Consider the following random sample from a normal population: 14, 10, 13, 16, 12, 18, 15, and 11. What is the 95% confidence interval for the population variance?

Answer:

The probability that all 70 messages are transmitted in less than 12 minutes is 0.117.

Step-by-step explanation:

Let X = the transmission time of each message.

The random variable X follows an Exponential distribution with parameter λ = 5 minutes.

The expected value of X is:

[tex]E(X)=\frac{1}{\lambda}=\frac{1}{5}=0.20[/tex]

The variance of X is:

[tex]V(X)=\frac{1}{\lambda^{2}}=\frac{1}{5^{2}}=0.04[/tex]

Now define a random variable T as:

T = X₁ + X₂ + ... + X₇₀

According to a Central limit theorem if a large sample (n > 30) is selected from a population with mean μ and variance σ² then the sum of random variables X follows a Normal distribution with mean, [tex]\mu_{s} = n\mu[/tex] and variance, [tex]\sigma^{2}_{s}=n\sigma^{2}[/tex].

Compute the probability of T < 12 as follows:

[tex]P(T<12)=P(\frac{T-\mu_{T}}{\sqrt{\sigma_{T}^{2}}}<\frac{12-(70\times0.20)}{\sqrt{70\times 0.04}})\\=P(Z<-1.19)\\\=1-P(Z<1.19)\\=1-0.883\\=0.117[/tex]

*Use a z-table for the probability.

Thus, the probability that all 70 messages are transmitted in less than 12 minutes is 0.117.

Following are the solution to the given question:

Let X signify the letter's transmission time  [tex]i^{th}[/tex] , and [tex]i =1,2,3,...,70.[/tex] this is assumed that the  [tex]X_i \sim Exp (\lambda =5)[/tex].

[tex]\to Mean =\frac{1}{\lambda} =\frac{1}{5}=0.2\\\\ \to Variance =\frac{1}{\lambda^2} =\frac{1}{5^2} =\frac{1}{25}=0.04\\\\[/tex]

               mean [tex]= E(T) = 70 \times 0.2=14[/tex]  

               Variance [tex]=V(T) =70 \times 0.04 = 2.8[/tex]  

       Therefore

             [tex]\to P(T<12) = P (Z < \frac{12-14}{\sqrt{2.8}})[/tex]

                                    [tex]= P(Z<-1.195) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (by sy memetry) \\\\=0.5-P(0<Z<1.19)\\\\=0.5-0.3830 \\\\=0.1170\\\\[/tex]

Therefore, the final answer is "0.1170".

Learn more about the Central Limit Theorem:

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Answer:

0.209

Step-by-step explanation:

Find the sample mean and standard deviation.

μ = 1050

s = 218 / √50 = 30.8

Find the z-score.

z = (1075 − 1050) / 30.8

z = 0.81

Find the probability.

P(Z > 0.81) = 1 − 0.7910 = 0.2090

Answer:

E (X) = 6.4

Step-by-step explanation:

SOLUTION:

A random variable x = number of cups of coffee consumed on an average day.

∴Let x = 4 represent four or more cups. Round your answers to four decimal places.

X          Probability (X)

0             0.1  

1              0.15

2             0.3

3             0.75

4             0. 25

5             0.21

∴ E (X) = Ux(Mean)

0x.0.1 + 1 x.15 + 2 x 0.3 + 3 x 0.75 + 4 x 0.25 + 5 x 0.21 =  6.4

Answer:

[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]

Step-by-step explanation:

We are given the following expression:

[tex](60)^{\frac{1}{2}}[/tex]

We are given the following options:

[tex]A.~\dfrac{60}{2}\\\\B.~\sqrt{60}\\\\C.~(\dfrac{1}{60})^{2}\\\\D.~\dfrac{1}{\sqrt{60}}[/tex]

Exponent Properties:

  [tex]x^{-a} = (\dfrac{1}{x})^{a}\\\\(x^m)^n = x^{mn}\\\\\dfrac{x^m}{x^n} = x^{m-n}[/tex]

Thus, the correct answer is

[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]

Answer:

B or [tex]\sqrt{60}[/tex]

Step-by-step explanation:

Answer:

The cost of goods sold is 65,850

Step-by-step explanation:

FIFO Perpetual chart is attached.

