Researchers measure the body temperature of 52 randomly selected adults. They find a mean temperature of 98.2 degrees with a standard deviation of 0.682 degrees. Which of the following is the correct t-test statistic and p-value for a test of the following hypotheses?
H_o: mu = 98.6 degrees
H_a: mu notequalto 98.6 degrees
The test statistic is negative 1. 039.21.and the p-value is less than 0.000001.
The test statistic is negative 0.46.and the p-value is 2 times P(t_51 > -0.46).
The t-test statistic is negative 0.315.and the p-value is P(t_51 < -0.315).
The t-test statistic is negative 3.33.and the p-value is two times P(t_51 < -3.33).

Answer: a) Food and performance b) Randomized Experiment c) Yes d) Age, type of food

Step-by-step explanation:

a) The explanatory variable in this case is the food and sleep hours given to the groups and the response variable is the performance or the reaction time.

b) It is a randomized experiment as the people are given food and water and is under the influence of the study head.

c) From the detailed summary given, we can conclude that eating late worsens the effects of alcep depreviation as the group A was not given food and performed better than the group B who which had access to food and water.

d.Confounding variables are the age 21-50, the gender of people who took part in study and the type of food. Because changing these will result in different results.

The study examines the effects of eating late at night on sleep-deprived individuals' reaction time and attention lapses. It was a randomized experiment that concluded that eating late at night worsens these effects. Confounding variables may exist.

a. The explanatory variable in this study is the access to food and drink at night, and the response variables are the participants' reaction time and attention lapses.

b. This is a randomized experiment because the participants were randomly assigned to either group A (water only) or group B (unlimited access to food and drink).

c. Based on the results of the study, we can conclude that eating late at night worsens some of the typical effects of sleep deprivation, specifically reaction time and attention lapses.

d. There are likely to be confounding variables in this study, such as individual differences in metabolism or other lifestyle factors that could impact the participants' performance.

The probability that exactly three out of six individuals have group A blood, with a 40 percent chance for each individual, is found using the binomial probability formula. The calculation yields approximately 27.648% as the probability for exactly three individuals having group A blood.

The question asks for the probability that exactly three out of six individuals have group A blood, given that 40 percent of all individuals have group A blood. This is a binomial probability problem because there are two outcomes (having group A blood or not) and a fixed number of trials (six individuals).

Let's denote X as the random variable representing the number of individuals with group A blood. The probability of success on any given trial is p = 0.40 (having group A blood). The probability of failure is q = 1 - p = 0.60 (not having group A blood).

The binomial probability formula is:

[tex]P(X = k) = C(n, k) * p^k * q^(n-k)[/tex]

where:

For our problem:

Plugging these values into the binomial formula gives:

First, calculate the combination:

C(6, 3) = 6! / (3! * (6-3)!) = 20

Then, calculate the probability:

[tex]P(X = 3) = 20 * (0.40)^3 * (0.60)^3 = 20 * 0.064 * 0.216 = 0.27648[/tex]

Hence, the probability that exactly three of the individuals have group A blood is approximately 0.27648 or 27.648%.

Answer:

[tex]z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9[/tex]  

[tex]p_v =P(z>1.9)=0.0287[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=12.95[/tex] represent the sample mean

[tex]\sigma=3[/tex] represent the population standard deviation for the sample  

[tex]n=36[/tex] sample size  

[tex]\mu_o =12[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 12[/tex]  

Alternative hypothesis:[tex]\mu > 12[/tex]  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9[/tex]  

P-value  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>1.9)=0.0287[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

Answer:

Yes, we will reject the claim population mean = 12.

Step-by-step explanation:

It is provided that Standard deviation, [tex]\sigma[/tex] = 3 and sample mean,Xbar = 12.95 .

Let,         Null Hypothesis,[tex]H_0[/tex] : mean, [tex]\mu[/tex] = 12

      Alternate Hypothesis,[tex]H_1[/tex] : mean, [tex]\mu[/tex] > 12

Since Population is Normal so our Test Statistics will be:

                 [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows standard normal,N(0,1)

Here, sample size, n = 36.

     Test Statistics = [tex]\frac{12.95-12}{\frac{3}{\sqrt{36} } }[/tex] = 1.9

So, at 5% level of significance z % table gives the critical value of 1.6449 and our test statistics is higher than this as 1.6449 < 1.9. So,we have sufficient evidence to reject null hypothesis or accept [tex]H_1[/tex] as our test statistics falls in the rejection region because it is more than 1.6449.

