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2 years ago

9

The average of a set of 18 consecutive integers is 22.5. what is the smallest integer in the set?

1 answer:

quester [9]2 years ago

6 0

The sum of the 18 numbers is 22.5 x 18 = 405.
Let the numbers be x, x + 1, x + 2, . . ., x + 17

Sum of n term of an arithmetic sequence = n/2(a + l)
18/2(x + x + 17) = 405
9(2x + 17) = 405
2x + 17 = 405/9 = 45
2x = 45 - 17 = 28
x = 28/2 = 14

Therefore, the smallest integer is 14.

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Which function has a simplified base of 4^3√4?

IrinaVladis [17]

Answer:

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Step-by-step explanation:

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The function is $f(x) = 4(\sqrt[3]{16} )^{2x}$ .

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1 year ago

When you select a random person, the probability that this person will go to Ches

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Step-by-step explanation:

n=8

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Or 2.31%

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1 year ago

What single decimal multiplier would you use to decrease by 6% followed by a 19% increase?

Kryger [21]

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2 years ago

The average lifetime of a certain new cell phone is 3.4 years. The manufacturer will replace any cell phone failing within three

melisa1 [442]

Answer:

68% of these phones last 3.87 years.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The average lifetime of a certain new cell phone is 3.4 years.

This means that $m = 3.4, \mu = \frac{1}{3.4} = 0.2941$

So

$P(X \leq x) = 1 - e^{-0.2941x}$

68% of these phones last how long (in years)?

This is x for which:

$P(X \leq x) = 0.68$

$P(X \leq x) = 1 - e^{-0.2941x}$

Then

$0.68 = 1 - e^{-0.2941x}$

$e{-0.2941x} = 0.32$

$\ln{e{-0.2941x}} = \ln{0.32}$

$-0.2941x = \ln{0.32}$

$x = -\frac{\ln{0.32}}{0.2941}$

$x = 3.87$

68% of these phones last 3.87 years.

4 0

1 year ago