FIFO Perpetual chart shows purchases, sales and balance of  the period.

The cost of goods sold is:

3,090 units x $5=$15,450

6,300 units x $8=$50,400

Total=65,850

Answer:

(a) the dependent variable here are the Participants.

(b) the dependent variable is discrete.

(c) The scale measurement of measurement used is interval/ratio to measure the dependent variable.

(d) the independent variable is Vitamin C

(e) The research method being used is experimental.

Step-by-step explanation:

The dependent variable (sometimes known as the responding variable) is what is being studied and measured in the experiment.

Examples of continuous dependent variable may include costs, profits and sales.But some dependent variables are discrete – that is, they take on a relatively small number of integer values.

Interval scale and ratio scale are the two variable measurement scales where they define the attributes of the variables quantitatively.A ratio scale is a measurement scale which has more or less all the properties of an interval scale. Ratio data on this scale has measurable intervals.

Experimental research is a study that strictly adheres to a scientific research design. It includes a hypothesis, a variable that can be manipulated by the researcher, and variables that can be measured, calculated and compared.

Answer: 1. The dependent variable is the common cold.

2. It is a discrete variable.

3. The interval/ratio scale.

4. The independent value is the large doses of Vitamin C administered.

5. The research method used is Experimental.

Step-by-step explanation:

1. The dependent variable is the factor that we are trying to understand. In this case, it is the common cold and how it is affected by large doses of vitamins.

2. A discrete value is computed by counting. So the dependent variable - the common cold was computed by counting the number of occurrences.

3. The Interval/ratio scale is used because both the order of measurement and the differences between them are observed.

4. The Independent variable is the control in the experiment which can be compared to changes in the dependent variable. The large doses of vitamin C affects the common cold.

5. Correlational research observes patterns in variables that occur naturally while Experimental research introduces a change and monitors its effect. The research is Experimental because a change in the form of the placebo is introduced and observed.

Answer:

E. Change in the central tendency of the process output.

Step-by-step explanation:

The x-bar chart is a control chart for the central tendency of the process output. Therefore it's purpose is to determine whether there has been a change in the central tendency of the process output.

Answer:

The change in the central tendency of the process output.

Step-by-step explanation:

Lets examine each of the listed options

The change in the percent defective in a sample is associated with the change in average outgoing quality (AOQ).

The change in the dispersion of the process output is associated with R-chart since the function of R-chart is to detect changes in the dispersion.

The change in the number of defects in a sample is associated with C-chart since the function of a C-chart is to show the number of flaws per unit in a sample.

The change in the central tendency of the process output is associated with X-bar chart since the function of X-bar chart is to check whether the values are within the appropriate limits (central tendency) of the process output or not.

Therefore, the purpose of an X-bar chart is to determine whether there has been a change in the central tendency of the process output.

Answer:

V₂=1061.85 ft³

Step-by-step explanation:

To determine where more wood is found, just find the volume of each log and see which one has the largest volume

knowing that the volume of a straight cylinder is

V=πR²h, where R=radius and h=height

[tex]V_{1}=\pi (2.2)^{2}60= 912.31ft^{3}\\V_{2}=\pi (2.6)^{2}50= 1061.85ft^{3}[/tex]

As we can observe

V₂>V₁

Answer:

x=-0.75

Step-by-step explanation:

Combine Like terms

2/3x= -5/2+2

2/3x= -2.5+2

2/3x= -0.5

Multiply both sides by 3

2x= -1.5

Divide both sides by 2

x= -0.75

Answer:

x = -4/3

Step-by-step explanation:

The LCM for the fractions is 6x.

