Indicate in standard form the equation of the line passing through the given points.
L(5.0), M(0,5)

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:

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:

1) a = -⅙, b = ⅙, c = 1

2) 6 units

Step-by-step explanation:

1) f(x) = ax² + bx + c

Given the roots, we can write this as:

f(x) = a (x + 2) (x − 3)

We know that f(13) = -25, so we can plug this in to find a:

-25 = a (13 + 2) (13 − 3)

-25 = 150a

a = -⅙

Therefore, the factored form is:

f(x) = -⅙ (x + 2) (x − 3)

Distributing:

f(x) = -⅙ (x² − x − 6)

f(x) = -⅙ x² + ⅙ x + 1

Graph: desmos.com/calculator/6m6tjoodvb

2) Volume of a right prism is area of the base times the height.

V = Ah

The base is an equilateral triangle.  Area of a triangle is one half the base times height.

V = ½ ab h

The height of the triangle is the same as the height of the prism.

V = ½ bh²

In an equilateral triangle, the height is equal to half the base times the square root of 3.

V = ½ b (½√3 b)²

V = ⅜ b³

Given that V = 81, solve for b.

81 = ⅜ b³

216 = b³

b = 6

In the given study the response variable is the exam performance, the explanatory variable is the light level, and the blocking variable is the gender of the students.

In this study, the response variable is the exam performance of the students. This is what is being measured as an outcome. The explanatory variable is the light level. The different light levels (fluorescent overhead lighting, yellow overhead lighting, or no overhead light and only desk lamps) constitute the treatments are its levels. The blocking variable in this case is the gender of the students. By making sure that both males and females are equally represented in each treatment, the researcher is controlling for the effect of gender. The levels of this blocking variable are male and female.

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The study measures the effect of light level (explanatory variable) on student exam performance (response variable), with gender as a blocking variable. Lurking variables and study design elements like random assignment and blinding are crucial for maintaining the validity of the results.

a) In the study described, the response variable is the exam performance of the students. This variable will be measured to assess the impact of different lighting conditions on students' ability to perform on an exam.

b) The explanatory variable, or independent variable, is the type of lighting. The levels of this variable are fluorescent overhead lighting, yellow overhead lighting, and no overhead lighting (only desk lamps).

c) The blocking variable is gender. The researcher wants to make sure that both males and females are equally represented in each treatment to test the hypothesis that light levels might affect genders differently. The levels of this variable are simply male and female.

When selecting participants, it is important to consider random assignment to ensure that each treatment group is similar in all respects other than the treatment itself. The idea of dividing participants to drive without distraction and to text and drive could be problematic due to ethical considerations and the introduction of confounding variables.

Lurking variables that could interfere with the study on light levels might include the time of day the exam is taken, students' prior knowledge and preparation levels, or even the difficulty of the exam itself.

Blinding could be used by ensuring that the person measuring exam performance does not know which lighting condition the student was exposed to, thus preventing any bias in the evaluation of the exam performance.

Answer:

For this case is wrong use the multiplication for P(X=2):

0.42*0.42*0.58*0.58*0.58 = 0.0344

Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:

[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]

And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=5, p=0.42)[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

For this case is wrong use the multiplication for P(X=2):

0.42*0.42*0.58*0.58*0.58 = 0.0344

Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:

[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]

And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.

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:

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.

Answer:

0.9506

Step-by-step explanation:

Pr(A) = 0.02

Pr(B) = 0.03

Pr(both) = 0.007

So,

Pr(Not A) = 1 - Pr(A)

               = 1 - 0.02

               = 0.98

Pr(Not B) = 1 - Pr(B)

               = 1 - 0.03

                = 0.97

Pr(Not by both) = 1 - Pr(both)

                          = 1 - 0.007

                          = 0.993

Thus,

Pr(Not B) or Pr(Not A) = 0.97 × 0.98

                                    = 0.9056

∴ the probability that the house would not be damaged by a hurricane or an earthquake during the year is 0.9506.

