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TiliK225 [7]
4 years ago
6

While selecting candy for students in his class, Professor Murphy must choose between gummy candy and licorice nibs. Gummy candy

packets come in five sizes, while packets of licorice nibs come in two. If he chooses gummy candy, he must select gummy bears, gummy worms, or gummy dinos. If he chooses licorice nibs, he must choose between red and black. How many choices does he have
Mathematics
1 answer:
d1i1m1o1n [39]4 years ago
6 0

Answer and explanation:

First Prof. Murphy has to choose between two alternatives: gummy candy or licorice nibs

To find total choices, we only need to multiply number of sizes for each alternative and then add totals for each alternative together

So if Prof Murphy chooses gummy candy. He has sizes= 5 choices and types =3 choices, therefore he has 5×3= 15 choices for gummy candy

If he chooses licorice nibs, he has sizes = 2 choices and types= 2 choices, therefore he has 2×2= 4 choices for licorice nibs

Total choices: 15+4=19

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In a bag there are 3 red marbles, 2 yellow marbles, and 1 blue marble. what is the probability in simplest form of a yellow marb
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Answer:

The probability is 1/3, or in percent form, 33.3%.

Step-by-step explanation:

First, add up all the marbles. 3 plus 2 plus 1 equals 6 total marbles.

There are 2 yellow marbles out of 6 total marbles. Make it a fraction: 2/6.

Simplify the fraction by dividing by 2, to get 1/3, which is 33.3%.

Hope this helps! Feel free to give me Brainliest if you feel this helped. Have a good day, and good luck on your assignment. :)

5 0
3 years ago
Question a) Compare the Variability of Expenditure of Families in Two Towns given as follows: Number of families Expenditure (Ru
Karolina [17]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Number of families __________Expenditure

______midpoint (x) ____Town A __Town B

21-30 __25.5____________3 _______2

31-40 __35.5____________61 ______14

41-50 __45.5___________132 _____ 20

51-60 __55.5__________ 153 ______27

61-70 __65.5__________ 140 ______ 28

71-80 __75.5___________ 51 _______ 7

81-90 __85.5___________ 2 _______ 2

The variability is the variance and standard deviation :

The variance (s):

√Σ(x - m)²/Σf(x)

Town A:

m = Σ(X * f(x)) / Σf ; Σf = 542

Σ x * f(x) = (25.5*3) + (35.5*61) + (45.5*132) + (55.5*153) + (65.5*140) + (75.5*51) + (85.5*2)

= 29931 / 542

= 55. 22

Σf(x - m)² / Σf - 1 = [3(25.5-55.22)^2 + 61(35.5-55.22)^2 + 132(45.5-55.22)^2 + 153(55.5-55.22)^2 + 140(65.5-55.22)^2 + 51(75.5-55.22)^2 + 2(85.5-55.22)^2] / 541

= 76458. 4928 / 541

Variance = 141.32808

Standard deviation = √variance

Standard deviation = √141.32808

Standard deviation = 11.89

TOWN B:

m = Σ(X * f(x)) / Σf ; Σf = 100

Σ x * f(x) = (25.5*2) + (35.5*14) + (45.5*20) + (55.5*27) + (65.5*28) + (75.5*7) + (85.5*2)

= 5490 / 100

= 54.90

Variance :

Σf(x - m)² / Σf - 1 = [2(25.5-54.90)^2 + 14(35.5-54.90)^2 + 20(45.5-54.90)^2 + 27(55.5-54.90)^2 + 28(65.5-54.90)^2 + 7(75.5-54.90)^2 + 2(85.5-54.90)^2] / 99

= 16764 / 99

= 169.33

Standard deviation = √variance

Standard deviation = √169.3333

Standard deviation = 13.013

From the result above, Town A has lesser variability than Town B due to the slightly lower variance and standard deviation values.

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