Hence we conclude after testing that we will reject claim of Population mean, μ = 12.

Answer:

the probability to be accepted is 0.302 (30.2%) (many will be rejected)

Step-by-step explanation:

assuming that the rate of 4% applies to the 60 tablets then since each tablet behaves independently, the random variable X= number of tablets with defects out of 60 tablets has a binomial distribution , where:

p(X)=n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of tablets tested = 60

x = number of defective tablets

p= probability to be defective = 0.04

then in order to be accepted x≤0 , then the probability that the batch is accepted Pa is

Pa=P(x≤1) = P(0) + P(1) = (1-p)^n + n*p*(1-p)^(n-1)

replacing values

Pa= (1-p)^n + n*p*(1-p)^(n-1) =  0.96^60 + 60*0.04*0.96^59 = 0.302 (30.2%)

then the probability to be accepted is  0.302 (30.2%) (many will be rejected)

The answer is a) [tex]X=X_1 +X_2 +X_3+..X_n[/tex]  b) The expression  is

[tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex].

Given:

An urn contains n+m balls of which n is red and m is black

a)

Expressing X in terms of the defined random variables:

Let be the indicator random variable for the event that red ball i is taken before any black ball is chosen. It takes the value 1 if this event occurs and 0 otherwise.

Now, the value of X is the number of red balls removed before the first black ball is chosen. This can be expressed as the sum of the individual indicator random variables for each red ball:

[tex]X=X_1 +X_2 +X_3+..X_n[/tex]

b)

Find E[X] (expected value of X):

The expected value of a sum of random variables is the sum of the expected values of those random variables. To find E[X],

Find the expected value of each indicator random variable and then sum them up.

Therefore, the expected value   [tex]X_i[/tex] is:

[tex]E[X_i] = 1. \dfrac{1}{n+m-i+1} + 0\cdot \dfrac{m}{n+m-i+1}[/tex]

[tex]= \dfrac{1}{n+m-i+1}[/tex]

Now, to find  E[X], we sum up the expected values of the indicator random variables for all red balls:

[tex]E[X] = E[X_1]+E[X_2]+E[X_3]+.....+E[X_n][/tex]

Substitute the values of [tex]E[X_i][/tex] into the sum and simplify:

[tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex]

a) [tex]X=X_1 +X_2 +X_3+..X_n[/tex]  b) [tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex]

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X can be expressed as X = R_1 + 2R_2 + 3R_3 + ... + nR_n. To find E[X], use the linearity of expectation and the probabilities of R_i.

To determine E[X], we need to express X in terms of the random variables and then find the expected value. Let R_i denote the event that red ball i is taken before any black ball is chosen. The expression for X is:

To find the expected value, we use the linearity of expectation. Since the probabilities for each R_i are the same, we have:

Now, we need to determine the probabilities of R_i. Since the balls are drawn without replacement, the probability that red ball i is drawn before any black ball is chosen is:

Substituting this probability into the expression for E[X], we get:

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Answer:

a) [tex] n =0,  \frac{(-1)^0}{2*0+1} x^{2*0+1}= x[/tex]

[tex] n =1,  \frac{(-1)^1}{2*1+1} x^{2*1+1}= -\frac{x^3}{3}[/tex]

[tex] n =2,  \frac{(-1)^2}{2*2+1} x^{2*2+1}= \frac{x^5}{5}[/tex]

[tex] n =3,  \frac{(-1)^3}{2*3+1} x^{2*3+1}= -\frac{x^7}{7}[/tex]

b) n=0

[tex] arctan(\pi/6) \approx \pi/6 = 0.523599[/tex]

The real value for the expression is [tex] arctan (\pi/6) = 0.482348[/tex]

And if we replace into the formula of relative error we got:

[tex] \% error= \frac{|0.523599 -0.482348|}{0.482348} * 100= 8.55\%[/tex]

n =1

[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} = 0.47576[/tex]

[tex] \% error= \frac{|0.47576 -0.482348|}{0.482348} * 100= 1.37\%[/tex]

n =2

[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5} = 0.483631[/tex]

[tex] \% error= \frac{|0.483631 -0.482348|}{0.482348} * 100= 0.27\%[/tex]

n =3

[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5}-\frac{(pi/6)^7}{7} = 0.48209[/tex]

[tex] \% error= \frac{|0.48209 -0.482348|}{0.482348} * 100= 0.05\%[/tex]

[tex] \arctan (\pi/6) = 0.48[/tex]