2/3x = -5/2 + 2

Multiply through by 6x

(6x) 2/3x = (6x) -5/2 + (6x)2

4 = -15x + 12x

4 = -3x

x = -4/3

Enjoy Maths!

Answer: Jeremy drove 84 miles.

Step-by-step explanation:

Let x represent the number of miles that Brenda drove.

If Jeremy drove twice

as far as Brenda, it means that the distance covered by Jeremy would be 2x miles

When they stopped after some time, they were already

126 miles apart. This means that the total distance covered by both of them is 126 miles. Therefore,

x + 2x = 126

3x = 126

x = 126/3

x = 42 miles

The number of miles that Jeremy drove is

42 × 2 = 84 miles

Answer:

The cutoff for an A grade is 82.8.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 70, \sigma = 10[/tex]

a. What is the cutoff for an A grade?

The top 10% of the class get an A grade. So the cutoff is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 70}{10}[/tex]

[tex]X - 70 = 1.28*10[/tex]

[tex]X = 82.8[/tex]

The cutoff for an A grade is 82.8.

Answer:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

For this case the value of r = -0.66

Now we can calculate the determination coeffcient:

[tex] r^2 = (-0.66)^2 = 0.4356[/tex]

And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable

And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable

Step-by-step explanation:

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

[tex]\bar x= \frac{\sum x_i}{n}[/tex]

[tex]\bar y= \frac{\sum y_i}{n}[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x[/tex]

And the model obtained for this case is:

[tex] y = -3.4 +5.2 x[/tex]

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

For this case the value of r = -0.66

Now we can calculate the determination coeffcient:

[tex] r^2 = (-0.66)^2 = 0.4356[/tex]

And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable

And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable

The Coefficient of determination gives the proportion of variation which can be explained by the regression line.

The proportion of variation that cannot be explained by the given regression line is 56.44%

From the question, the correlation Coefficient, R = - 0.66

This means that :

Therefore, proportion of the variation in y that cannot be explained is 56.44%

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Answer:

Part a: The probability is 0.3436

Part b: The probability is 0.5381

Part c: The probability is 0.7600

Step-by-step explanation:

As per the given data

[tex]p_{T | SP} = e^{0.1}\\p_{T | Good} = e^{0.3}\\[/tex]

(a)

Probability that train is delayed by more than 1 minute = P(T > 1) = P(SP) * P(T > 1 | SP) + P(Good) * P(T > 1 | Good)

[tex]= 0.3 e^{-.1 \times 1 }+ 0.7 * e^{-.3 \times 1}\\ = 0.3 e^{-.1} + 0.7 e^{-.3}[/tex]

Probability that after 1 minute of waiting, probability that train has signal problems = P(SP | T > 1)

= P(T > 1 | SP) * P(SP) / P(T > 1) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.1 }}{ 0.3 e^{-.1 }+ 0.7 e^{-.3 }}\\\\= \dfrac{0.2714512}{0.790024}\\\\= 0.3435987[/tex]

(b)

Probability that train is delayed by more than 5 minutes = P(T > 5) = P(SP) * P(T > 5 | SP) + P(Good) * P(T > 5 | Good)

[tex]= 0.3 e^{-.1 \times 5 }+ 0.7 * e^{-.3 \times 5}\\ = 0.3 e^{-.5} + 0.7 e^{-1.5}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 5)

= P(T > 5 | SP) * P(SP) / P(T > 5) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.5 }}{ 0.3 e^{-.5 }+ 0.7 e^{-1.5 }}\\\\= \dfrac{0.1819592}{0.3381503}\\\\= 0.5381015[/tex]

(c)

Probability that train is delayed by more than 10 minutes = P(T > 10) = P(SP) * P(T > 10 | SP) + P(Good) * P(T > 10 | Good)

