For each gym class a school has 10 soccer balls and 6 volleyballs all of the classes share 15 basketballs. The expression 10c+6c+15 represents the total number of balls the school has for c classes what is a simpler form of the expression

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:

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:

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:

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:

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).

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

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:

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.

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: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:

[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:

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.

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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:

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:

$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 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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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:

b)continous

Step-by-step explanation:

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

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:

[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)

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.

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:

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:

(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:

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

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.