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1 year ago

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A telemarketer is paid a fixed amount of $15 in addition to $2 per call. If the average cost per call is $3, find the number of

calls made by the telemarketer

1 answer:

Schach [20]1 year ago

6 0

$10 because 2 and 3 add together and subtract 15 equals 10

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{4x − 3y = −32x + 6y = −4 Solve the system of equations.

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G is the circumcenter. Find BG if BG =(2x-15)

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i) Baraka bought ksh.20,000 worth of shares of a company each month. During the first three months, he bought the shares at a pr

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The average price paid by him for the shares after 3 months is ksh. 163.33

<h3>Average</h3>

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1 year ago

Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel

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The unusual $X$ values for this model are: $X = 0, 1, 2, 7, 8$

Step-by-step explanation:

A binomial random variable $X$ represents the number of successes obtained in a repetition of $n$ Bernoulli-type trials with probability of success $p$ . In this particular case, $n = 8$ , and $p = 0.53$ , therefore, the model is ${8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}$ . So, you have:

$P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024$

$P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215$

$P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848$

$P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912$

$P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695$

$P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431$

$P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371$

$P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442$

$P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062$

The unusual $X$ values for this model are: $X = 0, 1, 7, 8$

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