[tex]= 0.3 e^{-.1 \times 10 }+ 0.7 * e^{-.3 \times 10}\\ = 0.3 e^{-1.0} + 0.7 e^{-3.0}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 10)

= P(T > 10 | SP) * P(SP) / P(T > 10) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-1.0 }}{ 0.3 e^{-1.0 }+ 0.7 e^{-3.0 }}\\\\= \dfrac{0.1103638}{0.1452148}\\\\= 0.7600038[/tex]

(a) The probabilities recalculated based on the waiting time are: 21.4% after 1 minute,

(b) 54.0% after 5 minutes, and

(c) 75.9% after 10 minutes.

These calculations use Bayes' theorem and the exponential distribution functions provided.

The question pertains to conditional probability and involves exponential distributions.

Let’s solve this step-by-step:

Part (a)

Given:
pT|SP(t) = [tex]0.1e^(-0.1t)[/tex]
pT|Good(t) = [tex]0.3e^(-0.3t)[/tex]
P(Signal Problems) = [tex]0.30[/tex]
P(Good Service) = [tex]0.70[/tex]

We need to find P(Signal Problems | T ≥ 1). Using Bayes’ theorem:

Using the law of total probability:

P(T ≥ 1) = P(T ≥ 1 | Signal Problems)P(Signal Problems) + P(T ≥ 1 | Good Service)P(Good Service)
= [tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70[/tex]

Now applying Bayes’ theorem:
P(Signal Problems | T ≥ 1) = [tex][[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70][/tex]
[tex]\approx [0.9048 * 0.30] / [0.9048 * 0.30 + 0.7408 * 0.70][/tex]
[tex]\approx 0.214[/tex]

Part (b)

Similarly for T ≥ 5, the probabilities become:

Using the law of total probability:
P(T ≥ 5) = [tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex][tex]* 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 5) = [tex][[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex] [tex]* 0.70][/tex]
≈ [tex][0.6065 * 0.30] / [0.6065 * 0.30 + 0.2231 * 0.70][/tex]
≈ [tex]0.540[/tex]

Part (c)

For T ≥ 10:

Using the law of total probability:
P(T ≥ 10) = [tex]e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 10) = [tex]\frac{e^{-1} \cdot 0.30}{e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70}[/tex]
≈ [tex][0.3679 * 0.30] / [0.3679 * 0.30 + 0.0498 * 0.70][/tex]
≈ [tex]0.759[/tex]

Answer:

The volume of a pyramid is 1/3 Sh, where S is the area of the base and h is the height.  Since the area of base (Square) is , S = [tex]s^{2}[/tex], where s is the side of the base.  So the volume is

V = 1/3 [tex]s^{2}[/tex]h.

Further solution is on paper (Pictures attached)

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ([tex]v_0[/tex]) = 40 ft/sec

Deceleration of the car ([tex]\frac{dv}{dt}[/tex]) = -10 ft/sec²

Final speed of the car ([tex]v_x[/tex]) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, [tex]\frac{dv}{dt}=-10\ ft/sec^2[/tex]

Negative sign means the velocity is decreasing with time.

Now, [tex]\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})[/tex] using chain rule of differentiation. Therefore,

[tex]\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx[/tex]

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

[tex]\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft[/tex]

Therefore, the car travels a distance of 80 feet before stopping.

