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Elena-2011 [213]
3 years ago
10

The middleweight division in wrestling is centered at 140 pounds. Deviation from that must be less than 15 pounds either way. Wr

ite and solve the Absolute value inequality. What is the range of the possible weights for the lightweight division?
Mathematics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer: 125lb < M < 155 lb

Step-by-step explanation:

The mean mass is 140 lb, and the deviation must be less than 15 pounds (exactly 15 is not accepted).

Then (140lb + 15lb) is not alowed, and (140lb - 15lb) is not alowed.

Then we can define M = mass.

We have the two equations:

M < 140lb + 15lb

M < 155lb

And

M > (140lb - 15lb)

M > 125lb

If we write these two relations togheter, we get:

125lb < M < 155 lb

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The Probability distribution is the function which describes the likelihood of possible values assuming a random variable. The cost of flowers for a wedding is $698. The 95% of all samples of size is 40 and the confidence interval will be mean cost of flowers at wedding. There is confidence that mean cost of wedding flowers is between $701 to $767.

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What does two and five fiths equal to
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2 and five fifths = 3


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Describe the motion of a particle with position (x, y) as t varies in the given interval. (For each answer, enter an ordered pai
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The motion of the particle describes an ellipse.

Step-by-step explanation:

The characteristics of the motion of the particle is derived by eliminating t in the parametric expressions. Since both expressions are based on trigonometric functions, we proceed to use the following trigonometric identity:

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A manufacturer of processing chips knows that 2\%2%2, percent of its chips are defective in some way. Suppose an inspector rando
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The data in the question seems a bit erroneous. I am writing the correct question below:

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it.

Answer:

The probability that at least one of the selected chips is defective is 0.0776.

Step-by-step explanation:

The question states that the probability of defective chips is 2% i.e. 0.02. Let p denote the probability of selecting a defective chip so, p = 0.02

An inspector selects 4 chips, which means n=4 and we need to compute the probability that at least one of the selected chips is defective. Let X be the number of defective chips selected. We need to compute P(X≥1) which means either 1, 2, 3 or 4 chips can be defective.

We will use the binomial distribution formula to solve this problem. The formula is:

<u>P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ</u>

where n = total no. of trials

          p = probability of success

          x = no. of successful trials

          q = probability of failure = 1-p

we have n=4, p=0.02 and q=1-0.02=0.98.

We need to compute P(X≥1) which is equal to:

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

A shorter method to do this is to use the total probability theorem:

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ (0.02)⁰(0.98)⁴⁻⁰

          = 1 - (0.98)⁴

          = 1 - 0.9224

P(X≥1) = 0.0776

4 0
3 years ago
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