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djverab [1.8K]
3 years ago
13

A cashier will randomly choose two coins one at a time without replacement. What is the probabilty that the cashier will chose a

penny first, followed by a dime?
Mathematics
1 answer:
Tasya [4]3 years ago
4 0

Complete question is;

A cash register contains coins. 10 pennies, 11 nickels, 13 dimes, and 6 quarters. A cashier will randomly choose two coins one at a time without replacement. What is the probability that the cashier will choose a penny first, followed by a dime?

Answer:

1/12

Step-by-step explanation:

We are told there are 10 pennies, 11 nickels, 13 dimes, and 6 quarters.

Thus total number of coins = 10 + 11 + 13 + 6 = 40 coins

Probability of picking a penny first is;

10/40 = 1/4

After which we will have 39 coins left.

Probability of a dime after that will be 13/39 = 1/3

Thus,probability of picking a penny first then a dime will be; ¼ × ⅓ = 1/12

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3 years ago
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lubasha [3.4K]

Answer:

CE = 17

Step-by-step explanation:

∵ m∠D = 90

∵ DK ⊥ CE

∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK

In the two Δ KDE and KCD:

∵ m∠KDE = m∠KCD

∵ m∠DKE = m∠CKD

∵ DK is a common side

∴ Δ KDE is similar to ΔKCD

∴ \frac{KD}{KC}=\frac{DE}{CD}=\frac{KE}{KD}

∵ DE : CD = 5 : 3

∴ \frac{KD}{KC}=\frac{5}{3}

∴ KD = 5/3 KC

∵ KE = KC + 8

∵ \frac{KE}{KD}=\frac{5}{3}

∴ \frac{KC+8}{\frac{5}{3}KC }=\frac{5}{3}

∴ KC + 8 = \frac{25}{9}KC

∴ \frac{25}{9}KC - KC=8

∴ \frac{16}{9}KC=8

∴ KC = (8 × 9) ÷ 16 = 4.5

∴ KE = 8 + 4.5 = 12.5

∴ CE = 12.5 + 4.5 = 17

7 0
3 years ago
There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl
Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

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FrozenT [24]

Answer:

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Step-by-step explanation:

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so

36+81=c^2

117=c^2

117/2=10.82

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3 years ago
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