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NNADVOKAT [17]
3 years ago
12

Help me please please

Mathematics
1 answer:
harina [27]3 years ago
3 0

Answer:

a)

  • EH = \sqrt{HG^2+EG^2} = \sqrt{4^2+6^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}

b)

  • EG² = GH*GF
  • EG² = 4*12
  • EG² = 48
  • EG = √48 = 4√3
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Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation o
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Answer:

46.18% of the items will weigh between 6.4 and 8.9 ounces.

Step-by-step explanation:

We are given that the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.

<em>Let X =  weight of items produced by a machine</em>

The z-score probability distribution for is given by;

                Z = \frac{  X -\mu}{\sigma}  ~ N(0,1)

where, \mu = mean weight = 8 ounces

            \sigma = standard deviation = 2 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of items that will weigh between 6.4 and 8.9 ounces is given by = P(6.4 < X < 8.9) = P(X < 8.9 ounces) - P(X \leq 6.4 ounces)

   P(X < 8.9) = P( \frac{  X -\mu}{\sigma} < \frac{  8.9-8}{2} ) = P(Z < 0.45) = 0.67364  {using z table}

   P(X \leq 70) = P( \frac{  X -\mu}{\sigma} \leq \frac{  6.4-8}{2} ) = P(Z \leq -0.80) = 1 - P(Z < 0.80)

                                                 = 1 - 0.78814 = 0.21186

<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.45 and x = 0.80 in the z table which has an area of </em>0.67364<em> and </em>0.78814<em> respectively.</em>

Therefore, P(6.4 < X < 8.9) = 0.67364 - 0.21186 = 0.4618 or 46.18%

<em>Hence, 46.18% of the items will weigh between 6.4 and 8.9 ounces.</em>

8 0
3 years ago
In baseball, a player pitches a ball from the mound to a catcher behind the plate. A pitch that passes over the plate above the
matrenka [14]

To solve this question, you just need to count all the probability of the options.

The probability that a pitch not over the plate is a strike is zero. So, P(A | D) = 0.

True. It is 0/0+20= 0

The probability that a pitch not over the plate is a ball is 1. So, P(B | D) = 1.

True, it is 20/20+0= 1

The probability that a pitch over the plate is a strike is 10:15. So, ...

Incomplete but it sounds to be true. It should be 10/10+5= 10/15 = 2/3

The probability that a pitch over the plate is a ball is 5:10. So, P(B | C) = 0.5.

7 0
3 years ago
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If I know what you're asking, the answer is theoretical probability.

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Step-by-step explanation:

1 part of 3 sections of a wall takes 1/4 of an hour, (.25 or 25% of an hour), (15 minutes is 1/4 of 60 minutes), so if it takes 15 minutes for one section to be painted & you have 3 sections,

(15+15+15) or (15×3)

= your answer

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3 years ago
Chec
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Answer:

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