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Vilka [71]
2 years ago
6

Here are 15,000 students at college x. of those students, 1,700 are taking both ethics and metaphysics this semester. there are

2,200 total students taking ethics. 9,500 students are taking neither of these classes. how many students are taking metaphysics this term?
Mathematics
2 answers:
tekilochka [14]2 years ago
7 0
I think the answer is 1,600 totals students taking metaphysics
mars1129 [50]2 years ago
3 0

This can be solved using formula for Venn Diagrams:

Total = n(A)+ n(B) - n(both A and B) + n(neither A nor B)

Where, the n in front indicates number of items <u><em>[ n(A) is number of elements in set A ]</em></u>

We can take:

  • set A as students taking Ethics
  • set B as students taking Metaphysics

Given:

Tota = 15,000

n(A) = 2200

n(both A and B) = 1700

n(neither A nor B) = 9500

Substituting these into our formula, we get:

15,000=2200+n(B)-1700+9500\\15,000=n(B)+10,000\\n(B)=15,000-10,000\\n(B)=5000

Hence, 5000 students are taking metaphysics this term.

<u><em>*** If we want to know students taking </em></u><u><em>metaphysics ONLY</em></u><u><em>, we can subtract n(both A and B) from 5000. So we have 5000-1700=3300 ***</em></u>


ANSWER:

Total students taking Metaphysics = 5000

Students taking Metaphysics only = 3300


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which shape had a larger area: a rectangle that is 7 inches by 3/4 inch, or a square with side length of 2.5 inches? Show your r
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What is the GCF of x2 and x9?​
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Step-by-step explanation:

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Read 2 more answers
A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100 lb and 5 lb and standard deviation
Bas_tet [7]

Answer:

Option b. None is the correct option.

The Answer is 63%

Step-by-step explanation:

To solve for this question, we would be using the z score formula

The formula for calculating a z-score is given as:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

We have boxes X and Y. So we will be combining both boxes

Mean of X = 100 lb

Mean of Y = 5 lb

Total mean = 100 + 5 = 105lb

Standard deviation for X = 1 lb

Standard deviation for Y = 0.5 lb

Remember Variance = Standard deviation ²

Variance for X = 1lb² = 1

Variance for Y = 0.5² = 0.25

Total variance = 1 + 0.25 = 1.25

Total standard deviation = √Total variance

= √1.25

Solving our question, we were asked to find the percent of filled boxes weighing between 104 lb and 106 lb are to be expected. Hence,

For 104lb

z = (x-μ)/σ,

z = 104 - 105 / √25

z = -0.89443

Using z score table ,

P( x = z)

P ( x = 104) = P( z = -0.89443) = 0.18555

For 1061b

z = (x-μ)/σ,

z = 106 - 105 / √25

z = 0.89443

Using z score table ,

P( x = z)

P ( x = 106) = P( z = 0.89443) = 0.81445

P(104 ≤ Z ≤ 106) = 0.81445 - 0.18555

= 0.6289

Converting to percentage, we have :

0.6289 × 100 = 62.89%

Approximately = 63 %

Therefore, the percent of filled boxes weighing between 104 lb and 106 lb that are to be expected is 63%

Since there is no 63% in the option, the correct answer is Option b. None.

3 0
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