She earns in 6 days= $50\times 6$ =$300

Total earning of Mikayla in one week = $300+0.2t$

$f(t)=300+0.2 t$

Hence, option D is true.

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Step-by-step explanation:

Let <em>S</em> = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = <em>p</em> = 0.30.

The number of sample selected is, <em>n</em> = 25.

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According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

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(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

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The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

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Compute the probability that the number of successes is at most 9 as follows:

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Thus, the probability that the number of successes is at most 11 is 0.9357.

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Answer:

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