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murzikaleks [220]

3 years ago

12

Factor x^4+xy^3+x^3+y^3 out completely. Hint: first factor out a monomial.

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

(x + y)(x^2 - xy + y^2)(x + 1)

Step-by-step explanation:

x(x^4/x + xy^3/x) + (x + y)(x^2 - xy + y^2)

x(x^4-1 + y^3) + (x + y)(x^2 - xy + y^2)

x(x^3 + y^3) + (x + y)(x^2 - xy + y^2)

x(x + y)(x^2 - xy + y^2) + (x + y)(x^2 - xy + y^2)

(x + y)(x^2 - xy + y^2)(x(x + y)(x^2 - xy + y^2/(x + y)(x^2 - xy + y^2 + x(x + y)(x^2 - xy + y^2/x(x + y)(x^2 - xy + y^2)

(x + y)(x^2 - xy + y^2)(x + 1)

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

GarryVolchara [31]

Answer:

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: $X = 0, 1, 7, 8$

6 0

4 years ago

Read 2 more answers