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-BARSIC- [3]
3 years ago
8

After a lesson on exponents,Tia went home and said to her mom,"I learned that 10 to the power of 4 is the same as 40,000." She h

as made a mistake in her thinking. Use words,numbers, or a place value chart to help Tia correct her mistake.
Mathematics
2 answers:
White raven [17]3 years ago
6 0

Tia went home and said to her mom,"I learned that 10 to the power of 4 is the same as 40,000."

Tia made a mistake in her thinking. Exponent represents the number of zeros.

10^4 means we multiply 10 four times

10^4= 10*10*10*10=10000

For exponent 4, we put 4 zeros. So answer is 10,000.


Julli [10]3 years ago
6 0

Answer:

10^{4} = 10 × 10 × 10 × 10 = 10,000

Step-by-step explanation:

After a lesson on exponents, Tia said to her mom, "I learned that 10 to the power of 4 is the same as 40,000."

In Mathematics, 10 to the power of 4,conventionally is written as 10^{4} which means ten multiplied by itself a certain number of times (given exponent).

She is saying 10^{4} that is 10 × 10 × 10 × 10 this would be equal to 10,000.

Thus, we can write

10 to the power of 4 = 10^{4} = 10 × 10 × 10 × 10 = 10,000

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Answer:

There is a 44.12% probability that the defective product came from C.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of the customer receiving a defective product. For this probability, we have:

P(A) = P_{1} + P_{2} + P_{3}

In which P_{1} is the probability that the defective product was chosen from plant A(we have to consider the probability of plant A being chosen). So:

P_{1} = 0.35*0.25 = 0.0875

P_{2} is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

P_{2} = 0.15*0.05 = 0.0075

P_{3} is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

P_{3} = 0.50*0.15 = 0.075

So

P(A) = 0.0875 + 0.0075 + 0.075 = 0.17

P(B) is the probability the product chosen being C, that is 50% = 0.5.

P(A/B) is the probability of the product being defective, knowing that the plant chosen was C. So P(A/B) = 0.15.

So, the probability that the defective piece came from C is:

P = \frac{0.5*0.15}{0.17} = 0.4412

There is a 44.12% probability that the defective product came from C.

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The chair of the operations management department at Quality University wants to construct a p-chart for determining whether the
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Answer:

Prof B and Prof D

Step-by-step explanation:

Step-by-step explanation:

From the question  we are told that

Sample size 100  

Instructor    Number of Failures  

Prof. A         13

Prof. B         0

Prof. C         11

Prof. D         16

Confidence level= 0.95

From Z table

Z=1.96

Generally proportion for failure is mathematically given as

  Proportion\ of\ Failure\ P=\frac{Number\ of\ Failure\ N}{Sample\ size\ S}

  P=\frac{N}{S}

For Prof A

    P_A=\frac{N_A}{S}

    P_A=\frac{13}{100}

    P_A=0.13

For Prof B

    P_B=\frac{N_B}{S}

   P_B=\frac{0}{100}

    P_B=0

For Prof C

    P_C=\frac{N_C}{S}

    P_C=\frac{11}{100}

   P_C=0.11

For Prof D

     P_D=\frac{N_D}{S}

   P_D=\frac{16}{100}

   P_D=0.16

Generally the average proportional failure is mathematically given as

    P_a_v_g=\frac{0.13+0+0.11+0.16}{4}

   P_a_v_g=0.10

Therefore  having this we calculate for the control limits

Generally the upper control limit UCl is mathematically given as

Upper control limits:

  UCL =P_a_v_g +Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }

  UCL =0.1 +1.96\sqrt{\frac{0.1(1-0.1) }{100}}

  UCL =0.1 +1.96*0.03

  UCL =0.1 +0.0588

  UCL =0.1588

Generally the upper control limit UCl is mathematically given as

Lower limit control limits:

 UCL =P_a_v_g _Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }

 UCL =0.1 -1.96\sqrt{\frac{0.1(1-0.1) }{100}}

  UCL =0.1 -1.96*0.03

  UCL =0.1 -0.0588

  UCL =0.0412

Therefore the Range is 0.0412 to 0.1588

Given the range 0.0412 to 0.1588 Prof B and Prof D are outside the Range making them the correct options

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3 years ago
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