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elena-14-01-66 [18.8K]
3 years ago
11

A manufacturing firm just received a shipment of 20 assembly parts, of slightly varied sizes, from a vendor. The manager knows t

hat there are only 15 parts in the shipment that would be suitable. He examines these parts one at a time.
a. Find the probability that the first part is suitable.
Mathematics
1 answer:
kaheart [24]3 years ago
6 0

Answer:

75% probability that the first part is suitable.

Step-by-step explanation:

A probability is the desired outcomes bivided by the number of total outcomes.

a. Find the probability that the first part is suitable.

Desired outcomes:

There are 20 parts, of which 15 are suitable, so D = 15

Total outcomes:

20 parts, so T = 20

Probability:

P = \frac{D}{T} = \frac{15}{20} = 0.75

75% probability that the first part is suitable.

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Jax is proving that the base angles of an isosceles triangle are congruent. He joins a line from
Sliva [168]

Answer: there you go

Step-by-step explanation:

4 0
3 years ago
Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0
Svetlanka [38]

<u>Answer:</u>

The equation that has solutions 6 and -6 is x^2 - 36 = 0

<u>Solution:</u>

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

x^{2}-12 x+36=0  --- eqn 1

x^{2}+12 x-36=0 -- eqn 2

x^{2}-36=0  ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.  

LHS:

72-72=0

RHS:  0  

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS:  0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2  

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is x^2 - 36 = 0

3 0
3 years ago
Read 2 more answers
Gabe made a rectangular sandbox for his brother to play in. Gabe wants to paint only the sides of the sandbox. A prism has a len
alukav5142 [94]

Answer: The surface area of the sides of the sandbox that Gabe wants to paint is 47ft2

Step-by-step explanation:

Hi to answer this we have to apply the formula:

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Where:

l= length

w= width

h= height

Since the sandbox is open , it has one less surface (length x width), we have to add only one lw term.

Surface Area of the sandbox = lw + 2wh + 2lh

A = (4x5) + 2( 5x1.5) + 2( 4x1.5)

A = 20+15+12

A= 47 ft2

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4 years ago
What is the area of this parallelogram?
Alenkasestr [34]

Answer:

The answer is 28, because 7x4=28

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Step-by-step explanation:

what are you trying to do?

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