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What is the opportunity cost of one gun in this nation

serious [3.7K]

To produce 5 guns they lose 5 butter so I think opportunity cost of one gun would be one butter.

4 0

3 years ago

Your ship sinks and you swim to a nearby island. On this island, the inhabitants do not use money. The island residents use isla

natita [175]

Answer:

240 bananas

Step-by-step explanation:

You would need 8 sets of figs. 10 coconuts is equal to 25 figs so you need 80 coconuts. You would need enough bananas to get 4 sets of coconuts. 60*4 = 240

5 0

3 years ago

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The Empirical Rule The following data represent the length of eruption for a random sample of eruptions at the Old Faithful geys

ad-work [718]

Answer:

(a) Sample Standard Deviation approximately to the nearest whole number = 6

(b) The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is.

(c) The percentage of eruptions that last between 92 and 116 seconds using the empirical rule is 95%

(d) The actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) The percentage of eruptions that last less than 98 seconds using the empirical rule is 16%

(f) The actual percentage of eruptions that last less than 98 seconds is 15.866%

Step-by-step explanation:

(a) Determine the sample standard deviation length of eruption.

Express your answer rounded to the nearest whole number.

Step 1

We find the Mean.

Mean = Sum of Terms/Number of Terms

= 90+ 90+ 92+94+ 95+99+99+100+100, 101+ 101+ 101+101+ 102+102+ 102+103+103+ 103+103+103+ 104+ 104+104+105+105+105+ 106+106+107+108+108+108 + 109+ 109+ 110+ 110+110+110+ 110+ 111+ 113+ 116+120/44

= 4582/44

= 104.1363636

Step 2

Sample Standard deviation = √(x - Mean)²/n - 1

=√( 90 - 104.1363636)²+ (90-104.1363636)² + (92 -104.1363636)² ..........)/44 - 1

= √(199.836777 + 199.836777 + 147.2913224+ 102.7458678+ 83.47314049+ 26.3822314+ 26.3822314+ 17.10950413+17.10950413+ 9.836776857+ 9.836776857, 9.836776857+9.836776857+ 4.564049585+ 4.564049585+ 4.564049585+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 0.01859504133+ 0.01859504133+ 0.01859504133+ 0.7458677685+ 0.7458677685+ 0.7458677685+ 3.473140497+ 3.473140497+ 8.200413225+ 14.92768595+ 14.92768595+ 14.92768595+ 23.65495868+ 23.65495868+ 34.38223141+ 34.38223141+34.38223141+ 34.38223141+ 34.38223141+47.10950414+ 78.56404959+ 140.7458677+ 251.6549586) /43

= √1679.181818/43

= √39.05073996

= 6.249059126

Approximately to the nearest whole number:

Mean = 104

Standard deviation = 6

(b) On the basis of the histogram drawn in Section 3.1, Problem 28, comment on the appropriateness of using the Empirical Rule to make any general statements about the length of eruptions.

The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is .

(c) Use the Empirical Rule to determine the percentage of eruptions that last between 92 and 116 seconds.

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

Mean = 104, Standard deviation = 6

For 68% μ - σ = 104 - 6 = 98, μ + σ = 104 + 6 = 110

For 95% μ – 2σ = 104 -2(6) = 104 - 12 = 92

μ + 2σ = 104 +2(6) = 104 + 12 = 116

Therefore, the percentage of eruptions that last between 92 and 116 seconds is 95%

(d) Determine the actual percentage of eruptions that last between 92 and 116 seconds, inclusive.

We solve for this using z score formula

The formula for calculating a z-score is is z = (x-μ)/σ

where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean = 104, Standard deviation = 6

For x = 92

z = 92 - 104/6

= -2

Probability value from Z-Table:

P(x = 92) = P(z = -2) = 0.02275

For x = 116

z = 92 - 116/6

= 2

Probability value from Z-Table:

P(x = 116) = P(z = 2) = 0.97725

The actual percentage of eruptions that last between 92 and 116 seconds

= P(x = 116) - P(x = 92)

= 0.97725 - 0.02275

= 0.9545

Converting to percentage = 0.9545 × 100

= 95.45%

Therefore, the actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) Use the Empirical Rule to determine the percentage of eruptions that last less than 98 seconds

The empirical rule formula:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

For 68% μ - σ = 104 - 6 = 98,

Therefore, 68% of eruptions that last for 98 seconds.

For less than 98 seconds which is the Left hand side of the distribution, it is calculated as

= 100 - 68/2

= 32/2

= 16%

Therefore, the percentage of eruptions that last less than 98 seconds is 16%

(f) Determine the actual percentage of eruptions that last less than 98 seconds.

The formula for calculating a z-score is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For x = 98

Z score = x - μ/σ

= 98 - 104/6

= -1

Probability value from Z-Table:

P(x ≤ 98) = P(x < 98) = 0.15866

Converting to percentage =

0.15866 × 100

= 15.866%

Therefore, the actual percentage of eruptions that last less than 98 seconds is 15.866%

4 0

3 years ago

Let Ebe the set of all even positive integers in the universe Zof integers, and XE : Z R be the characteristic function of E.

AnnZ [28]

Answer:

$\mathbf{X_E (2) = 1}$

$\mathbf{X_E (-2) = 0 }$

$\mathbf{\{ x \in Z: X_E(x) = 1\} = E}$

Step-by-step explanation:

Let E be the set of all even positive integers in the universe Z of integers,

i.e

E = {2,4,6,8,10 ....∞}

$X_E : Z \to R$ be the characteristic function of E.

∴

$X_E(x) = \left \{ {{1 \ if \ x \ \ is \ an \ element \ of \ E} \atop {0 \ if \ x \ \ is \ not \ an \ element \ of \ E}} \right.$

For XE(2)

$\mathbf{X_E (2) = 1}$ since x is an element of E (i.e the set of all even numbers)

For XE(-2)

$\mathbf{X_E (-2) = 0 }$ since - 2 is less than 0 , and -2 is not an element of E

For { x ∈ Z: XE(x) = 1}

This can be read as:

x which is and element of Z such that X is also an element of x which is equal to 1.

∴

$\{ x \in Z: X_E(x) = 1\} = \{ x \in Z | x \in E\} \\ \\ \mathbf{\{ x \in Z: X_E(x) = 1\} = E}$

E = {2,4,6,8,10 ....∞}

5 0

3 years ago

Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa

4vir4ik [10]

Using the z-distribution, it is found that since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is tested if the mean length of jail time is still of 2.5 years, that is:

$H_0: \mu = 2.5$

At the alternative hypothesis, it is tested if it has increased, that is:

$H_1: \mu > 2.5$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
$\sigma$ is the standard deviation of the sample.
n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 3, \mu = 2.5, \sigma = 1.5, n = 26$

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}$

$z = 1.7$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05, is of $z^{\ast} = 1.645$

Since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

A similar problem is given at brainly.com/question/24166849

5 0

3 years ago