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3 years ago

5

For population parameter μ = true average resonance frequency for all tennis rackets of a certain type, each of these is a confi

dence interval. (114.4, 115.6) and (114.1, 115.9) What is the value of the sample mean resonance frequency?
a. 114.5
b. 115.8
c. 114.1
d. 115.0

1 answer:

Andreyy893 years ago

5 0

Answer:

X[bar]= 115

Step-by-step explanation:

Hello!

Every Confidence interval to estimate the population mean are constructed following the structure:

"Estimator" ± margin of error"

Wich means that the intervals are centered around the sample mean. To know the value of the sample mean you have to make the following calculation:

$X[bar]= \frac{Upper bond + Low bond}{2}$

$X[bar]= \frac{115.6+114.4}{2}$ = 115

Since both intervals were calculated with the information of the same sample, you can choose either to calculate the sample mean.

I hope it helps!

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Raina must choose a number between 55 and 101 that is a multiple of 2, 8 and 10. Write all the numbers that she could choose. If

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Answer:

The only possible number is $80$ .

Step-by-step explanation:

The number in question needs to be a multiple of all three of $2$ , $8$ , and $10$ . As a result, it must also be a multiple of the least common multiplier (lcm) of the three number.

Start by finding the least common multiplier of the three numbers.

Factor each number into its prime components:

$2$ is a prime number itself.
$8 = 2^3$ .
$10 = 2 \times 5$ .

The only prime factors are $2$ and $5$ .

The greatest power of $2$ among the three numbers is $3$ .
The greatest power of $5$ among the three numbers is $1$ .

Therefore, the least common multiplier of the three number should be the product of $2^3$ and $5$ . That's equal to $2^3 \times 5 = 8 \times 5 = 40$ .

In other words, the number (or numbers) in question could be written in the form $40\, k$ , where $k$ is an integer.

The question requires that this number be between $55$ and $101$ . In other words,

$55 \le 40\, k \le 101$ .

The goal is to find the possible values of $k$ . Note that from integer division by $40$ ,

$55 = 1 \times 40 + 15$ , and
$101 = 2 \times 40 + 21$ .

The inequality becomes:

$1 \times 40 + 15 \le 40\, k \le 2 \times 40 + 21$ .

However,

$1 \times 40 < 55 = 1 \times 40 + 15$ , and
$2 \times 40 + 21 = 101 < 3 \times 40$ .

Hence,

$1 \times 40 < 1\times 40 + 15 \le 40\, k \le 2 \times 40 + 21 < 3 \times 40$ .

$1 \times 40 < 40\, k < 3 \times 40$ .

Divide by the positive number $40$ to obtain:

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Since $k$ is an integers, $k = 2$ .

Indeed, $40 \, k = 80$ is between $55$ and $101$ .

Therefore, $80$ is the number in question.

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4 years ago

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attashe74 [19]

Answer:

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(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable $X\sim Poisson(\lambda = 0.62)$ .

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...$

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

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Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: $\lambda=7\times0.62=4.34$

Compute the probability of no deaths in a corps over 7 years as follows:

$P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130$

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

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