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murzikaleks [220]
3 years ago
8

Alex rolls a six-sided number cube 570 times. Approximately how many times can he expect to roll a 5?

Mathematics
2 answers:
Annette [7]3 years ago
8 0

1)First determine the probability of rolling a 5

2)Since you can only roll a 5 once in a 6-sided cube the probability is ⅙ or about 0.1667

3)Multiply the probability and the total number of rolls

.1667 * 570= 95

Answer : 95 times

madreJ [45]3 years ago
5 0

Answer:

95 times

Step-by-step explanation:

Probability of rolling a 5 =1/6

So no. of times 5 is expected when the number cube is rolled out 570 times = 570*1/6 = 95 times

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Find f(-4) if f(x) = 3x - 6.<br><br> A 6<br> B -6<br> C 18<br> D -18
brilliants [131]

Answer:

f(-4) = -18

Step-by-step explanation:

f(-4)  means x = - 4

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3 years ago
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Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
3 years ago
7. The largest of the Great Pyramids in Egypt was built with a square base,
strojnjashka [21]

Answer:

Difference between the original and current volumes = 4013140.30 cubic feet

Step-by-step explanation:

Largest pyramid has the square base with one edge = 754 feet

Area of the square base = (Side)²

                            = (754)²

                            = 568516 square feet

Volume of the pyramid = \frac{1}{3}(\text{Area of the base})(\text{Height})

                                       = \frac{1}{3}\times 568516\times 481

                                       = 91152065.3 cubic feet

After erosion,

Area of the base = (745)² = 555025 square feet

Volume of the pyramid = \frac{1}{3}(555025)(471)

                                       = 87138925 cubic feet

Difference between the original and current volumes of the pyramid

= 91152065.30 - 87138925

= 4013140.30 cubic feet

5 0
3 years ago
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