Step-by-step explanation:

Part a

the general term is given by:

[tex] a_n = \frac{(-1)^n}{2n+1} x^{2n+1}[/tex]

And if we replace n=0,1,2,3 we have the first four terms like this:

[tex] n =0,  \frac{(-1)^0}{2*0+1} x^{2*0+1}= x[/tex]

[tex] n =1,  \frac{(-1)^1}{2*1+1} x^{2*1+1}= -\frac{x^3}{3}[/tex]

[tex] n =2,  \frac{(-1)^2}{2*2+1} x^{2*2+1}= \frac{x^5}{5}[/tex]

[tex] n =3,  \frac{(-1)^3}{2*3+1} x^{2*3+1}= -\frac{x^7}{7}[/tex]

Part b

If we use the approximation [tex] arctan x \approx x[/tex] we got:

n=0

[tex] arctan(\pi/6) \approx \pi/6 = 0.523599[/tex]

The real value for the expression is [tex] arctan (\pi/6) = 0.482348[/tex]

And if we replace into the formula of relative error we got:

[tex] \% error= \frac{|0.523599 -0.482348|}{0.482348} * 100= 8.55\%[/tex]

If we add the terms for each value of n and we calculate the error we see this:

n =1

[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} = 0.47576[/tex]

[tex] \% error= \frac{|0.47576 -0.482348|}{0.482348} * 100= 1.37\%[/tex]

n =2

[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5} = 0.483631[/tex]

[tex] \% error= \frac{|0.483631 -0.482348|}{0.482348} * 100= 0.27\%[/tex]

n =3

[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5}-\frac{(pi/6)^7}{7} = 0.48209[/tex]

[tex] \% error= \frac{|0.48209 -0.482348|}{0.482348} * 100= 0.05\%[/tex]

And thn we can conclude that the approximation is given by:

[tex] \arctan (\pi/6) = 0.48[/tex]

Rounded to 2 significant figures

Final answer:

The Maclaurin series for arctan x includes the first four terms: x, -x^3/3, x^5/5, and -x^7/7. To estimate arctan(π/6), we incrementally add these terms, leading to progressively better approximations.

Explanation:

The Maclaurin series for the arctangent function is given by:

arctan x = ∑ n=0 [infinity] (−1)n/(2n +1) x2n+1

(a) Writing out the first 4 terms for n = 0 to 3, we get:

For n = 0: x

For n = 1: −x3/3

For n = 2: x5/5

For n = 3: −x7/7

The series starts with x and alternates between subtracting and adding subsequent odd-powered terms, each divided by the respective odd number.

(b) To estimate arctan(π/6), also known as arctan(1/(√3)), we add terms of the series one by one:

Simplest estimate: arctan(π/6) ≈ π/6

Adding the second term: arctan(π/6) ≈ π/6 - (π/6)3/3

Including the third term: arctan(π/6) ≈ π/6 - (π/6)3/3 + (π/6)5/5

Including the fourth term: arctan(π/6) ≈ π/6 - (π/6)3/3 + (π/6)5/5 - (π/6)7/7

By computing these sums, we get increasingly accurate estimates for the value of arctan(π/6).

Check the picture below.

as we can see, the triangular prism is really just 3 rectangles and 2 triangles stacked up to each other at the edges, so if we simply get the area of each of those five figures and sum them up, that's the surface area of the triangular prism.

[tex]\bf \stackrel{\textit{three rectangles' area}}{(17\cdot 11)+(16\cdot 11)+(17\cdot 11)}~~+~~\stackrel{\textit{two triangles' area}}{\cfrac{1}{2}(16)(15) + \cfrac{1}{2}(16)(15)} \\\\\\ 187+176+187+120+120\implies 790[/tex]

Answer:

95%

Step-by-step explanation:

Upper limit = 8.09 cm

Lower limit = 6.41 cm

Distribution mean = 7.25 cm

Standard deviation = 0.42 cm

The number of standard deviations from the mean of the upper and lower limits are, respectively:

[tex]N_U=\frac{U-M}{SD} =\frac{8.09-7.25}{0.42}=2 \\N_L=\frac{M-L}{SD} =\frac{7.25-6.41}{0.42}=2[/tex]

Both limits are two standard deviations away from the mean.

According to the empirical rule, in normal distributions, 95% of the data falls within two standard deviations of the mean. Therefore, 95% of the apples have diameters that are between 6.41cm and 8.09 cm.

