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lara [203]
3 years ago
11

imes \sqrt{12}" alt="\sf \sqrt{12} \times \sqrt{12}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

12

Step-by-step explanation:

\sqrt{12} \times\sqrt{12}

\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}\sqrt{a}=a,\:\quad \:a\ge 0

\sqrt{12}\sqrt{12}=12

=12

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The correct answer would be -3/8 or in decimal form -0.375 :)

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Help me? please?????????
bagirrra123 [75]
8/63 is the simplest form of the fraction
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The largest taco contained approximately 1 kg of onion for every 6.8 kg grilled steak. The total weight of these two ingredients
erastovalidia [21]

Answer: Onion = 79.17 kg

Grilled steak = 538.356 kg

Step-by-step explanation:

Hi, to answer this question we have to write an equation with the information given:

1 k + 6.8 k = 617.6  

Where “k” is a constant value, a multiplier.

Solving for k:

7.8 k = 617.6

k = 617.6 /7.8  

k= 79.17

So:

Amount of onion used:

1 (79.17) = 79.17 kg

Amount of grilled steak used:

6.8 (79.17) = 538.356 kg

Feel free to ask for more if needed or if you did not understand something.

6 0
3 years ago
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The function y = 2x2 + 4x - 6 is graphed.
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Answer:

2 for the first drop down

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Step-by-step explanation:

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7 0
3 years ago
A manufacturing process produces semiconductor chips with a known failure rate of . If a random sample of chips is selected, app
AleksandrR [38]

Answer:

The probability that at least 14 of the chips will be defective is 0.6664.

Step-by-step explanation:

The complete question is:

A manufacturing process produces semiconductor chips with a known failure rate of 5.4%. If a random sample of 300 chips is selected, approximate the probability that at least 14 will be defective. Use the normal approximation to the binomial with a correction for continuity .

Solution:

Let <em>X</em> = number of defective chips.

The probability that a chip is defective is, <em>p</em> = 0.054.

A random sample of <em>n</em> = 300 chips is selected.

A chip is defective or not is independent of the other chips.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 300 and <em>p</em> = 0.054.

But the sample selected is too large.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=300\times 0.054=16.2>10\\n(1-p)=300\times (1-0.054)=283.8>10

Thus, a Normal approximation to binomial can be applied.

So,  X\sim N(\mu =16.2,\ \sigma^{2}=15.3252).

Compute the probability that at least 14 of the 300 chips will be defective as follows:

Use continuity correction:

P (X ≥ 14) = P (X > 14 + 0.50)

               = P (X > 14.50)

               =P(\frac{X-\mu}{\sigma}>\frac{14.50-16.20}{\sqrt{15.3252}})

                =P(Z>-0.43)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that at least 14 of the chips will be defective is 0.6664.

5 0
2 years ago
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