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2 years ago

What has one set of parallel lines and two right angles

Mathematics

2 answers:

velikii [3]2 years ago

7 0

Quadrilateral is the answer

Andrei [34K]2 years ago

5 0

The answer to your question is Quadrilateral

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Please help with this anyone

tresset_1 [31]

You distribute the 4 to both x and 5. 4 times x is 4x and 4 times 5 is 20. Divide both sides by 4. X=5

5 0

2 years ago

The tree diagram represents an experiment consisting of two trials.<br><br>P(A and C) = [?]

schepotkina [342]

Yo sup??

To find P(A and C) we have to simply multiply the probabilities of getting an A with that of getting a C

P(A)=0.6

P(C|A)=0.3

P(A and C)=0.6*0.3

=0.18

Hope this helps.

7 0

3 years ago

(5x-4)(x+1) multiply then simplify

viva [34]

Answer:

5x^2+x-4

Step-by-step explanation:

Multiply 5x into (x+1)

to get 5x^2+5x

Multiply -4 into (x+1)

to get -4x-4

Combine like terms

5x^2+5x-4x-4

Answer:

5x^2+x-4

3 0

1 year ago

Please help me with this one

Sindrei [870]

Answer:

Step-by-step explanation:

5 0

3 years ago

An assembled product is built by 10 worker who coordinate their task. A total of 100 units will be made. Time records indicate t

Brut [27]

Answer:

a) 86%

b) 2nd unit = 82 hs.

3rd unit = 75 hs

c) 100th unit = 36 hs

Step-by-step explanation:

We can model the learning curve for manufacturing the units as:

$t=aX^b$

where t is the time for the Xth unit, and a and b are parameters that we will calculate from the data,

We know that t(1)=95. Then, we have:

$t(1)=a\cdot 1^b=95\\\\a=95$

And we know that the fourth unit (X=4) take 71 hours to be completed (t(4)=71). Then, we can calculate the other parameter as:

$t(4)=95\cdot4^b=71\\\\4^b=71/95\approx 0.7473\\\\b\cdot ln(4)=ln(0.7473)\\\\b=ln(0.7473)/ln(4)=-0.291/1.386\\\\b=-0.21$

We have the model for the learning curve:

$t=95X^{-0.21}$

The learning rate percentage is calculated from the b parameter:

$b=\dfrac{ln(LRP)}{ln(2)}=-0.21\\\\\\ln(LRP)=-0.21*ln(2)=-0.21*0.693=-0.1455\\\\LRP=e^{-0.1455}=0.86$

The learning rate percentage is 86%.

b) The most likely times required for the 2nd and 3rd units are calculated with the model:

$t(2)=95\cdot2^{-0.21}=95*0.864=82\\\\t(3)=95\cdot3^{-0.21}=95*0.794=75$

c) If we use the model to calculate the time required for the 100th unit, we have:

$t(100)=95\cdot100^{-0.21}=95*0.38=36$

8 0

2 years ago