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3 years ago

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an urn contains 10 blue marbles and 8 yellow marbles a marble is drawn and dropped back into the urn. Then a second marble is dr

awn and dropped back into the urn both marbles are yellow if another marble is drawn. what is the probability that it will be blue

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

Answer:

80/729

Step-by-step explanation:

8/18x8/18x10/18=640/5832

80/729

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Vinil7 [7]

The answer is -8. Do you need an explanation?

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4 years ago

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Use Gauss's approach to find the following sum (do not use formulas):5+11+17+23...+83

LUCKY_DIMON [66]

Denote the sum by <em>S</em>. So

<em>S</em> = 5 + 11 + 17 + 23 + ... + 83

There's a constant difference of 6 between consecutive terms in <em>S</em>, so the 3 terms before 83 are 77, 71, and 65. So

<em>S</em> = 5 + 11 + 17 + 23 + ... + 65 + 71 + 77 + 83

Gauss's approach involves inverting the sum:

<em>S</em> = 83 + 77 + 71 + 65 + ... + 23 + 17 + 11 + 5

If we add terms in the same position in the sums, we get

2<em>S</em> = (5 + 83) + (11 + 77) + ... + (77 + 11) + (83 + 5)

and we notice that each grouped term on the right gives a total of 88. So the right side consists of several copies <em>n</em> of 88, which means

2<em>S</em> = 88<em>n</em>

and dividing both sides by 2 gives

<em>S</em> = 44<em>n</em>

<em />

Now it's a matter of determining how many copies get added. The terms in the sum form an arithmetic progression that follows the pattern

11 = 5 + 6

17 = 5 + 2*6

23 = 5 + 3*6

and so on, up to

83 = 5 + 13*6

so <em>n</em> = 13, which means the sum is <em>S</em> = 44*13 = 572.

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3 years ago

.find the distance between the points (4, 3) and (0, 3).<br> a) 2 <br> b) 4 <br> c) 10 <br> d) 12

kap26 [50]

The distance is 4 from

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3 years ago

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The mean salary of actuaries is LaTeX: \mu=\$100,000????=$100,000 and the standard deviation is LaTeX: \sigma=\$36,730????=$36,7

krek1111 [17]

Answer:

The lower limit of 95% confidence interval is 99002.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $100,000

Sample mean, $\bar{x}$ = $111,000

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = $36,730

First, we design the null and the alternate hypothesis

$H_{0}: \mu = \$100,000\\H_A: \mu \neq \$100,000$

We have to find the lower limit of the 95% confidence interval.

95% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$111000 \pm 1.96(\dfrac{36730}{\sqrt{36}} )\\\\ = 111000 \pm 11998.4667 \\= (99001.5333,122998.4667)\\ \approx (99002,122999)$

The lower limit of 95% confidence interval is 99002.

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4 years ago

Solve for x.<br> 30<br> 18<br> х<br> A. 35<br> B. 32<br> OC. 22<br> D. 24

Nikitich [7]

Answer:

option d is the correct answer

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3 years ago