A 40% antifreeze solution is to be mixed with a 70% antifreeze
solution to get 240 liters of a 50% solution. How many liters of the
40% solution and how many liters of the 70% solution will be used?

Answer:

a.) A= [RRR, LLL, SSS]

b.) B=[RLS, RSL, SRL, SLR, LRS, LSR]

c.) C=[RRL, RRS, RSR, RLR, SRR, LRR]

e.) D'= C' = [RRR, LLL, SSS, RLS, RSL, SRL, SLR, LRS, LSR, LLR, LLS, LSL, LRL, SLL, RLL, SSL, SSR, SLS, SRS, RSS, LSS]

Step-by-step explanation:

If three consecutive vehicles are considered with their direction as R, L and S, then

a.) when the three vehicles go in same direction, A= [RRR, LLL, SSS]

b.) When the three vehicles take the different direction, B=[RLS, RSL, SRL, SLR, LRS, LSR]

c.) when exactly two vehicles turn right, C=[RRL, RRS, RSR, RLR, SRR, LRR]

d.) repeated question of C above.

e.) D'= C' = [RRR, LLL, SSS, RLS, RSL, SRL, SLR, LRS, LSR, LLR, LLS, LSL, LRL, SLL, RLL, SSL, SSR, SLS, SRS, RSS, LSS]

Answer:

The mean number of adults who would have bank savings accounts is 32.

Step-by-step explanation:

For each adult surveyed, there are only two possible outcomes. Either they have bank savings accounts, or they do not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

In this problem, we have that:

[tex]p = 0.64[/tex]

If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.

This is E(X) when [tex]n = 50[/tex].

So

[tex]E(X) = np = 50*0.64 = 32[/tex]

The mean number of adults who would have bank savings accounts is 32.

Answer:

37 and 26

Step-by-step explanation:

Let the bigger number be x and the smaller number be y.

Then the two numbers total 63 will give us:

y+x=63------>eqn1

The two numbers have a difference of 11.

y-x=11------->eqn2

We add equations 1 and 2 to obtain:

2y=74

Divide both sides to get:

y=37

Put y=37 into eqn2 to get:

37-x=11

x=37-11

x=26

This larger number is 37 and the smaller number is 26

Answer:

114.3

Step-by-step explanation:

If a 70kg person ingests 2L of water per day containing 8 mg/L of  dinitrotoluene, the concentration of  dinitrotoluene on that person's body is:

[tex]C=2\frac{L}{day}*8\frac{mg}{L} *\frac{1}{70\ kg}\\C=0.22857\frac{mg}{kg-day}[/tex]

The hazard ratio is defined by dividing the intake dosage (C) by the reference dose (RfD)

[tex]H=\frac{0.22857}{2*10^{-3}}\\H=114.3[/tex]

The hazard ratio is most nearly 114.3.

Answer:

The answer to your question is

Step-by-step explanation:

Inequality 1

                             4(3x - 4) < 32

                             12x - 16 < 32

                             12x < 32 + 16

                              12x < 48

                                 x < 48/12

                                 x < 4

Inequality 2

                           2x + 1 ≤ 8x + 25

                           2x - 8x ≤ 25 - 1

                               - 6x ≤ 24

                                   x ≥ 24/-6

                                   x ≥ - 4

- See the graph below

- Interval [-4, 4)

To solve the inequality system, we divide both sides of each inequality by the respective coefficient to isolate the variable. The solutions are x < 4 and -4 ≤ x. The graph of the solution is a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

To solve the inequality 4(3x - 4) < 32, we divide both sides of the inequality by 4 to isolate the variable. This gives us 3x - 4 < 8. Adding 4 to both sides of the inequality gives us 3x < 12.

Finally, dividing both sides of the inequality by 3 gives us x < 4.