Answer:

a. 15.9%

b. 1.2%

Step-by-step explanation:

using the normal distribution we have the following expression

[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{5-M}{SD}) \\[/tex]

Where  the first expression in the right hand side is the z-scores, M is the mean of value 3.77 and SD is the standard deviation of value 1.23.

if we simplify and substitute values, we arrive at

[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{a-M}{SD}) \\P(X\geq 5)=P(Z \geq \frac{5-3.77}{1.23})\\P(X\geq 5)=P(Z \geq 1)\\P(X\geq 5)=1- P(Z < 5)\\P(X\geq 5)=1-0.841\\P(X\geq 5)=0.159[/tex]

in percentage, we arrive at 15.9%

b. for the percentage rated the eyes most a 1

[tex]P(X\leq a)=P(\frac{X-u}{\alpha } \leq \frac{a-M}{SD})\\for a=1\\P(X\leq 1)=P(Z \leq \frac{1-3.77}{1.23})\\P(X\leq 1)=P(Z \leq -2.25})\\P(X\leq 1)=0.012\\[/tex]

In percentage we have 1.2%

Answer:

The total possible number of complete councils that could be selected is 24,174,150.

Step-by-step explanation:

The order of the men and the women is not important. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

What is the total possible number of complete councils that could be selected

3 women from a set of 15

5 men from a set of 25

[tex]T = C_{15,3}*C_{25,5} = \frac{15!}{3!12!}*\frac{25!}{5!20!} = 455*53130 = 24174150[/tex]

The total possible number of complete councils that could be selected is 24,174,150.

Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

[tex]\frac{dB}{dt} = rB-k[/tex]

[tex]\Rightarrow \frac{dB}{dt} - rB=-k[/tex].......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor [tex]=e^{\int p(t) dt[/tex]

                                     [tex]=e^{\int (-r)dt[/tex]

                                     [tex]=e^{-rt}[/tex]

Multiplying the integrating factor the both sides of equation (1)

[tex]e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}[/tex]

[tex]\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt[/tex]

Integrating both sides

[tex]\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt[/tex]

[tex]\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C[/tex]        [ where C arbitrary constant]

[tex]\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}[/tex]

Initial condition B=7864 when t =0

[tex]\therefore 7864= \frac{k}{r} - Ce^0[/tex]

[tex]\Rightarrow C= \frac{k}{r} -7864[/tex]

Then the general solution is

[tex]B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}[/tex]

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

[tex]B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}[/tex]

[tex]\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0[/tex]

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

                                                 =$1573.17

The payment rate that is required to pay off the loan in 3 years is approximately 2949.51 dollars/year. The interest paid during the 3-year period is about 984.53 dollars.

To solve this problem, the formula for continuously compounded interest should be used which is A = P * e^(rt), where A is the value of the investment at a future time t, P is the principal amount, r is the annual interest rate, and t is the time the money is invested or borrowed for.

First, set up the equation: 7864 = k * (1/(0.13)) * (e^(0.13 * 3) - 1), then solve for k: k = 7864 * (0.13) / (e^(0.13 * 3) - 1) ≈ 2949.51 dollars/year.

Using the formula A = P * e^(rt), the total amount paid is A = 2949.51 * 3 = 8848.53 dollars. The interest paid during the 3-year period is calculated by subtracting the loan amount from the total amount paid, that is 8848.53 - 7864 = 984.53 dollars.

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Answer:  

Step-by-step explanation:  

so from the question i have here, i will be giving a step by step analysis of the question.  

(a). Here we are showing a relationship, i.e.  

P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

from the LHS;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) = P(Granite ║  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Granite)P(R₁ ∠ R₂ ∠ R₃ ║  Granite) / [ P(Granite  R₁ ∠ R₂ ∠ R₃) + P(Basaltic  R₁ ∠ R₂ ∠ R₃) ]  

= 0.25 × 0.6 / [(0.25×0.6)+(0.75×0.1)] = 0.667  

from the RHS;  

P (Basalt ║  R₁ ∠ R₂ ∠ R₃) = P(Basalt  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Basalt)P(R₁ ∠ R₂ ∠ R₃ ║  Basalt) / [ P(Basalt  R₁ ∠ R₂ ∠ R₃) + P(Granite  R₁ ∠ R₂ ∠ R₃) ]  

= 0.75 × 0.1 / [(0.25 × 0.6)+(0.75 × 0.1)] = 0.333  

Therefore from this we can infer that;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

(b). here we are asked to classify the rocks considering the measurement yield.  

