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Nostrana [21]
3 years ago
6

Write the product in simplest form: 3w^4 * (-5w^2)

Mathematics
1 answer:
hichkok12 [17]3 years ago
8 0

remember that (ab)(cd)=(a)(b)(c)(d)

also remember that (x^a)(x^b)=x^{a+b}

so

(3w^4)(-5w^2)=

(3)(w^4)(-5)(w^2)=

(3)(-5)(w^4)(w^2)=

(-15)(w^{4+2})=

(-15)(w^6)=

-15w^6

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Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles  carry just one person.    

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

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Now, we are given n = 10

We have to evaluate:

P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

\mu = np = 92(0.64) = 58.88

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We have to evaluate the probability that more than 47 cars carry just one person.

P(x \geq 47)

After continuity correction, we will evaluate

P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)

= 1 - P(z \leq -2.6913)

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

P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%

0.996 is the probability that more than half out of 92 vehicles carry just one person.

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3 years ago
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