Answer:

$568.148

Step-by-step explanation:

We will relate the loan (X) with their nominal annual rate converted semiannually:

Jennifer's loan, [tex]X = 800(1+{\frac{j}{2}})^{-10*2}[/tex]

Brian's loan, [tex]X = 1,120(1+{\frac{2j}{2}})^{-10*2}[/tex]

Since Brain and Jennifer took the same amount loan, the two equations of semi annual rates can be combined thus:

[tex]X = 800(1+{\frac{j}{2}})^{-20}} = 1,120(1+{\frac{2j}{2}})^{-20}\\= {\frac{800}{1,120}}(1+{\frac{j}{2}})=(1+{\frac{2j}{2}})[/tex]

For simplicity, we will use "Y" to represent 0.714  (i.e:  [tex]{\frac{800}{1,120}} = 0.714[/tex] )

Therefore, continuing with the equation above:

[tex]Y + {\frac{Yj}{2}}=1+j\\2Y+Yj=2+2j\\2Y+Yj-2j=2\\Yj-2j=2-2Y\\j(Y-2)=2-2Y\\j={\frac{2-2Y}{Y-2}}[/tex]

substituting the real value of Y (0.714) into the equation, we have:

[tex]j = {\frac{2-(2*0.714)}{0.714-2}}\\={\frac{2-1.428}{-1.286}}\\={\frac{0.572}{-1.286}}\\ =-0.445[/tex]

Solving for the value of X using j, we have:

[tex]X(1+{\frac{j}{2}})=800\\X=568.148[/tex]

This is a mathematical problem about compound interest and loan repayment. It creates two equations based on the given scenario. These equations can be solved to find the initial loan amount X.

This question pertains to the concepts of compound interest and loan repayment in mathematics. Given the terms, we can use the formula for the future value of a loan: FV = P(1 + r/n) ^(nt). Here, FV is the future value of the loan, P is the principal loan amount, r is the annual interest rate, n is the number of times that interest is compounded per time t, and t is the time the money is invested for.

According to the problem, the nominal semi-annual rate charged to Jennifer is exactly half of the rate charged to Brian. Hence, if we denote the rate for Brian as 2r, the rate for Jennifer would be r. We know Jennifer's payment is 800 and Brian's is 1,120; these are the future values of their respective loans.

So, we get two equations from this problem:

1. X(1 + r/2)^(2*10) = 800

2. X(1 + 2r/2)^(2*10) = 1,120

You can solve these equations to find the value of X, which represents the amount they borrowed initially.

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The answer & explanation for this question is given in the attachment below.

The number of aces in the first game and the number of spades in the 5th game follow a Hypergeometric Distribution while the number of games receiving at least one ace can be modeled by a Binomial distribution. The event of all cards being from the same suit can be thought of as a Uniform distribution.

a) The number of aces you get in the first game follows a Hypergeometric Distribution. In such a distribution, you are drawing cards without replacement. The parameters are N=52 (the population size), K=4 (the number of success states in the population i.e., the number of aces in a deck), and n=13 (the number of draws).

b) The number of games in which you receive at least one ace can be modeled by a Binomial distribution. Each game you play (out of 50) is a single trial, with the probability of success (getting at least one ace) being the same for every trial. The parameters are n=50 (the number of trials/games) and p (the probability of getting at least one ace).

c) The likelihood of all your cards being from the same suit in a game is heavily reliant on chance, can be modeled as a Uniform distribution given its rare occurrence. Essentially, the parameters would be minimum = 0 and maximum = 1. However, determining the parameters would require calculation of the specific probabilities, which is complex due to the nature of the game.

d) The number of spades you receive in the 5th game also follows a Hypergeometric distribution, similar to the situation in the first game. The parameters in this case are N=52, K=13 (number of spades in a deck), and n=13 (the number of drawn cards).

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Answer:

[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]

[tex] 2.939 \leq \sigma \leq 5.379[/tex]

Step-by-step explanation:

Data given and notation

s represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=23 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

The sample standard deviation for this case was s = 3.8

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=23-1 =22[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabele to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,22)" "=CHISQ.INV(0.975,22)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=36.78[/tex]

[tex]\chi^2_{1- \alpha/2}=10.98[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(22)(3.8)^2}{36.78} \leq \sigma^2 \leq \frac{(22)(3.8)^2}{10.98}[/tex]

[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]

Now we just take square root on both sides of the interval and we got:

[tex] 2.939 \leq \sigma \leq 5.379[/tex]

Answer:

Option A The linear model on tar content accounts for​ 92.4% of the variability in nicotine content.

