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2 years ago

1.2, 3, 7.5, 18.75, ...

Mathematics

1 answer:

Svetradugi [14.3K]2 years ago

3 0

Answer:

1.2(2.5)x - 1

Step-by-step explanation:

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Murljashka [212]

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3 years ago

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An oil exploration firm is formed with enough capital to finance ten explorations. The probabil- ity of a particular exploration

tensa zangetsu [6.8K]

Answer:

μ = 1 The firm expects that one oil exploration will be successful.

v(x)= 0.9

Step-by-step explanation:

The first step is to define the random variable x as:

x: number of oil explorations being succesful

Then x can be take this values:

x = 0 , x =1 ... x =10

x is a binomially distributed random variable with parameters.

p = 0.1 and n=10

And the mean or the expected value of x is:

μ = E(x) = np

Then μ = 10*0.1 = 1

And the variance of x is:

V(x) = np(1-p)

V(x) = 10(0.1)(1-0.1)= 0.9

6 0

2 years ago

Alenkasestr [34]

Answer:

0.36 = 36/100 = 9/25

so the simplest form is 9/25

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2 years ago

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Chang wants to rent a boat and spend at most $44. the boat cost $8 per hour, and chang has a discount coupon for $4 off. what ar

blagie [28]

8t-4 is less than or equal to 44

t=6

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3 years ago

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Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average int

meriva

Answer:

b) 95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14

sample standard deviation 'S' = 9.3

95 percent confidence interval for this single-sample t test

The values are $(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })$ the 95 percent confidence interval for the population mean 'μ'

Degrees of freedom γ=n-1=62-1=61

t₀.₀₅ = 1.9996 at 61 degrees of freedom

$(14 - 1.9996\frac{9.3}{\sqrt{62} } ,14+1.9996\frac{9.3}{\sqrt{62} })$

(14-2.361 , 14 + 2.361)

[(11.64 , 16.36]

Conclusion:-

95 percent confidence interval for this single-sample t test

[11.64, 16.36]

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3 years ago