For the inequality 2x + 1 ≤ 8x + 25, we subtract 2x from both sides of the inequality to isolate the variable. This gives us 1 ≤ 6x + 25. Subtracting 25 from both sides of the inequality gives us -24 ≤ 6x.

Finally, dividing both sides of the inequality by 6 gives us -4 ≤ x.

The solutions to the inequality system are x < 4 and -4 ≤ x. The graph of the solution would be a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

The interval notation for the solution is (-∞, 4) and [-4, ∞).

Answer:

Step-by-step explanation:

given that we  have an urn with m green balls and n yellow balls. Two balls are drawn at random.

a) Assume that the balls are sampled without replacement.

m green and n yellow balls

For 2 balls to be drawn at the same colour

no of ways = either 2 green or 2 blue = mC2+nC2

Total no of ways = (m+n)C2

Prob =

= [tex]\frac{mC2 +nC2}{(m+n)C2} \\=\frac{m(m-1)+n(n-1)}{(m+n)(m+n-1)}[/tex]

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

B) Assume that the balls are sampled with replacement

In this case, probability for any draw for yellow or green will be constant as

n/M+n or m/m+n respectively

Reqd prob = [tex](\frac{m}{m+n} )^2 +(\frac{n}{m+n} )^2[/tex]

=[tex]\frac{m^2+n^2}{(m+n)^2}[/tex]

c) Part B prob will be more than part a because with replacement prob is more than without replacement.

II time drawing same colour changes to m-1/.(m+n-1) if with replacement but same as m/(m+n) without replacement

[tex]\frac{m}{m+n} >\frac{m-1}{m+n-1} \\m^2+mn-m>m^2+mn-m-n\\n>0[/tex]

Since n>0 is true always, b is greater than a.

The question explores the concept of probability within scenarios of drawing balls of different colors from an urn, with and without replacement. It explores how the number of balls left in the urn changes the likelihood of drawing two balls of the same color. The answer is calculated using mathematical odds and conditions, showing that replacement affects probability especially when the total number of items (balls in this case) is small.

The subject of this question is probability, specifically conditional probability and probability with and without replacement. Here are the calculations needed to answer the question:

In a nutshell, the calculation for probability tells us how likely an outcome is, given the mathematical odds and conditions (in this case, if the balls are replaced or not after drawing).

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Answer:

a. 6

b. 18-21.9

c. 4

Step-by-step explanation:

a)

The number of classes can be assessed by counting the number of class intervals

S.no Class interval

1             10-13.9

2            14-17.9

3            18-21.9

4            22-25.9

5            26-29.9

6            30-33.9

Hence, there are 6 number of classes

b)

The class limits for third class is 18-21.9.

Upper class limit= 18

Lower class limit=21.9

c)

The class width can be calculated by taking difference of two consecutive upper class limits or two consecutive lower class limits.

Class width=14-10=4

The data represent the top speeds of players in a soccer league with 7 classes, the class limits for the third class are 18 to 21.9 km/hr, and the class width is 3.9 km/hr.

The data provided represents the top speed of soccer players in specific intervals known as classes. We can analyze the information to answer the student's queries as follows:

Based on the data provided, if we assume that the sequence goes like 10-13.9, 14-17.9, 18-21.9, and so on, we can derive the following:

You can't deduce the length of a side from the perimeters of a rectangle.

Say that [tex]s[/tex] and [tex]S[/tex] are, respectively, the short and long side of the rectangle.

So, we know that

[tex]2s+2S=130 \iff 2(s+S)=130 \iff s+S=65[/tex]

But we can't solve exactly for [tex]s[/tex] nor for [tex]S[/tex], unless more information is given.

The length of one of the shorter sides of the window cannot be explicitly determined without additional information. However, considering the window has a rectangular shape, the length of the shorter side should be less than half of the total perimeter, in this case, less than 65 inches.

To find the length of one of the shorter sides of the window, we must first understand that the perimeter of a rectangle is calculated by the formula: 2*length + 2*width = Perimeter.