Measurement yielded R1 < R3 < R2  

P(Granite ║  R₁ ∠ R₃ ∠ R₂) = P(Granite  R₁ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Granite)P(R₁ ∠ R₃ ∠ R₂ ║  Granite) / [ P(Granite  R₁ ∠ R₃ ∠ R₂) + P(Basaltic  R₁ ∠ R₃ ∠ R₂) ]  

= 0.25 × 0.25 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.294  

also for RHS;  

P (Basalt ║  R₁ ∠ R₃ ∠ R₂) = P(Basalt ║  ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Basalt)P(R₁ ∠ R₃ ∠ R₂ ║  Basalt) / [ P(Basalt  R₁ ∠ R₃∠ R₂) + P(Granite  R₁ ∠ R₃ ∠ R₂) ]  

= 0.75 × 0.2 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.706  

from this we can infer that;

P(Granite ║  R₁ ∠ R₃ ∠ R₂) ∠ P (Basalt ║  R₁ ∠ R₃ ∠ R₂)  

Also considering measurements yielded R₃ ∠ R₁ ∠ R₂  

P(Granite ║  R₃ ∠ R₁ ∠ R₂) = P(Granite  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Granite)P(R₃ ∠ R₁ ∠ R₂ ║  Granite) / [ P(Granite  R₃ ∠ R₁ ∠ R₂) + P(Basaltic  R₃ ∠ R₁ ∠ R₂) ]  

= 0.25 × 0.15 / [(0.25×0.15)+(0.75×0.7)] = 0.067

from the RHS;

P (Basalt ║  R₃ ∠ R₁ ∠ R₂) = P(Basalt  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Basalt)P(R₃ ∠ R₁ ∠ R₂ ║  Basalt) / [ P(Basalt  R₃ ∠ R₁ ∠ R₂) + P(Granite  R₃ ∠ R₁ ∠ R₂) ]  

1 - 0.067 = 0.933  

from this we can infer that;

P(Granite ║  R₃ ∠ R₁ ∠ R₂) ∠ P (Basalt ║  R₃ ∠ R₁ ∠ R₂)  

cheers i hope this helps  

Final answer:

Measurements yielding R1 < R2 < R3 suggest a granitic composition, while the pattern R1 < R3 < R2 leans granitic, and R3 < R1 < R2 lean towards basaltic. The rock type can be further supported by the appearance of the rock, with granitic rocks being coarse and light-colored due to feldspar and quartz, while basaltic rocks are fine-grained and dark with ferromagnesian minerals. If granite contains basaltic inclusions, the granite is younger.

Explanation:

In analyzing rock compositions from infrared spectrum measurements, if measurements yield R1 < R2 < R3, based on typical patterns, the rock would more likely be classified as granitic because, for granite, the intensity measurements usually increase in that order. Conversely, when measurements produce R1 < R3 < R2, it does not fit neatly into the categories provided for basalt or granite, but it leans closer to a granitic composition given that R3 is not the least among the three, which is a characteristic of basaltic rocks. Similarly, if R3 presents a lower intensity compared to R1 and R2, and you have to classify without further information, it might suggest a basaltic nature following the typical pattern for basalt (R3 < R1 < R2), although it is important to note that real-world applications may require additional contextual information to accurately determine the rock type.

When studying rock samples, you should also consider differences in mineral size (coarse-textured like granite versus fine-textured like basalt), mineral content (dark, mafic minerals vs light, felsic minerals), and elemental content of the igneous rock types. For instance, granitic rock contains feldspar and quartz, which is reflected in its coarse and light-colored appearance. In contrast, basaltic rock, which is fine-grained and dark-colored, includes ferromagnesian minerals and feldspars. Additionally, if a granitic rock has inclusions of basalt (xenoliths), this implies that the granitic rock would typically be younger than the basalt since the inclusions must have been present already for the granite to incorporate them during its formation process.