Answer:

a) decreases at interval (-∞,5) and  increases at (5,∞)

b)  is convave down at interval (-∞,6) an up at interval (6,∞)

c) f(x) has an inflexion point at x=6

Step-by-step explanation:

a) for the function

f(x) = (4 − x)*e^(−x)

then the derivative of f(x) indicates if the function decreases or increases. Thus

f'(x) =df(x)/dx = -e^(−x) -(4 − x)*e^(−x)= (x-5)*e^(−x)

since e^(−x) is always positive , then

f'(x) < 0 for x<5 → f(x) decreases when x<5 ( interval (-∞,5) )

f'(x) > 0 for x>5 → f(x) increases when x>5 ( interval (5,∞) )

f'(x) = 0 for x=5 → f(x) has a local minimum ( since first decreases and then increases)

b) the concavity is found with the second derivative of f(x) , then

f''(x) =d²f(x)/(dx)² = e^(−x) - (x-5)*e^(−x) = (6-x)*e^(−x)

then

f''(x) < 0 for x>6 → f(x) is convave up for x>6 ( interval (6,∞) )

f'(x) > 0 for x<6 → f(x) is concave down  when x<6 ( interval (-∞,6) )

f'(x) = 0 for x=6 → f(x) has an inflection point at x=6

The function f(x) = (4 - x)e^-x is decreasing on the interval (-∞, 1), increasing on (1, ∞), concave up on (2, ∞), concave down on (-∞, 2), and has its point of inflection at [tex](2, 2e^-^2).[/tex]

To find the intervals of increase or decrease, we first need to find the derivative f'(x) of the function f(x) = (4 - x)e-x. Using the product rule and chain rule, we find [tex]f'(x) = e-x - (x - 4)e-x.[/tex]

Setting f'(x) to zero and solving for x, we find the critical points x = 1 and x = 4. Using test points, we determine that the function is decreasing on (-∞, 1) and increasing on (1, ∞).

Next, we find the second derivative [tex]f''(x) = -2e-x + 2(x - 4)e-x[/tex] to determine concavity. Setting f''(x) equal to zero and solving for x, we find x = 2. Using test points, we find that the function is concave up on (2, ∞) and concave down on (-∞, 2). Since the function changes concavity at x = 2, this is the point of inflection.

Substituting x = 2 into the original function f(x), we find y = 2e-2, so the point of inflection is (2, 2e-2).

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Answer:

No

explanation:

given:

n=9

[tex]\alpha[/tex]=0.05

see the attachment

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

d=0.9-2.4+0.8+...+6.5+2.2+0.8/9

 =1.0556

The variance is the sum of squared deviations from the mean divided by n-1. The standard deviation is the square root of the variance. Determine the sample standard deviation of the differences:  

s_d=√(0.9-1.0556)^2+...+(0.8-1.0556)^2/9-1

     =2.6

CLASSICAL APPROACH :

Given claim: new procedure reduces [tex]u_{d}[/tex] > 0  

The claim is either the null hypothesis or the alternative hypothesis The null hypothesis and the alternative hypothesis state the opposite of each other The null hypothesis needs to contain an equality  

[tex]H_{0}:u_{d}=0\\ H_{1}:u_{d}>0[/tex]

Determine the value of the test statistic

t=d-[tex]u_{d}[/tex]/s_d/√n

=1.220

Determine the critical value from the Student T distribution table in the appendix in the row with d_f = n- 1 = 9-1 = 8 and in the column with [tex]\alpha[/tex] = 0.05  t =1.860  

The rejection region then contains all values larger than 1.860  

If the value of the test statistic is within the failed region, then the null hypothesis is failed

1.220 < 1.860 failed H_0

There is not sufficient evidence to support the claim that the new loading/unloading procedure is effective in reducing the wait time.  

P VALUE APPROACH:

Given claim: new procedure reduces [tex]u_{d}[/tex]  > 0  

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.  

[tex]H_{0}:u_{d}=0\\ H_{1}:u_{d}>0[/tex]

Determine the value of the test statistic:  

 t=d-[tex]u_{d}[/tex]/s_d/√n

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number (or interval) in the column title of the Students T distribution in the appendix containing the t-value in the row d_f = n-1 = 9-1 = 8

0.10 < P < 0.15

If the P-value is less than the significance level, reject the null hypothesis.  

P > 0.05   failed H_0

There is not sufficient evidence to support the claim that the new loading unloading procedure is effective in reducing the wait time.  