However, the problem doesn't specify the dimensions of the box, but it does tell us that one pair of sides (the length and the width) are not equal. This suggests that the window is a rectangle. If we assume that the length of the window is longer than the width (length > width), then the shorter sides of the rectangle (the widths) will be of equal length.

Without more details, we can't calculate the exact measurement of the length of one of the shorter sides, but we can say that it should be less than half of the total perimeter, which is less than 65 inches.

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Answer: 1

Step-by-step explanation:

If a random variable x is uniformly distributed in [a,b] the

Mean = [tex]\dfrac{a+b}{2}[/tex]

Standard deviation : [tex]\sqrt{\dfrac{(b-a)^2}{12}}[/tex]

Let x =  Times required for a cable company to fix cable problems

As per given.

x is  uniformly distributed between 40 minutes and 65 minutes.

Then , mean = [tex]\dfrac{65+40}{2}=52.5[/tex] minutes

Standard deviation : [tex]\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22[/tex]minutes

Consider , P (mean- 2(Standard deviation) < X < mean+2(Standard deviation) )

= P(52.5-2(7.22)< X <  52.5+2(7.22))

=P(38.06 <X < 66.94 ).

But x lies between 40 minutes and 65 minutes.

Also,   [40 minutes,  65 minutes]⊂ [38.06 minutes , 66.94 minutes]

Therefore ,P(38.06 <X < 66.94 ) =1

∴ The probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean is 1.

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean,

Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

z = (x - μ)/σ

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

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean.

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Answer:

When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically (0.0392 to 0.0198), to almost the half of its original value too.

Step-by-step explanation:

Poisson distribution formula

P(X=x) = f(x) = (λˣe^(-λ))/x!

λ = 0.04.

And the probability that a products one or more defects is the same thing as 1 minus the probability that a product has no defect.

P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)

P(X ≥ 1) = 1 - (0.04⁰e^(-0.04))/0! = 1 - 0.9608 = 0.0392

When the occurrence rate of defect is cut in half, that is, λ = 0.02,

P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)

P(X ≥ 1) = 1 - (0.02⁰e^(-0.02))/0! = 1 - 0.9802 = 0.0198

When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically, to almost the half of its original value too.

Hope this helps!

Answer:

variance = (27*24656384-20182^2)/(27*26) =368104.3

standard devaition SD= sqrt(368104.3) =606.716

maximum possible amount that could be awarded under the two-standard-deviation rule = mean +2*SD

= (20182/27)+(2*606.716)

= 1960.913

=$1960913

Answer:

y = - x + 5

Step-by-step explanation:

L(5.0), M(0,5)

y = mx + b

m = (5 - 0) / (0 - 5) = 5 / -5 = - 1

b = y - mx = 5 - ((-1) x 0) = 5

y = - x + 5

Answer:

Step-by-step explanation:

given is a vector as (5,9,3)

a = (5,9,3)

To find out direction cosines

First let us calculate modulus of vector a

[tex]||a|| =\sqrt{5^2+9^2+3^2} \\=\sqrt{25+81+9} \\=\sqrt{115}[/tex]

Direction ratios are (5,9,3)

Magnitude of vector a = [tex]\sqrt{115}[/tex]

So direction cosines would be

[tex](\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

Angles would be

[tex](\alpha, \beta, \gamma) = arccos ((\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

=cos inverse  (0.4662, 0.8393, 0.2798)

= (62.21, 32.93,32,94)

Answer:

Option C and D are false

Step-by-step explanation:

All the mentioned option are correct in the given scenario except option C and D.

The reason is that dose is categorized as nil, low and high so, dose is categorical variable. Also, number of tumors is quantitative variable because it can be meaningfully interpreted in numerical form. The number if tumors is discrete quantitative variable.

Now consider all options

A) Gender is categorical ; dose is ordinal

This option is true because gender can be categorized into male and female and also dose is ordinal because it has order i.e. nil,low and high.