Answer:

Step-by-step explanation: see attachment

Answer:

Question is not clear please post question clearly lots of question marks.

Your differential equation is not displayed well. It though looks like this:

2d²y/dx² - 20dy/dx + 136y = 0

If this is not the differential equation, the method of solving this would still be used in solving the correct one.

We first write an auxiliary equation to the differential equation.

The auxiliary equation is:

2m² - 20m + 136 = 0

Dividing by 2, we have

m² - 10m + 68 = 0

Next, we solve the auxiliary equation to obtain the values of m.

Solving using the quadratic formula

m = [-b ± √(b² - 4ac)]/2a

Where a = 1, b = -10, and c = 68

m = [10 ± √(100 - 272)]/2

= 5 ± (1/2)√(-172)

= 5 ± (1/2)i√172

= 5 ± 6.6i

For solutions of the form a ± ib, the complimentary solution is

y = e^(ax)[C1cosbx + C2sinbx]

Therefore, the complimentary solution is

y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]

Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.

The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.

In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

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Answer:

y= 65 degrees. and x=30 degrees

Step-by-step explanation:

i haven't done this for a while so i may or may not be correct i tried to help though :)

Answer: y = 52°. x = 64°

Step-by-step explanation:

y = 26 + 26 = 52 degrees ( exterior angle of a triangle is the sum of the two opposite interior angle of the triangle)

x = 180-(52+64)(sum of angles in a triangle is 180°)

x = 180-116

x = 64degrees ( base angles of an isosceles triangle are equal)

Answer:

Step-by-step explanation:

Given that a college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section.

Quantitative reasons scores are N (429, 148)

Hence Z score for the college senior = [tex]\frac{530-429}{148} \\=0.6824[/tex]

The mean score for Verbal Reasoning section was 460 with a standard deviation of 105

Verbal reasoning score was N(460, 105)

The score of college senior = 530

Z score for verbal reasoning =[tex]\frac{530-460}{105} \\=0.6667[/tex]

comparing we can say he scored more on quantiative reasoning.

Thus comparison was possible only by converting to Z scores.

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, [tex]\mu[/tex] is given by;

P(-2.032 < [tex]t_3_4[/tex] < 2.032) = 0.95

P(-2.032 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.032) = 0.95

P(-2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95

P(X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]

                                            = [ 1080 - 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] , 1080 + 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] ]

                                             = [ 990.70 , 1169.30 ]

No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.

Therefore, the store manager believe is not correct.

The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.

In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:

The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.

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Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

√V(X)= 6.45

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

There are [tex]\binom{52}3=\frac{52!}{3!(52-3)!}[/tex] (or "52 choose 3") ways of drawing any 3 cards from the deck.

There are 13 hearts in the deck, and 26 cards with a black suit. So there are [tex]\binom{13}3[/tex] and [tex]\binom{26}3[/tex] ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is

[tex]P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}[/tex]

and the probability of drawing 3 black cards is

[tex]P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}[/tex]

All other combinations can be drawn with probability [tex]1-\frac{11}{850}-\frac2{17}=\frac{739}{850}[/tex].

Let [tex]W[/tex] be the random variable for one's potential winnings from playing the game. Then

[tex]P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\$50\\\frac2{17}&\text{for }w=\$25\\\frac{739}{850}&\text{otherwise}\end{cases}[/tex]

a. For a single game, one can expect to win

[tex]E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\$50\cdot11}{850}+\frac{\$20\cdot2}{17}+\frac{\$0\cdot739}{850}=\$3[/tex]

b. For a single game, one's winnings have a variance of

[tex]V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]

where

[tex]E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\$50^2\cdot11}{850}+\frac{\$20^2\cdot2}{17}+\frac{\$0^2\cdot739}{850}=\$^2\frac{1350}{17}\approx\$^279.41[/tex]

so that [tex]V[W]=\$^2\frac{1197}{17}\approx\$^270.41[/tex]. (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.

c. With a $5 buy-in, the expected value of the game would be

[tex]E[W-\$5]=E[W]-\$5=-\$2[/tex]

i.e. a player can expect to lose $2 by playing the game (on average).

d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:

[tex]V[W-\$5]=V[W][/tex]

so the standard deviation is the same, roughly $8.39.

e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.