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:

Step-by-step explanation:

We are provided that IQs of 16 students from one area of a city had a mean of 107 and a standard deviation of 10 while the IQs of 14 students from another area of the city had a mean of 112 and a standard deviation of 8.

And we have to check that is there a significant difference between the IQs of the two groups.

Firstly let,    Null Hypothesis, [tex]H_0[/tex] : The two groups have same IQs { [tex]\mu_1 = \mu_2[/tex] }

          Alternate Hypothesis, [tex]H_1[/tex] : The two groups have different IQs{ [tex]\mu_1 \neq \mu_2[/tex]}

Since we don't know about population standard deviations;

The test statistics we will use here will be ;

            [tex]\frac{(X_1bar - X_2bar)- (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] follows t distribution with [tex](n_1 + n_2 -2)[/tex] degree

                                                    of freedom { [tex]t_n__1 +n_2 - 2[/tex] }

 Here, [tex]X_1bar[/tex] = 107      [tex]X_2bar[/tex] = 112         [tex]s_1[/tex] = 10          [tex]s_2[/tex] = 8

           [tex]n_1[/tex] = 16                [tex]n_2[/tex] = 14

           [tex]s_p[/tex] = [tex]\sqrt{\frac{(n_1 - 1)*s_1^{2} + (n_2 -1)*s_2^{2} }{(n_1 + n_2 -2)} }[/tex] = 9.1261

  Test statistics = [tex]\frac{(107-112) - 0}{9.1261*\sqrt{\frac{1}{16} +\frac{1}{14} } }[/tex]  follows [tex]t_2_8[/tex]

                           = -1.50

Now at 1% level of significance t table is giving the critical value of -2.467 and our test statistics is higher than this means it does not fall in the rejection region so we will accept our null hypothesis and conclude that there is no significant difference between the IQs of the two groups.

This question can be addressed by conducting a two-sample t-test to determine if there is a significant difference between the mean intelligence quotients of students from two areas of a city. The steps include calculating pooled standard deviation, followed by standard error, calculating the t-score, and comparing it to a critical value. You can either reject or fail to reject the null hypothesis based on these results.

The question is asking if there is a significant difference in the mean intelligence quotients (IQs) of students between two areas of a city. Conducting a two-sample t-test can address this. First, let's define the null and alternative hypotheses.

To perform this test, follow these steps:

Remember, depending on the specific results, we may or may not find enough evidence to support the alternative hypothesis.

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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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Answer: D. Excess Kurtosis

Step-by-step explanation:

A fat tailed distribution is a kind of probability distribution that exhibits excess kurtosis because it means the resulting numbers from the probability distribution are on a large scale power increment or very small/ slow decreeing order. This makes the graph on the distribution literally fat tailed and makes skewness in such distribution data extremely difficult to ascertain.

Fat tails in a distribution signify excess kurtosis, which signifies more extreme values or outliers than in a normal distribution. Neither positive skewness, negative skewness, nor a kurtosis of zero signify a distribution's 'fat tails'.

If a distribution has 'fat tails', it represents 'excess kurtosis'. This term is used to describe a distribution of data that features tails that are fatter and longer than in a normal distribution. This often means the distribution exhibits more extreme values or outliers. When a distribution has excess kurtosis, it has strong outliers.

Positive skewness, negative skewness, and a kurtosis of zero have no correlation with 'fat tails'. While skewness refers to the asymmetry of a distribution, and a kurtosis of zero refers to a normal distribution, neither of these refer to the concept of 'fat tails'.

So, Fat tails in a distribution signify excess kurtosis, not positive skewness, negative skewness, or a kurtosis of zero.

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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:

[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:

x = 14.48

Step-by-step explanation:

first we have to see that we have the measurements from all sides

and we know that the angle between side 21 and 20 is 90 degrees

well to start we have to know the relationships between angles, legs and the hypotenuse.

a: adjacent

o: opposite

h: hypotenuse

sin α = o/h

cos α= a/h

tan α = o/a

let's take the left angle as α

sin α = 21/29

α = sin^-1 (21/29)

α = sin^-1 (0.7241)  

α = 46.397

Now we do the same with the smaller triangle

tan α = o/a

sin 46.397  = x/20

0.724 = x/20

0.724 * 20 = x

14.48 = x

x = 14.48

if we want to check it we can do the same procedure with the other angle

Answer:

The passengers have an average of 8.15 minutes to wait in line

Step-by-step explanation:

Using Little's law

Average Inventory = Average Flow Time * Average Flow Rate

Average Inventory = 220 passengers

Average Flow Rate = 27

Average Flow time =?

