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vodomira [7]
3 years ago
15

Logarithmic and Exponential Forms of Equations In Exercise,Write the logarithmic equation as an exponential equation,or vice ver

sa.
In 0.2 = - 1.6094...
Mathematics
1 answer:
lesantik [10]3 years ago
4 0

Answer:

e^-1.6094 = 0.2

Step-by-step explanation:

The inverse of In is e

If In 0.2 = -1.6094

Then e^-1.6094 = 0.2

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Plzzzz help! i will give brainliest 100 points! Let f(x) = 8x3 + 16x2 − 15 and g(x) = 2x + 1. Find f of x over g of x. A. 4 time
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Answer:

C. 4 times x squared plus 6 times x minus 3 minus 12 over the quantity 2 times x plus 1

Step-by-step explanation:

f(x) = 8x^3 + 16x^2 − 15

g(x) = 2x + 1.

f(x) / g(x) = (8x^3 + 16x^2 − 15) / (2x + 1)

Using long division

       

         

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Answer:

1/4 <u>divided by</u> 1/3

1/3 <u>divided into</u> 1/4

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The equation y = 5x represents the amount of miles Fran runs in x days. When a graph of this situation is drawn, what does the p
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Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,
Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that p = 0.15

Sample of 6:

This means that n = 6

What is the probability that there will be 0 or 1 defects in a sample of 6?

P(X \leq 1) = P(X = 0) + P(X = 1)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771

P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764

77.64% probability that there will be 0 or 1 defects in a sample of 6.

5 0
3 years ago
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