The expected winning for a single game defined is : $3.59

The standard deviation of winning is : $10.11

Expected winning if game costs $5 to play is :

- $0.79

The standard deviation of winning if game costs $5 to play is : $11.33

The game should not be played with a game play fee of -$5 as the expected winning value is negative.

Recall : selection is done without replacement :

Number of hearts in a deck = 13

Probability of drawing 3 hearts :

P(drawing 3 Hearts)

First draw × second draw × third draw

13/52 × 12/51 × 11/50 = 1716/132600 = 858/66300

Probability of selecting 3 black cards :

Number of black cards in a deck = 26

P(drawing 3 black cards) :

First draw × second draw × third draw

26/52 × 25/51 × 24/50 = 15600 / 132600 = 7800/66300

Probability of making any other draw :

P(3 hearts) + P(3 blacks) + P(any other draw) = 1

858/66300 + 7800/66300 + P(any other draw) = 1

P(any other draw) = 57642/66300

For a single game :

X _______ $50 ________ $25 _______ $0

P(X)_ 858/66300__ 7800/66300_57642/66300

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+0]

E(X) = $3.588

The standard deviation = √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+0] = 105.88235

Var(X) = 105.88235 - 3.588 = 102.29435

Standard deviation of winning = √102.29435 = $10.114

If the game cost $5 to play :

Net amount won if :

3 hearts are drawn = $50 - $5 = $45

3 blacks are drawn = $25 - $5 = $20

Any other combination are drawn= $0 - $5 = -$5

The distribution becomes :

X _______ $45 ________ $20 _______ -$5

P(X)_ 858/66300__ 7800/66300_57642/66300

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+ (-5 × 57642/66300)]

E(X) = - $0.7588

Standard deviation of winning :

The standard deviation = √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+ (-5² × 57642/66300)] = 127.61764

Var(X) = 127.61764 - (-0.7588) = 128.37644

Standard deviation of winning :

Std(X) = √Var(X) = √128.37644 = $11.330

With a game cost of - $5 ; the expected winning for a single game gives a negative value, therefore you should not play the game.

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The probability that none of their four children has the disease = 0.316

Step-by-step explanation:

Step 1 :

The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25

So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75

Step 2 :

There are four children. The probability that each child has the disease is independent of the condition of the other children

The probability of more than one independent events happening together can be obtained by multiplying  probability of individual events

So, the probability that none of their 4 children has the sickle cell disease  = [tex](0.75)^{4}[/tex] = 0.316

Step 3 :

Answer :

The probability no children has the specified cell disease = 0.316

Answer:

Let x is the number of the cannon and y is the number of the row boat.Thus the equation will become

8x+5y=110

Answer: 8C+5R≤110

Step-by-step explanation: Let hours of fabrication of canoes = C

Hours of fabrication of rowboat = R

If a canoe requires 8 hours of fabrication. And a row boat requires 5 hours of fabrication. Then

8C + 5R

The fabrication department has at most 110 hours to labor each week. 

Therefore

8C+5R≤110

At most means less than or equal to.

Answer:

Range = The difference between the highest and lowest which is

The mean is the sum of all values divided by the number of values

= 59+19+30+75+42+49+57+81+11+87+91 divided by n (which is 11)

601÷11

= 54.6

Variance = sum of squared deviations from the mean divided by n-1

=679.65

Standard Deviation = This is the square root of variance which gives us;

26.07