B) Gender is discrete; weight is continuous

This option is false because gender can be a discrete variable and weight is continuous variable because it is measurable. So, the statement is true.

Option C and D are already discussed an option E is discussed in option B.

The false statement is C) Number of tumors is categorical. The number of tumors is a discrete variable that involves countable values, not a categorical variable.

In this experiment, we have different types of variables. The statement C) Number of tumors is categorical is false. The number of tumors is actually a discrete variable as it involves countable values. The other statements are correct: A) Gender is categorical; dose is ordinal, as gender is divided into male and female, which is a categorical classification, and the dose is ranked as nil, low, high which makes it an ordinal variable. Statement B) Gender is discrete; weight is continuous is also correct because gender is a discrete variable (only two possible values, male or female), and weight is a continuous variable as it can take any value within a certain range. Statement D) Dose is discrete is correct, as the dose can only take certain values (nil, low, high) it is considered a discrete variable. Lastly, E) Weight is continuous is accurate as Weight can take any value within a range and it involves measurement.

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Final answer:

The negations of the given predicate logic statements are expressed by switching the quantifiers (∀ and ∃) and adding negation symbols (¬) directly before the predicates, resulting in new statements that oppose the original ones.

Explanation:

The goal is to express the negation of each of the given predicate logic statements such that all negation symbols immediately precede predicates.

Answer:

A. Control group  

Step-by-step explanation:

A control group is a group in an experiment or study that does not receive experimental procedure during such that it is then used as a benchmark to measure how the other tested subjects do.

An experimental group is a group in an experiment or study that receives an experimental procedure. The values of gotten from the test are recorded and the effect of independent variables on the dependent variables are determined.

Control experiment can be used to determine whether or not the customers are friendly. Experimental group will be the customers whom the waiter is extra friendly toward. The control group will be the alternate customer whom the waiter is not extra friendly toward.          

Answer:

[tex] P(X>9) = 0.3593[/tex]

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=25, p=0.3089)[/tex]  

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

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

For this case we want this probability:

[tex] P(X >9)[/tex]

And we can use the complement rule like this:

[tex] P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)] [/tex]And we can find the individual probabilities like this:

[tex] P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974[/tex]  

[tex] P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011[/tex]  

[tex] P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584[/tex]  

[tex] P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02[/tex]  

[tex] P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049[/tex]  

[tex] P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092[/tex]  

[tex]P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138[/tex]

[tex] P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167[/tex]

[tex] P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168[/tex]

And in order to do the operations we can use the following excel code:

"=1-BINOM.DIST(8,25,0.3089,TRUE)"  

And we got:

[tex] P(X>9) = 0.3593[/tex]

Answer:

The correct option is (a).

Step-by-step explanation:

The hypothesis of the study can be defined as:

H₀: The divorce rate is 50% for couples who had a child within the first two years of marriage, i.e. p = 0.50

Hₐ: The divorce rate is different from 50% for couples who had a child within the first two years of marriage, i.e. p ≠ 0.50

The 90% confidence interval is: 0.402 ± 0.056 = (0.346, 0.458) ≈ (0.35, 0.46)

The confidence level is 90%, the significance level (α) is:

[tex]\alpha =1-\frac{Confidence\ level}{100}\\=1-\frac{90}{100}\\ =0.10\ or\ 10\%[/tex]

Decision Rule:

If the null hypothesis value is not contained in the 90% confidence interval then the null hypothesis will be rejected and vice-versa.

Interpretation of the Confidence interval:

Thus all the options are correct.

The 90% confidence interval for the proportion of couples who had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056. Based on this interval, we can conclude that the divorce rate is between 35% and 46%.

Based on the given information, the sociologist selected a random sample of 200 couples who had a child within the first two years of marriage. Out of these couples, 80 were found to be divorced within five years. The 90% confidence interval for the proportion of all couples that had a child within the first two years of marriage and are divorced within five years is given as 0.402 ± 0.056.