Step-by-step explanation:

R-square also known as coefficient of determination measures the variability in dependent variable explained by the linear relationship with independent variable.

The given scenario demonstrates that nicotine content is a dependent variable while tar content is an independent variable. So, the given R-square value 92.4% describes that 92.4% of variability in nicotine content is explained by the linear relationship with tar content. We can also write this as "The linear model on tar content accounts for​ 92.4% of the variability in nicotine content".

The Upper R squared or the coefficient of determination here represents the percentage of the variability in the nicotine content that can be explained by the tar content in the regression model, which in this case is 92.4%.

In this context, the meaning of Upper R squared is represented by option A. The linear model on tar content accounts for 92.4% of the variability in nicotine content. This indicates that 92.4% of the change in nicotine content can be explained by the amount of tar content based on the linear regression model used. This measure is also known as the coefficient of determination. Meanwhile, options B, C, and D are not correct interpretations of the R squared in this context. Both B and D wrongly relate the percentage to the predictability of the data points and option C incorrectly associates this percentage with the residual magnitude.

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Answer:

v_s,1 > v_s,2 > v_s,3

Step-by-step explanation:

Given:

- Material 1:

         modulus of elasticity = E_1

         density of material = p_1

- Material 2:

         modulus of elasticity = E_2

         density of material = p_2

- Material 3:

         modulus of elasticity = E_3

         density of material = p_3

- E_1 > E_2 > E_3

- p_1 < p_2 < p_3

Find:

- Which material has highest speed of sound from highest to lowest:

Solution:

- The relationship between velocity of sound in a material with its elastic modulus and density is:

                               v_s = sqrt ( E / p )

- Since , v_s is proportional to E^0.5 and inversely proportional to p^0.5, then we have:

                                E_1 > E_2 > E_3

                                E_1^0.5 > E_2^0.5 > E_3^0.5

and                         p_1 < p_2 < p_3

                                p_1^0.5 < p_2^0.5 < p_3^0.5

Divide the two:      (E_1 / p_1)^0.5 > (E_1 / p_1)^0.5 > (E_1 / p_1)^0.5      

Hence,                    v_s,1 > v_s,2 > v_s,3

Answer:

(E) (0.00 + or - 0.04) kg

Step-by-step explanation:

(0.00 + or - 0.04) kg gives the most appropriate form of mass measurement.

Lower bound of the mass = 0.00 - 0.04 = -0.04 kg

Upper bound of the mass = 0.00 + 0.04 = 0.04 kg

The mass lies between -0.04 kg and 0.04 kg

Answer:

0.10865 killograms

Step-by-step explanation:

calculating the liquid mass of ammonia removed through the bottom value from a rigid tank at constant temperature.

Given:

temperature: [tex]T=70 F[/tex]

quality : 50% = 0.5

initial mass: [tex]m1= 4.5 lbm[/tex]

to find the removed liquid mass first we have to find total volume from which we can find remaining mass. as the tang is rigid the temperature and volume remains constant.

by taking the difference of mass we can determine the mass of liquid removed.

 we have two phases at temperature [tex]T= 70 F[/tex] with specific volume for liquid  [tex]vf=0.02631 ft^3/lbm[/tex]  and specific volume for vapor is  [tex]vg=2.3098 ft^3/lbm[/tex] .

The Volume in the initial state is given by, (Using definition of specific volume)

                                          [tex]V= m1v1[/tex]

 using [tex]v1=x(vf+vg)[/tex]

                                    [tex]V= m1x(vf+vg)[/tex]

substituting [tex]m1= 4.5 lbm\\[/tex] , [tex]vf= 0.02631 ft^3/lbm[/tex] , [tex]vg=2.3098 ft^3/lbm[/tex]

we get

        [tex]V= (4.5 lbm)(0.5)(0.02631 ft^3/lbm +2.3098 ft^3/lbm)[/tex]  

finally          [tex]V=5.2625 ft^3[/tex]  

we know the formula to find liquid mass is

[tex]mass =density *volume[/tex]

density of ammonia is  [tex]0.73 kg/m^3[/tex]

inserting the values into the formula we get the value for liquid mass removed through the valve.

[tex]m = (0.73 kg/m^3)*(5.25625 ft^3)[/tex]

the final answer is

                           [tex]m= 0.10865 kg[/tex]

Answer:

Step-by-step explanation:

Given that in a high school graduating class of 202 students, 95 are on the honor roll. Of these, 71 are going on to college, and of the other 107 students, 53 are going on to college.

