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3 years ago

If 15 workers did a job is 8 hours, how long would it take 5 workers to do half of this job?

2 answers:

8 0

Since 15 workers did 8 hours of work, the total amount of work needed is 120 hours. Half of this job would be 60 hours. Then divide this amount by the 5 workers doing the job. This would make the job 12 hours

yarga [219]3 years ago

5 0

J=15w*8h=120(worker*hours)

5h=120/2

5h=60

h=12 hours

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ipn [44]

It will be b after you do the math

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3 years ago

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lozanna [386]

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3 years ago

Experiments with repeated independent trials will be described by the binomial distribution if

bija089 [108]

Answer:

d. each trial has exactly two outcomes whose probabilities do not change

Step-by-step explanation:

A binomial experiment is one where there are exactly two outcomes for each trial and probability for getting success is constant in each trial.

In other words, each trial is independent of the other.

The trials need not be continuous nor time between trials to be constant.

Since trials are to be independent, each trial cannot influence the next.

Only option d is right.

d. each trial has exactly two outcomes whose probabilities do not change

Examples are tossing of coins, throwing dice, drawing cards or balls with replacement, etc

7 0

3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago

Find the dimensions of a rectangle with perimeter 112 m whose area is as large as possible. step 1 if a rectangle has dimensions

Veronika [31]

Please use the solution below:
Let P = perimeter, A = area
As provided above, a = xy and
We know that the formula to solve for the perimeter of a rectangle is P = 2x + 2y. Using the given 112m, we can solve the perimeter using the formula:
112 = 2x + 2y
56 = x + y

x = 56-y or y = 56-x
Let's solve the perimeter in terms of y using the formula below:
A = (56-y)(y)
Find the derivative of A = 56-y^2 to get the value of y.
dA/dy = 56-2y = 0
y = 56/2
y = 28
To find X, substitute the value of y in the equation x = 56 - y.
x = 56 - 28
Therefore, x = 28.
We can conclude that the figure is not a rectangle but a square.

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3 years ago