So,

220 = Average Flow Time * 27

Average Flow Time = 220/27

Average Flow Time = 8.14814814814

Average Flow Time = 8.15 --------- Approximated

So the average wait time for a passenger is 8.15 minutes

The average passenger had to wait in line for approximately 0.123 minutes.

To find the average waiting time, we need to divide the total waiting time by the number of passengers. The total waiting time can be calculated by multiplying the average number of passengers waiting (27) by the time period (1 hour). The average waiting time is then found by dividing this total waiting time by the number of passengers checked in (220).

Average waiting time = (Average number of passengers waiting × Time period) / Number of passengers checked in

Therefore, the average passenger had to wait in line for approximately 0.123 minutes.

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Answer:

Step-by-step explanation:

A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your

b2n+32n+3 = 2 (2n+2)22 b2n+12n+1

your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:

1. Either placing a 1 in available n places (because you can't place it at the end)

2. or by placing a zero in available n+1 places.

0 ϵ P

x ϵ P → 1x ϵ P , x1 ϵ P

x' ϵ P,x'' ϵ P → xx'x''ϵ P

A recursive definition for the set of binary string palindromes starts with the base cases '0' and '1'. Other palindromes can be obtained by nesting a palindrome between '0' and '0' or '1' and '1'.

A recursive definition for the set S, consisting of all binary strings that are palindromes, would be defined by two rules:

For the base cases, both '0' and '1' are in S. This covers the palindromes of length 1.

The inductive step would be: If 'P' is a string in S, then both '0P0' and '1P1' are in S. This allows us to generate palindromes of increasing lengths all the way to infinity.

By this definition, a string is a palindrome if it is the same when read from left to right and right to left. It starts with the simplest cases (single digit palindromes) and then defines how to build larger examples based on smaller ones.

Learn more about Recursive Definition here:

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I think expanding means to remove the parentheses/brackets from a problem.

For example: Say we have the expression: 3 (4 + 5). I think expanding means to multiply 3, by every number in the parentheses. So that means:

(3 * 4) + (3 * 5) = 27.

Another way to think about it is to (if you're on paper) draw a line from 3, to all the numbers inside the parentheses. The line that connects from 3 to 4, is signaling for you to multiply 3 * 4 = 12. And the line from 3 to 5 = 3 * 5 = 15. And add them.

Final answer:

In mathematics, 'expand' refers to writing an expression in an extended form using distribution. This can result in a polynomial or an infinite series, as seen in binomial expansion or exponential arithmetic.

Explanation:

In mathematics, to expand means to increase the length of an expression by distributing multiplication over addition or subtraction. For example, expanding (a + b)(c + d) results in ac + ad + bc + bd. This does not change the value of the expression, but rather writes it in an alternative form that might be more useful for further operations, such as simplification or evaluation. Binomial expansion, specifically, refers to expressing a binomial raised to a power as a series of terms, using the binomial theorem, which can sometimes result in an infinite series or a polynomial of finite length. This expansion is applicable in situations like expanding (x + y)^n or when dealing with power series expansions of standard mathematical functions including exponential arithmetic where numbers are expressed as a product of a digit term and an exponential term such as in the notation 4.57 x 10^3.

Answer:

[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]

[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]

[tex] s = \sqrt{70.131}= 8.374[/tex]

Step-by-step explanation:

For this case we can calculate the expected value with the following table"

Class    Midpoint(xi)   Freq. (fi)       xi fi          xi^2 * fi

61-66     63.5                13             825.5      52419.5  

67-72     69.5               10              695        48302.5

73-78     75.5                3              226.5      17100.75

79-84     81.5                11              896.5      73064.75

85-90    87.5                3              262.5       22968.75

________________________________________________

Total                            40            2906          213856

For this case the midpoint is calculated as the average between the minimum and maximum point for each class.  