This means that we can be 90% confident that the true proportion of couples who had a child within the first two years of marriage and are divorced within five years lies between 0.402 - 0.056 and 0.402 + 0.056.

Therefore, the correct statement based on this interval is that the divorce rate is between 35% and 46%.

The proportion of sample wafers that had at least one particle is 96%, those with at least five particles is 71%. For wafers between five and ten particles (inclusive), the proportion is 64%, and strictly between five and ten particles it is 44%. The histogram based on the given data would appear roughly bell-shaped with a right skew.

(a.) To calculate the proportion of sampled wafers that had at least one particle, we sum the frequencies of all groups with one or more particles and divide by the total sample size. So, the proportion is given by: (2+3+12+11+15+18+10+12+4+5+3+1+2+1)/100 = 96/100 = 0.96 or 96%.

Similarly, for wafers with at least five particles, the calculation is: (15+18+10+12+4+5+3+1+2+1)/100 = 71/100 = 0.71 or 71%.

(b.) For wafers with between five and ten particles (inclusive), we sum frequencies from 5 to 10 particles, giving (15+18+10+12+4+5)/100 = 64/100 = 0.64 or 64%.

For wafers strictly between five and ten particles (i.e., 6 to 9 particles inclusive), the calculation is: (18+10+12+4)/100 = 44/100 = 0.44 or 44%.

(c.) A histogram with relative frequency would show the proportion of wafers (y-axis) against the number of particles (x-axis). Given the distribution of frequencies, it might be expected that the histogram appears roughly bell-shaped, though the peak would be skewed to the right considering higher numbers around 5 to 8 particles.

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Answer:

a) E(X) = -$0.0813 , s.d (X) = 3

b) E(X) = -$0.0813 , s.d (X) = 3

c) expected loss and higher stakes of loosing.

Step-by-step explanation:

Given:

- There are total 37 slots:

           Red = 18

           Black = 18

           Green = 1

- Player on bets on either Red or black

- Wins double the bet money, loss the best is lost

Find:

a) Expected value of earnings X if we place a bet of $3

b) Expected value and standard deviation if we bet $1 each on three rounds

c) compare the two answers in a and b and comment on the riskiness of the two games

Solution:

- Define variable X as the total winnings per round. We will construct a distribution tables for total winnings per round for bets of $3 and $ 1:

- Bet: $3

         X                   -3                          3                              E(X)

        P(X)           1-0.48 = 0.5135      18/37 = 0.4864    3*(.4864-.52135) = -0.08

-The s.d(X) = sqrt(9*(0.5135 + 0.4864) - (-0.08)^2) = 3.0

- Bet: $1

         X                   -1                          1                          E (X)      

        P(X)        1-0.48 = 0.5135      18/37 = 0.4864   1*(.4864-.5135) = -0.0271

-The s.d(X) = sqrt(1*(0.5135 + 0.4864) - (-0.0271)^2) = 0.999

- The expected value for 3 rounds is:

                       E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3)

- The above X winnings are independent from each round, hence:

                       E(3*X_1) = 3*E(X_1) = 3*(-0.0271) = -0.0813

- The standard deviation for 3 rounds is:

              sqrt(Var(X_1 + X_2 + X_3)) = sqrt(Var(X_1) + Var(X_2) + Var(X_3))

- The above X winnings are independent from each round, hence:

                       sqrt(Var(3*X_1)) = 3*Var(X_1) = 3*(0.999) = 2.9988

- For above two games are similar with an expected loss of $0.0813 for playing the game and stakes are very high due to high amount of deviation for +/- $3 of winnings.

In the game of European roulette, betting $1 in three different rounds is less risky than betting $3 in one round as it has a lower expected loss and lower variability.