Total students = 202

Honor roll = 95

College going out of honor = 71

Non college going out of honor = 24

Not in honor roll = 107

Not going to college from not in honour = 53

The probabilities that the person chosen is

(a) going to college,  = [tex]\frac{71+53}{202} \\=0.614[/tex]

(b) not going to college, =[tex]\frac{202-124}{202} \\=0.485[/tex]

(c) on the honor roll, but not going to college

=[tex]\frac{24}{202} \\=0.119[/tex]

Final answer:

The probabilities for a student selected at random from the class are: going to college, not going to college, and being on the honor roll but not going to college, calculated based on provided numbers for a graduating class of 202 students.

Explanation:

To solve this problem, we first need to understand the given information about the high school graduating class consisting of 202 students, with 95 on the honor roll and 107 not on the honor roll. Out of those on the honor roll, 71 are going to college, and of those not on the honor roll, 53 are going to college. Let's calculate the probabilities for each scenario.

Probability of Going to College

Total students going to college = Students on the honor roll going to college + Students not on the honor roll going to college = 71 + 53 = 124

Probability = (Total students going to college) / (Total students) = 124 / 202

Probability of Not Going to College

Total students not going to college = Total students - Total students going to college = 202 - 124 = 78

Probability = 78 / 202

Probability of Being on the Honor Roll but Not Going to College

Total students on the honor roll but not going to college = Total students on the honor roll - Students on the honor roll going to college = 95 - 71 = 24

Probability = 24 / 202

Answer:A) A

B) B

C) A

Step-by-step explanation:Experimental unit is a unit of statistical analysis.

It is also a member in a set entities.

b) Treatment is a combination of factor levels. It is an independent variable and can be manipulated by the one doing the experiment.

c) Response variables is an independent variable in which changes can be made to effect a change in the result.

Factor is a circumstance that engenders to a result.

e) Placebo is an inert substance or treatment which delivers a therapeutic value.

The experimental terms are defined as follows:

(a) The definition of an Experimental Unit is A. A​ person, object, or some other​ well-defined item upon which a treatment is applied.

(b) The definition of treatment is C. Any combination of the values of the factors​ (explanatory variables).

(c) The definition of a response variable is A. The quantitative or qualitative variable for which the experimenter wishes to determine how its value is affected by the explanatory variable.

(d) The definition of a factor is A. A variable whose effect on the response variable is to be assessed by the experimenter.

(e) The definition of a Placebo is C. An innocuous​ medication, such as a sugar​ tablet, that​ looks, tastes, and smells like the experimental medication.

(f) The definition of Confounding is when A. The effect of two factors​ (explanatory variables on the response​ variable) cannot be distinguished.

A confounding factor affects the dependent and independent variables with spurious effects.

Thus, the terms have been correctly defined with the right options.

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Answer:

2683

Step-by-step explanation:

Using the linear regression equation that predict the relationship between the weight of the luggage and the total number of passenger y = 103 + 30x, we can plug in the number of passenger x = 86 to predict the weight of the luggage on a flight:

y = 103 + 30*86 =  2683

Final answer:

To calculate the probability of the first failure of an electronic device occurring after 9 months and before 12 months, we use the exponential distribution formula. For part a, we raise the probability that one device surpasses 9 months to the power of five. For part b, we use the complementary probability that at least one device fails before 12 months.

Explanation:

The time-to-failure of an electronic device which follows an exponential distribution can be used to calculate probabilities for different time intervals. With a mean of 15 months, the rate [tex](\lambda\))[/tex] is the reciprocal of the mean, thus [tex]\(\lambda = 1/15\)[/tex]. Here's how we can calculate the requested probabilities:

For a sample of five such devices, assuming independence, we require all five devices to surpass 9 months for the first failure to occur after that time, thus we raise the probability found in part a to the power of five. Similarly, for the first failure to occur before 12 months, at least one device must fail before that time, so we use the complementary probability of all devices lasting longer than 12 months.

Let's calculate these probabilities using the exponential distribution formula.

