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Naily [24]
4 years ago
12

A standard number cube numbered one through six on each side is rolled five times. What is the probability of landing on a three

or four on all five rules?
Mathematics
1 answer:
Sav [38]4 years ago
5 0

Answer:

  • 1/243 ≈ 0.004 (which is a low probability)

Explanation:

The event landing on a three or four on a rolling is independent of the event  of landing on a three or four on any other rolling.

Thus, for independent events you calculate the joint probability as the product of the probabilities of each event:

  • P(A and B and C and D and E) = P(A) × P(B) × P(C) × P(D) × P(E)

Here, P(A) = P(B) = P(C) = P(D) = P(E) = probability of landing on a three or four.

Now to calculate the probability of landing on a three or four, you follow this reasoning:

  • Positive outcomes = three or four
  • Number of positive outcomes = 2
  • Possible outcomes = one, two, three, four, five, or six
  • Number of possible outcomes = 6
  • Probability = numer of positive outcomes / numer of possible outcomes
  • Probability = 2/6 = 1/3

Finally, probability of landing on a three or four on all five rolls: (1/3)⁵ = 1/243 ≈ 0.004 ← answer

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3 years ago
Given: ∆PQR, m∠R = 90° m∠PQR = 75°, m∠MQR = 60° M ∈ PR , MP = 18 Find: RQ
Darina [25.2K]

Answer:

RQ = 9.

Step-by-step explanation:

m < RPQ = 180 - 90 - 75 = 15 degrees

m < PQM = 75 - 60 = 15 degrees.

So triangle PMQ is isosceles so side PM is congruent to side MQ, therefore MQ = 18.

Triangle  RQM is a 30-60-90 triangle with RQ the shortest leg (because it is opposite the lowest angle) so that makes RQ = 1/2 of MQ.  (The sides in a 30-60-90 triangle are in the ratio 1:sqrt3:2)

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6 0
3 years ago
Based on a sample of 150 textbooks at the store, you find an average of 70.69 and a standard deviation of 29.6. The point estima
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Answer:

Step-by-step explanation:

Given that:

The sample size = 150

The sample mean = 70.69

The standard deviation = 29.60

The sample mean is equal  to the point estimate of the population = 70.69

At 99% confidence interval level is:

∝ = 1 - 0.99

∝ = 0.01

The critical value:

Z_{\alpha/2} = Z_{0.01/2} \\ \\  Z_{0.005} = 2.58  \ using \ Z \ tables

The Margin of error = Z_{\alpha/2} \times \dfrac{\sigma}{\sqrt{n}}

= 2.58 \times \dfrac{29.6}{\sqrt{150}}

= 2.58 × 2.4168

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At 99% confidence interval; the estimate of the population mean lies within the interval:

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3 years ago
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