In the game of roulette, the probability of winning (i.e. the ball landing on either red or black) is 18/37, and the probability of losing (the ball landing on green) is 1/37.

(a) When you bet $3 on a single round, the expected value of your winnings is given by (18/37 * $6) - (19/37 * $3) = -$0.081, and the standard deviation is sqrt([$6-$(-0.081)]^2 * 18/37 + [-$3-$(-0.081)]^2 * 19/37) = $2.899.

(b) When you bet $1 in three different rounds, the expected value of your winnings is 3 * [ (18/37 * $2) - (19/37 * $1) ] = -$0.243, and the standard deviation is sqrt(3 * [(($2-$(-0.081))^2 * 18/37) + (($-1-$(-0.081))^2 * 19/37)]) = $1.578.

(c) If we compare the two games, it is clear that betting $1 in three different rounds decreases both the expected loss and the variability of the results, indicating that the latter game is less risky.

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Answer: 24 feet

Step-by-step explanation:

By using Pythagoras rule:

Let x be the high up the house does the ladder reached.

X^2 + 5^2= 25^2

X^2 = 25^2 - 5^2

x^2 = 625 - 25

x^2 = 600

Square both side

x = sqrt(600)

x= 24.495

x = 24 feet

Answer:

Yes, one of the solutions of this differential equation is a constant solution equation.

Step-by-step explanation:

9xy' + 5y = 10.

If y' = 0, 9xy' = 0

5y = 10

y = 2 = c.

So, for all real values of x such that -∞ < x < ∞, 9xy' will be 0 and one of the solutions of the differential equation will be y = 2.

Hope this helps!

Answer:

The remainder is 9

Step-by-step explanation:

3^128 is divided by 17

find the value of 3^128 first.

3^128 = 1.17901845777E61

Then you divide by 17

1.17901845777E61 ÷ 17

= 6.93540269279E59

Approximately 6. 9340

6 remainder 9

Answer:

6 remander 1

Step-by-step explanation:

first you solve 3^128 which equals E61 then divide it by 17 which equals E59

Answer:

The critical value would be: [tex]\chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                      F          B         S      Bs     Total

home wins    39      156      25      83      303

Visitor wins   31       98        19      75       223

Total              70      254     44      158      526

We need to conduct a chi square test in order to check the following hypothesis:

H0: The number of home team and visiting team losses is independent of the sport.

H1: The number of home team and visiting team losses is dependent of the sport.

The level of significance assumed for this case is [tex]\alpha=0.01[/tex]

The statistic to check the hypothesis is given by:

[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{70*303}{526}=40.32[/tex]

[tex]E_{2} =\frac{254*303}{526}=146.32[/tex]

[tex]E_{3} =\frac{44*303}{526}=25.35[/tex]

[tex]E_{4} =\frac{158*303}{526}=91.02[/tex]

[tex]E_{5} =\frac{70*223}{526}=29.68[/tex]

[tex]E_{6} =\frac{254*223}{526}=107.68[/tex]

[tex]E_{7} =\frac{44*223}{526}=18.65[/tex]

[tex]E_{8} =\frac{158*223}{526}=66.98[/tex]

And the expected values are given by:

                        F           B            S          Bs     Total

home wins    40.32   146.32    25.35    91.02    303

Visitor wins   29.68   107.68    18.65    66.98    223

Total                70         254         44        158      526

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(39-40.32)^2}{40.32}+\frac{(156-146.32)^2}{146.32}+\frac{(25-25.35)^2}{25.35}+\frac{(83-91)^2}{91}+\frac{(31-29.68)^2}{29.68}+\frac{(98-107.68)^2}{107.68}+\frac{(19-18.65)^2}{18.65}+\frac{(75-66.98)^2}{66.98} =3.29[/tex]Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(4-1)(2-1)=3[/tex]

The critical value would be: [tex]chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Answer:

