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3 years ago

Trevor paints 1/6 of the fence surrounding his farm each day. How many days will it take him to paint 3/4 of the fence?

1 answer:

8 0

To do this you have to divide 3/4 by 1/6. To do this multiply 3/4 by the reciprocal of 1/6. The reciprocal of 1/6 is 6/1. So you have to multiply 3/4 and 6/1. This will give you 18/4. You now simplify it to 4 2/4, which can be simplified even more to 4 1/2. It will take Trevor 4 1/2 days to paint 3/4 of the fence.
BTW- In case you are wondering how long it'll would take him to paint the entire fence, it would be 6 days.

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Which is more, 285 inches or 8 yards? 285 inches 8 yards neither; they are equal submit?

Dima020 [189]

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3 years ago

Identify all of the problems that would result in a product of 2 5 ?

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3 years ago

What is the average rate of change for <br><br> f (x)=0.5 (-10x + 24)?

xenn [34]

Answer:

The rate of change of the function is -5.

Step-by-step explanation:

Given : Function $f(x)=0.5(-10x+24)$

To find : What is the average rate of change ?

Solution :

Function $f(x)=0.5(-10x+24)$

The rate of change is the first derivative.

$f(x)=-5x+12$

Derivate w.r.t. x,

$f(x)=-5$

Therefore, The rate of change of the function is -5.

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3 years ago

What is 4.5 divided by 2

pav-90 [236]

Answer:

2.25

Step-by-step explanation:

You could just use a calculator.

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4 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

4 years ago