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A project has three paths and a desired completion time of 43 days. The first path has a 70 percent probability of being complet

8090 [49]

Answer: The required probability is 0.82.

Step-by-step explanation:

Since we have given that

Number of projects = 3

So, probability of getting each project = $\dfrac{1}{3}$

Probability of first path being completed in 43 days = 0.7

Probability of second path being completed in 43 days = 0.85

Probability of third path being completed in 43 days = 0.90

So, Probability that the project will be completed on time would be

$\dfrac{1}{3}(0.7+0.85+0.90)\\\\=\dfrac{1}{3}\times 2.45\\\\=0.82$

Hence, the required probability is 0.82.

3 0

3 years ago

Which is the value of this expression when j = negative 2 and k = negative 1?

Natalija [7]

Answer:

Option A.

Step-by-step explanation:

It is given that

$j=-2, k=-1$

The given expression is

$\left(\dfrac{jk^{-2}}{j^{-1}k^{-3}}\right)^3$

Substitute the given values in the above expression.

$\left(\dfrac{(-2)(-1)^{-2}}{(-2)^{-1}(-1)^{-3}}\right)^3$

$\Rightarrow \left(\dfrac{(-2)(1)}{\dfrac{1}{-2}(-1)}\right)^3$

$\Rightarrow \left(\dfrac{-2}{\dfrac{1}{2}}\right)^3$

$\Rightarrow \left(-4\right)^3$

$\Rightarrow -64$

Therefore, the correct option is A.

3 0

4 years ago

Read 2 more answers

You have asked to design a rectangle box with a square base and an open top. The volume of the box must be620 cm to the 3rd powe

hammer [34]

Answer:

$69.21

Step-by-step explanation:

Since the box has a square base the length and breadth of the box will be equal. Let it be $x$

Let h be the height of the box

V = Volume of the box = $620\ \text{cm}^3$

$x^2h=620\\\Rightarrow h=\dfrac{620}{x^2}$

Now surface area of the box with an open top is given

$s=x^2+4xh\\\Rightarrow s=x^2+4x\dfrac{620}{x^2}\\\Rightarrow s=x^2+\dfrac{2480}{x}$

Differentiating with respect to x we get

$\dfrac{ds}{dx}=2x-\dfrac{2480}{x^2}$

Equating with zero

$0=2x-\dfrac{2480}{x^2}\\\Rightarrow 2x^3-2480=0\\\Rightarrow x^3=\dfrac{2480}{2}\\\Rightarrow x=(1240)^{\dfrac{1}{3}}\\\Rightarrow x=10.74$

Double derivative of the function

$\dfrac{d^2s}{ds^2}=2+\dfrac{4960}{x^3}=2+\dfrac{4960}{1240}\\\Rightarrow \dfrac{d^2s}{ds^2}=6>0$

So, x at 10.74 is the minimum value of the function.

$h=\dfrac{620}{x^2}\\\Rightarrow h=\dfrac{620}{10.74^2}\\\Rightarrow h=5.37$

So, minimum length and breadth of the box is 10.74 cm while the height of the box is 5.37 cm.

The total area of the sides is $4xh=4\times 10.74\times 5.37=230.7\ \text{cm}^2$

The area of the base is $x^2=10.74^2=115.35\ \text{cm}^2$

Cost of the base is $0.40 per square cm

Cost of the side is $0.10 per square cm

Minimum cost would be

$230.7\times 0.1+0.4\times 115.34=\$69.21$

The minimum cost of the box is 69.21 dollars.

8 0

3 years ago

John runs 18 miles in 4 hours. How many miles can John run per hour?

disa [49]

Answer:

4.5 miles per hour

Step-by-step explanation:

you start with 18 = 4x

divide both sides by 4

and you get x = 4.5

8 0

3 years ago

Read 2 more answers

Simplify the expression

uysha [10]

In extract form 23/20 with mixed numbers -1 3/20

8 0

3 years ago