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Paha777 [63]
4 years ago
13

Subtract 4 from 1/6 of 42

Mathematics
2 answers:
ivanzaharov [21]4 years ago
5 0
\frac{1}{6} \times 42 - 4= \frac{42}{6}-4 = 7-4=3
Luba_88 [7]4 years ago
5 0

Subtract 4 from \frac{1}{6} of 42, the answer is 3.

Subtract 4 from \frac{1}{6}  of 42 can also be expressed as: ( \frac{1}{6} x 42 ) - 4

<h2>Further Explanation</h2>

Therefore, you have:

= ( \frac{1}{6} x 42 ) – 4 (since 42 and 6 can divide each other, then we can evaluate the expression)

Therefore, you have:

= (1 x 7) – 4

= (7) – 4

= 7 – 4

= 3

Therefore, if you subtract 4 from \frac{1}{6} of 42, the answer is 3.

Notably, the question was solved using the rule of BODMAS.

It represents Bracket, order, division, multiplication, addition, subtraction.

Significantly we derived the solution by using multiplication, division, and subtraction. Also, it is important to know that 1 sixth can be expressed as \frac{1}{6} and 1 sixth of 42 is the same as \frac{1}{6} x 42.

LEARN MORE:

  • Subtract 4 from 1/6 of 42  brainly.com/question/282405
  • Subtract 4 from 1/6 of 42  brainly.com/question/1048099

KEYWORDS:

  • bodmas
  • 1/6 of 42
  • evaluate
  • multiply
  • subtraction
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borishaifa [10]

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A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of
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Answer:

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(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let <em>X</em> = time spent per week shopping online.

It is provided that the random variable <em>X</em> follows a Poisson distribution.

The probability function of a Poisson distribution is:

P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...

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(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

              =1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

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Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

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