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Anna71 [15]
4 years ago
15

Sally either walks, cycles or drives to work. She walks about 15% of the time and drives about 3/5 of the time. What percentage

of the time does she cycle? SHOW ME YOUR WORK! please :)
Mathematics
1 answer:
goblinko [34]4 years ago
8 0

Answer: She cycles 25% of the time

Step-by-step explanation:

From the question, Sally walks 15% of the time.

She drives 3/5 of the time.

We then convert 3/5 to percentage.

Converting to percentage means multiplying by a factor of 100.

Hence, Sally drives 3/5*100 = 60% of the time.

If Sally walks 15% of the time and drives 60% of the time, adding up the percentage time used by these two media of transportation will give us

60% + 15% = 75%

Hence if she cycles for the remainder of the time, expressing it as percentage would mean subtracting the time used in other means by 100%

Hence it will be 100% - 75% = 25%

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\root(3)(-3)\root(3)(-5)=\root(3)(15) true or false
Oksi-84 [34.3K]

The equation \sqrt[3]{-3} \cdot \sqrt[3]{-5} = \sqrt[3]{15} is true

<h3>How to determine the true statement?</h3>

The equation is given as

\sqrt[3]{-3} \cdot \sqrt[3]{-5} = \sqrt[3]{15}

Apply the law of indices

\sqrt[3]{-3 * -5} = \sqrt[3]{15}

Evaluate the product

\sqrt[3]{15} = \sqrt[3]{15}

Both sides of the equation are the same

Hence, the equation \sqrt[3]{-3} \cdot \sqrt[3]{-5} = \sqrt[3]{15} is true

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2 years ago
Answers should be exact decimals (please do not leave as fractions or unfinished calculations). A six-sided die (cube) with face
Ghella [55]

Answer:

(a) P = 0.0001

(b) P = 0.6561

(c) P = 0.2916

(d) P = 0.3439

(e) P = 0.2

Step-by-step explanation:

This is a probability problem.

The dice is rolled 4 times (n=4) and we calculate the probability of different outcomes.

The probability of a 6 in a roll is 0.5.

The probability of a 1, 2, 3, 4 or 5 in a roll is 0.5/5=0.1.

<u />

<u>a) Outcome: all the rolls are 2.</u>

The probability of having a 2 in a roll is 0.1, so we can calculate the probability of having a 2 in four consecutive rolls as

P(x1=2;x2=2;x3=2;x4=2)=P(x=2)^4=0.1^4=0.0001

<u>b) Outcome: none of the rolls is a 2.</u>

The probability of having any other number but 2 in 4 rolls is:

P(x1\neq2;x2\neq2;x3\neq2;x4\neq2)=P(x\neq2)^4=(1-0.1)^4=0.6561

<u>c) Outcome: exactly one roll is a 2</u>

This is the sum of the probability of having a 2 in the first, second, third or fouth roll, and others numbers in the rest of the rolls. These 4 combinations have the same probability, so we will multiply it by 4.

P(exactly \,one\,2)=4*P(x1=2;x2\neq2;x3\neq2;x4\neq2)\\\\P(exactly \,one\,2)=4*0.1*0.9*0.9*0.9=0.2916

<u>d) Outcome: at least one of the rolls is a 2</u>

In this case, is the probability of having at least one 2, is the sum of the probability of getting a 2 in the first roll, the probability of getting a 2 in the second roll, the probability of getting a 2 in the third roll and the probability of getting a 2 in the four roll:

P(x1=2)+P(x2=2)+P(x3=2)+P(x4=2)=0.1+0.9*0.1+0.9*0.9*0.1+0.9*0.9*0.9*0.1\\\\P(x1=2)+P(x2=2)+P(x3=2)+P(x4=2)=0.1+0.09+0.081+0.0729\\\\P(x1=2)+P(x2=2)+P(x3=2)+P(x4=2)=0.3439

<u>e) Outcome: either the first roll or the last roll is a 2</u>

The probability of getting a 2 in the first roll is equal to having it in a fourth roll, and its the probability of getting a 2 in a roll (multiplied by 2, beacuse there can be either in the first or in the last roll).

P(x1=2)+P(x4=2)=2*P(x=2)=2*0.1=0.2

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