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3 years ago

10

Evaluate the expression if x=4 and y=72x(3y-4x)^2

1 answer:

Pavlova-9 [17]3 years ago

7 0

I believe it would be 160000

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One hundred employees of a company are asked how they get to work and whether they work full time or part time. The table below

inna [77]

The probability is 56/100, or 14/25 = 0.56.

These events are not mutually exclusive, meaning they can happen at the same time. This means we use

P(A or B) = P(A) + P(B) - P(A and B)

P(carpool or full time) = P(carpool) + P(full time) - P(carpool & full time)

There are 6+9=15 people out of 100 that carpool.
There are 7+4+30+6=47 people out of 100 that work full time.
There are 6 people out of 100 that carpool and work full time.

This gives us
15/100 + 47/100 - 6/100 = 56/100

5 0

3 years ago

Ask questions slove for x 5(x-3)=15

Pavel [41]

Answer:

Step-by-step explanation:

Your answer is 6

4 0

3 years ago

For f(x) = 3x + 1 and g(x) = x2 - 6, find (f+ g)(x).

kirza4 [7]

Answer:

B

Step-by-step explanation:

(f+g)(x)=f(x)+g(x)

3x+1+x2-6 = x2+3x-5

5 0

4 years ago

The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probabili

Mariana [72]

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

$\lambda= 4.5$

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

a) No accidents occur in one week.

$P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111$

b) 5 or more accidents occur in a week.

$P( x \geq 5) = 1-\displaystyle \sum P(x$

c) One accident occurs today.

The mean number of accidents per day is given by

$\lambda = \dfrac{4.5}{7} = 0.64$

$P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747$

5 0

3 years ago

Which expression is equivalent to (4p^-4 q)^-2/ 10pq^-3?

Anit [1.1K]

Answer:

$\dfrac{p^{7} q}{160}$

Step-by-step explanation:

$\dfrac{(4p^-4 q)^-2}{10pq^-3} =$

$= \dfrac{4^{-2}p^{-4\times (-2)} q^{-2}}{10pq^-3}$

$= \dfrac{4^{-2}}{10}p^{8-1} q^{-2-(-3)}$

$= \dfrac{1}{4^2 \times 10}p^{7} q^{-2 + 3}$

$= \dfrac{1}{160}p^{7} q$

$= \dfrac{p^{7} q}{160}$

4 0

3 years ago

Read 2 more answers