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kogti [31]
2 years ago
7

Mr zuro finds the mean height of all 14 students in his class to be 69.0 inches. Just as Mr Zuro finishes explaining how to get

the mean, Danielle walks in late. Danielle is 70.5 inches tall. What is the mean height of the 15 students in the class?
Mathematics
1 answer:
ycow [4]2 years ago
4 0

Answer:

69.1

Step-by-step explanation:

<em>Mean height of 14 students = 69.0 inches</em>

<em>Mean = sum of data/number of data</em>

<em>69 = sum of data/14</em>

<em>sum of data = 966</em>

<u>FOR NEW MEAN:</u>

<em>Height of additional student = 70.5</em>

Mean = sum of data/ number of data

Mean = 966+70.5/14+1

Mean = 1036.5/15

Mean = 69.1

Therefore, the mean height of the 15 students in the class is 69.1

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7. During the Middle Ages, people often used grains, scruples, and drahms to measure the weights of different medicines. If 120
Serjik [45]

Answer:

5 drahms.

Step-by-step explanation:

From the question given above, the following data were obtained:

120 grains = 6 scruples

6 scruples = 2 drahms

300 grains (in drahms) =..?

From the above data,

120 grains = 6 scruples

6 scruples = 2 drahms

Therefore,

120 grains = 2 drahms

Thus, we can obtain 300 grains in drahms as follow:

120 grains = 2 drahms

Therefore,

300 grains = 300 grains × 2 drahms /120 grains

300 grains = (300 × 2)/120 drahms

300 grains = 5 drahms

Therefore, 300 grains is equivalent to 5 drahms.

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2 years ago
How do you do solve for this again? Picture above
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Answer:

Step-by-step explanation:

Do you remember the formula so that you can solve this problem??

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A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm
AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, P\left(\frac{E}{A}\right)=0.02

Now, from the conditional probability,

P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}

\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)

\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

P\left(\frac{E}{A}\right)=0.40

and \Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)

This is the probability of disregarding the partially defective parts by inspection machine.

P\left(\frac{E}{A}\right)=0.98

and \Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

=P(E\cap C)

= 0.0686 [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)

\Rightarrow P(E)=0.018+0.012+0.0686

\Rightarrow P(E)=0.0986

So, the total probability that the part produced get shipped

=1-P(E)=1-0.0986=0.9014

The probability that the part is good (either defect free or slightly defective)

=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)

=(0.9-0.018)+(0.03-0.012)

=0.9

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}

=\frac{0.9}{0.9014}

=0.9984

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

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Answer:

the correct answer is D

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