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mojhsa [17]
3 years ago
15

The Martian Colonies elect their government through a lottery. There are 100,000 people living on Mars, and every year, a counci

l of 99 co-equal leaders is randomly selected from the population. In how many ways can the leadership be elected? Give your answer in terms of permutations or combinations and explain your choice. You do not have to evaluate.
Mathematics
1 answer:
Rom4ik [11]3 years ago
4 0

Answer:

100,000C999

Step-by-step explanation:

Given that the Martian Colonies elect their government through a lottery. There are 100,000 people living on Mars

every year, a council of 99 co-equal leaders is randomly selected from the population.

Since these are members, order of selection of not important

So combinations woul dbe used here.

Total number of persons avaiiable = 100,000

No of persons to be selected = 999

No of ways this can be done =

999 out of 100,000

= 100,000C999

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Answer:

1/14

Step-by-step explanation:

5/15 × 3/14 = 1/14

Or 0.0714 (3sf)

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7. What are the solutions of the equation? x2 = 11x − 24
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1/4 X Plus x equals negative 3 plus 1/2 x
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3 years ago
The perimeter of ABCD is 66 and DC is twice as long as CB how long is a B
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<span>Length = x Width = 2x
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</span><span>Length = x Width = 2x
Perimeter = 2 * length + 2 * width
= 2x + 4x
= 6x 
6x = 66
x = 11</span>
3 0
3 years ago
A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)
skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

<u>Step-by-step explanation</u>:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

<u>Conclusion</u>:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0
3 years ago
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