Answer:

a) 1.4% of the samples break during shipment

b) the probability is 4/7 ( 57.14%)

Step-by-step explanation:

a) defining the event B= the sample of laboratory glass breaks , then the probability is:

P(B)= probability that sample is shipped in small packaging * probability that the sample breaks given that was shipped in small packaging +  probability that sample is shipped in large packaging * probability that the sample breaks given that was shipped in large packaging = 0.40* 0.02 + 0.60*0.01 = 0.014

b) we can use the theorem of Bayes for conditional probability. Then defining the event S= the sample is shipped in small packaging . Thus we have

P(S/B)= P(S∩B)/P(B) = 0.40* 0.02 / 0.014= 4/7 ( 57.14%)

where

P(S∩B)= probability that sample is shipped in small packaging and it breaks

P(S/B)= probability that sample was shipped in small packaging given that is broken

Answer:

b)continous

Step-by-step explanation:

A continuous data is a finite number within a chosen range example temperature range.

Answer:

There is a 24.3% probability that one of the calculators will be defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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

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

In which [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]

And p is the probability of X happening.

The probability of a defective calculator is 10 percent.

This means that [tex]p = 0.1[/tex]

If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

This is P(X = 1) when n = 3. So

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

[tex]P(X = 1) = C_{3,1}.(0.1)^{3}.(0.9)^{2} = 0.243[/tex]

There is a 24.3% probability that one of the calculators will be defective.

By definition, the half life is a quantity is the time it takes to lose half of that quantity.

So, if the half life is 60 days, it means that after 60 days you have lost half the initial amount, i.e. you're left with 100 grams.

After another 60 days, you've lost, again, half of that amount, so you're left with 50 grams.

In other words, every time a half life period passes, you're left with half the quantity you had at the beginning of that period.

So, after two half-life periods (i.e. 120 days), you'll have half of the half of the initial quantity, i.e. one quarter.

Answer:

1/2

Step-by-step explanation:

well imagine there are only 2 people who has a laptop or phone. so it would be 1/2 because there are 2 people with a laptop and phone, and they are asking who uses a cell phone to stay connected which is 1.

Answer:

Let [tex]t_i[/tex] be the time for the [tex]i[/tex]th task.

We know these times have a certain structure:

In the form of an equations we have

[tex]t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6[/tex]

So, we have the following system of equations:

[tex]t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10[/tex]

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

[tex]\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

Here is the reduced echelon form for the augmented matrix.

[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]

c) The additional rows are

[tex]\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right[/tex]

and the augmented matrix is

[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right][/tex]

d) To solve the system you must use these row operations.

[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right][/tex]

The solutions are: [tex](t_1,...,t_6)=(5,1,44,10,90,20)[/tex].

Answer:

Step-by-step explanation:

Fractions are used in telling time.

Fractions are used to determine discounts when there is sales going on.

Fraction is used to represent part of whole .

Fractions are used in baking to tell how much of an ingredient to use.

Answer:

A) 8.9%

B.) 17.35%

Step-by-step explanation:

Let X be the total number of persons age 25-plus

This means 48.7% of X are males.

Number of females become:

(100 - 48.7)x = 51.3%X

IF 23.8% of these males were graduates, then it means

23.8% * 48.7%X of these males are graduates.

0.238 * 0.487 X = 0.116X males are graduates.

That is 11.6% of persons are male College graduates.

If the question states that 20.5% of male and female were college graduates, then to answer question A, to get the number of female college graduates, it becomes:

20.5% - 11.6% = 8.9%

That means 8.9% of persons were female college graduates. .

If 53.1% of persons are females, let y be the percentage of these females that are College graduates.

Then it implies

Y * 51.3 = 8.9

Y = 8.9/51.3

Y = 0.1735 = 17.35%

This means the proportion of females that are College graduates = 17.35%

In this exercise we have to use the knowledge of statistics to describe the results in percentage, in this way we can describe as:

A) [tex]8.9\%[/tex]

B) [tex]17.35\%[/tex]

Knowing that the information given at the beginning of the utterance is:

Calculating the number of women:

[tex](100 - 48.7)X = 51.3\%X[/tex]

[tex]20.5\% - 11.6\% = 8.9\% [/tex]

If 53.1% of human being happen woman, let y exist the allotment of these woman that exist College graduates. Then it indicate that:

[tex]Y * 51.3 = 8.9\\ Y = 8.9/51.3\\ Y = 0.1735 = 17.35\%[/tex]

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Answer:

Part 1) [tex]AC=40\ units[/tex]

Part 2) [tex]DC=29\ units[/tex]

Part 3) [tex]m\angle ABE=39^o[/tex]

Part 4) [tex]m\angle BCE=51^o[/tex]

Step-by-step explanation:

we know that

A kite has two pairs of consecutive, congruent sides the diagonals are perpendicular and the non-vertex angles are congruent

