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Sonbull [250]
3 years ago
8

Determine if each scenario is either a permutation or combination. Do NOT solve these scenarios. a) An art gallery displays 125

different pieces of artwork at a time. However, 15 pieces are selected to be displayed prominently throughout in the gallery. For instance, the most popular piece is displayed in a location that can be seen as soon as you walk in the door. Determine how many ways these more popular 15 pieces can be displayed throughout the gallery. b) An art gallery has a total of 320 different pieces of art. However, only 125 pieces can be displayed at a time. Determine how many ways those 125 pieces can be selected.
Mathematics
1 answer:
Serjik [45]3 years ago
3 0

Answer:

a) Permutations

b) combination

Step-by-step explanation:

a)

Since the order of 15 most important artwork pieces matter, with most popular at first and then at N0.2, 3,4, and so on. Whenever we are dealing with "order of placement" the question at hand is of permutations.

b)

Out of 320 pieces, 125 pieces are to be "selected" and displayed, the process of selection incurs Combinations in which the order in which we select does not matter.  

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The Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
kirza4 [7]

Answer:

Step-by-step explanation:

given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging

Clerk          I           II           III    total

   

Pack      0.33          0.23    0.44 1

   

Defective 0.02 0.025    0.015  

   

Pack&def 0.0066 0.00575 0.0066 0.01895

a)  probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate

= P(II clerk) -P(II clerk and defective) = 0.23-0.00575=0.22425

b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)

=0.01895

c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3

=P(clerk 3 and def)/P(defective)

=\frac{0.0066}{0.01895} \\=0.348285

8 0
3 years ago
Leila received a $90 gift card for a coffee store. She used it in buying some coffee that cost $8.64 per pound. After buying the
klasskru [66]
She bought 5 pounds of coffee
3 0
3 years ago
If AC equals 48 meters, what is the perimeter of the field
Alinara [238K]

Remark

A kite is constructed such that AB = BC and AD = DC. AB = sqrt( (1/2)AC + 18^2) see diagram. AD = sqrt(24^2 + 32^2)

Step One

Solve for AB

1/2 AC = 24 (AC is given as 48)

18 is a given length

AB = sqrt(24^2 + 18^2) = sqrt(576 + 324) = sqrt(900) = 30

Step Two

Find the length of AD

AD = sqrt(32^2 + 24^2) = sqrt(1024 + 576) = sqrt(1600) = 40

Step Three

Find the Perimeter.

P = 2 * 30 + 2*40 = 60 + 80 = 140

P = 140 <<<<< Answer

5 0
3 years ago
Read 2 more answers
Will mark Brainliest!! :)
Georgia [21]

Step-by-step explanation:

The system of equations for eq 1 which is 3x + y = 118 represents the Green High School which filled three buses(with a specific number of students identified as x) and a van(with a specific number of students identified as y) with a total of 118 students.

for eq 2; 4x + 2y = 164; represents Belle High School which filled four buses(with a specific number of students identified as x) and two vans(with a specific number of students identified as y) with a total of 164 students.

The solution represents the specific number of students in the buses and vans in eq1 and eq 2 with x being 36 students and y being 10 students.

substituting 36 for x and 10 for y in eq 1;

3(36) + 10 = 108 + 10 = 118 total students for Green High School

substituting 36 for x and 10 for y in eq2;

4(36) + 2(10) = 144 + 20 = 164 total students for Belle High school

6 0
2 years ago
Jasmine bought 32 kiwi fruit for $16. IfJasmine has $4, how many kiwi can shebuy?
Sav [38]

Answer:

8 kiwis

Step-by-step explanation:

(÷4)32=$16(÷4)

8 =$4

hope this helps

7 0
3 years ago
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