The expected value can be calculated with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]

For this case n =40 represent the total number of obervations given,  

And for the sample variance we can use the following formula:

[tex] s^2 = \frac{\sum x^2_i f_i -\frac{\sum x_i f_i}{n}}{n-1}[/tex]

And if we replace we got:

[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]

And for the deviation we take the square root:

[tex] s = \sqrt{70.131}= 8.374[/tex]

To calculate the sample standard deviation and sample variance, first calculate the sample mean, then calculate the sample variance, and finally find the square root of the sample variance to get the sample standard deviation.

To calculate the sample standard deviation and sample variance for the given frequency distribution of heart rates, we need to follow these steps:

Using the provided data, the sample standard deviation and sample variance can be calculated as follows:

Sample mean = (65 * 13 + 69.5 * 10 + 75.5 * 3 + 81.5 * 11 + 87.5 * 3) / (13 + 10 + 3 + 11 + 3) ≈ 72.74

Sample variance = [(65 - 72.74)² * 13 + (69.5 - 72.74)² * 10 + (75.5 - 72.74)² * 3 + (81.5 - 72.74)² * 11 + (87.5 - 72.74)² * 3] / (13 + 10 + 3 + 11 + 3 - 1) ≈ 54.21

Sample standard deviation = √(54.21) ≈ 7.36

Answer:

b)continous

Step-by-step explanation:

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

Answer: i think its B

Step-by-step explanation:

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

Step-by-step explanation:

The median rent for a one-bedroom apartment stands at $3,460 a month.

Also The estimated cost of annual necessities for a single person is $43,581 — or $3,632 a month, making it the most expensive city for single people to settle down in.

And For a family of four, expect to pay about $91,785 a year for necessities — that's $7,649 per month.

For a family of four, expect to pay about $91,785 a year for necessities — that's $7,649 per month.

Although exact data isn't provided, information on related costs such as average salary and gasoline prices suggest that the average cost of living in San Francisco is high.

The average cost of living in San Francisco is significantly higher than the national average. According to Numbeo, San Francisco's overall cost of living index is 176.89, which is 76.89% higher than the U.S. average of 100. This means that you can expect to pay about 77% more for goods and services in San Francisco than you would in the average American city.

However, we can infer that the cost of living is high, considering the mean starting salary for San Jose State University graduates, nearby to San Francisco, is at least $100,000 per year. This suggests that a significant income is required to support oneself in the Bay Area.

Other factors indirectly hint at the costs associated with San Francisco living. For instance, the average cost of unleaded gasoline in the Bay Area was once $4.59, which is notably high. These pieces of information, though incomplete, indicate a high cost of living.

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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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The zeroth derivative of a function [tex]y(x)[/tex] is simply the function itself, so the zeroth order linear ODE takes the general form

[tex]y(x)=f(x)[/tex]

whose solution is [tex]f(x)[/tex].

Answer:

D < 40.2 inches

Step-by-step explanation:

The maximum width of the TV must be 36 inches. Since TVs are approximately twice as wide as they are tall, the maximum height is 18 inches.

The diagonal of a TV can be determined as a function of its width (w) and height (h) as follows:

[tex]d^2=h^2+w^2\\d=\sqrt{18^2+36^2}\\d= 40.2\ in[/tex]

Therefore, the diagonal must be at most 40.2 inches.

Since the answer choices were not provided with the question, you should choose the biggest value that is under 40.2 inches.

The maximum diagonal size of the widescreen TV that can fit between two windows 36 inches apart is slightly more than 40 inches, given that the TV has an aspect ratio where the width is about twice the height.

Let's denote the TV's width as w and the height as h. Given that widescreen TVs are about twice as wide as they are tall, we can express the width as w = 2h.

The diagonal d of the TV can be found using Pythagoras' theorem where d² = w² + h².

Substituting 2h for w, we get d² = (2h)² + h² which simplifies to d² = 4h² + h² and further to d² = 5h².

Thus, d = h√5.

If the space between windows is 36 inches, this would be the maximum width of the TV. Therefore, 36 = 2h which means that h = 18 inches. Using this height in the diagonal equation, we get d = 18√5 which is approximately 40.2 inches. This means the longest diagonal of the widescreen TV that can fit between the windows is slightly more than 40 inches.

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).