Part 1) Find AC

we know that

BD is the axis of symmetry, bisects the diagonal AC

so

[tex]AE=EC[/tex]

we have

[tex]EC=20\ units[/tex]

[tex]AC=AE+EC[/tex] ----> by segment addition postulate

therefore

[tex]AC=20+20=40\ units[/tex]

Part 2) Find CD

we know that

CDE is a right triangle (the diagonals are perpendicular)

so

Applying Pythagorean Theorem

[tex]DC^2=EC^2+ED^2[/tex]

substitute the values

[tex]DC^2=20^2+21^2[/tex]

[tex]DC^2=841\\DC=29\ units[/tex]

Part 3) Find m∠ABE

we know that

In the right triangle ABE

[tex]51^o+m\angle ABE=90^o[/tex] ----> by complementary angles

[tex]m\angle ABE=90^o-51^o=39^o[/tex]

Part 4) Find m∠BCE

we know that

[tex]m\angle BCE=m\angle BAE[/tex] ----> diagonal BD is the axis of symmetry

we have

[tex]m\angle BAE=51^o[/tex]

therefore

[tex]m\angle BCE=51^o[/tex]

The measure of AC is 40 units.

The measure of the length DC is 29 units.

The measure of the angle ABE is 39 degrees.

The measure of the angle BCE is 51 degrees.

The missing lengths and angle measures in kite ABCD.

According to the question

The rhombus is a four-sided quadrilateral with all its four sides equal in length.

Rhombus is a kite with all its four sides congruent.

A kite is a special quadrilateral with two pairs of equal adjacent sides.

1. The measure of the length of AC is;

In the figure, BD is the axis of symmetry, bisects the diagonal AC.

Then,

[tex]\rm AE = EC[/tex]

And the measure of AC is,

[tex]\rm AC = AE+EC \\ \\ AC = 20+20 \\ \\ AC = 40 \ units[/tex]

The measure of AC is 40 units.

2. In the figure, CDE is a right triangle (the diagonals are perpendicular)

Then,

By applying the Pythagoras Theorem

[tex]\rm DC^2=EC^2+ED^2\\ \\ DC^2=20^2+21^2\\\\ DC^2 = 400+441 \\ \\ DC^2 = 841\\ \\ DC = 29 \ \rm units[/tex]

The measure of the length DC is 29 units.

3. In the right triangle ABE by the complementary angles;

[tex]\rm 51+ \angle ABE = 90\\ \\ \angle ABE=90-51\\ \\ \angle ABE=39 \ degrees[/tex]

The measure of the angle ABE is 39 degrees.

4. By the axis of symmetry the diagonal BD is;

[tex]\rm m \angle BCE = m \angle BAE\\\\ m \angle BCE = m \angle BAE = 51 \ degrees[/tex]

The measure of the angle BCE is 51 degrees.

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The probability that each of the balls will be of the same color can be calculated using combinations, while the probability that each of the balls will be of all different colors can be calculated using permutations. When sampling with replacement, the probabilities of selecting each color remain the same in each selection.

(a) Probability that each of the balls will be of the same color:

This can be calculated using the concept of combinations. There are a total of 19 balls in the urn. Let's calculate the number of ways to choose all 3 balls of the same color:

Summing up the number of ways for each color, we get the total number of ways to pick 3 balls of the same color. The probability will be this number divided by the total number of ways to pick any 3 balls from the urn, which is C(19, 3).

(b) Probability that each of the balls will be of all different colors:

This can be calculated using the concept of permutations. There are a total of 19 balls in the urn. Let's calculate the number of ways to choose 1 ball of each color:

Since these events are independent, we multiply the number of ways for each color. The probability will be this number divided by the total number of ways to pick any 3 balls from the urn, which is C(19, 3).

(c) Probability that each of the balls will be of the same color (sampling with replacement):

When sampling with replacement, each ball has the same probability of being chosen in each selection. So the probability of selecting 1 red ball, 1 blue ball, and 1 green ball is the product of the probabilities of selecting each color. The probability for each color is the number of balls of that color divided by the total number of balls in the urn. Therefore, the probability will be (5/19) * (6/19) * (8/19).

(d) Probability that each of the balls will be of all different colors (sampling with replacement):

When sampling with replacement, the probability of each color being chosen in each selection remains the same. Therefore, the probability of selecting 1 red ball, 1 blue ball, and 1 green ball is the product of the probabilities of selecting each color. The probability for each color is the number of balls of that color divided by the total number of balls in the urn. Therefore, the probability will be (5/19) * (6/19) * (8/19).

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