Let G be the number of eggs in the meal

V be the number of servings of mixed vegetables in the meal

R be the number of servings of brown rice in the meal

Objective function = Minimize 3.75G + 3.50V + 2R

Constraints:

2G + 14V + 40R ≥ 35(Carbohydrates)

18G + 15V + 6R ≥ 30(Protein)

12G + 8V + R ≤ 45(Fat)

G, M, B ≥ 0

Answer:

[tex]P(X<480)=P(\frac{X-\mu}{\sigma}<\frac{480-\mu}{\sigma})=P(Z<\frac{480-412}{68})=P(z<1)[/tex]

[tex]P(z<1)=0.841[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(412,68)[/tex]  

Where [tex]\mu=412[/tex] and [tex]\sigma=68[/tex]

We are interested on this probability

[tex]P(X<480)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<480)=P(\frac{X-\mu}{\sigma}<\frac{480-\mu}{\sigma})=P(Z<\frac{480-412}{68})=P(z<1)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(z<1)=0.841[/tex]

Answer:

a) 16% of students have an SAT math score greater than 615.

b) 2.5% of students have an SAT math score greater than 715.

c) 34% of students have an SAT math score between 415 and 515.

d) [tex]Z = 1.05[/tex]

e) [tex]Z = -1.10[/tex]

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the empirical rule.

Normal probability distribution

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

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

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

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

Empirical rule

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

[tex]\mu = 515, \sigma = 100[/tex]

(a) What percentage of students have an SAT math score greater than 615?

615 is one standard deviation above the mean.

68% of the measures are within 1 standard deviation of the mean. The other 32% are more than 1 standard deviation from the mean. The normal probability distribution is symmetric. So of those 32%, 16% are more than 1 standard deviation above the mean and 16% more then 1 standard deviation below the mean.

So, 16% of students have an SAT math score greater than 615.

(b) What percentage of students have an SAT math score greater than 715?

715 is two standard deviations above the mean.

95% of the measures are within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. The normal probability distribution is symmetric. So of those 5%, 2.5% are more than 2 standard deviations above the mean and 2.5% more then 2 standard deviations below the mean.

So, 2.5% of students have an SAT math score greater than 715.

(c) What percentage of students have an SAT math score between 415 and 515?

415 is one standard deviation below the mean.

515 is the mean

68% of the measures are within 1 standard deviation of the mean. The normal probability distribution is symmetric, which means that of these 68%, 34% are within 1 standard deviation below the mean and the mean, and 34% are within the mean and 1 standard deviation above the mean.

So, 34% of students have an SAT math score between 415 and 515.

(d) What is the z-score for student with an SAT math score of 620?

We have that:

[tex]\mu = 515, \sigma = 100[/tex]

This is Z when X = 620. So

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

[tex]Z = \frac{620 - 515}{100}[/tex]

[tex]Z = 1.05[/tex]

(e) What is the z-score for a student with an SAT math score of 405?

We have that:

[tex]\mu = 515, \sigma = 100[/tex]

This is Z when X = 405. So

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

[tex]Z = \frac{405 - 515}{100}[/tex]

[tex]Z = -1.10[/tex]

a. Approximately 15.87% of students have an SAT math score greater than 615.

b. Approximately 2.28% of students have an SAT math score greater than 715.

c. 68% percentage of students have an SAT math score between 415 and 515

d. 1.05 is the z-score for student with an SAT math score of 620

e. -1.1 is the z-score for a student with an SAT math score of 405

To answer these questions, we can use the properties of a bell-shaped distribution and the empirical rule. The empirical rule states that for a bell-shaped distribution:Approximately 68% of the data falls within one standard deviation of the mean. Approximately 95% of the data falls within two standard deviations of the mean.Approximately 99.7% of the data falls within three standard deviations of the mean.

Given information: Mean (μ) = 515 and Standard Deviation (σ) = 100

(a) To find this, we need to calculate the z-score for a score of 615 and then find the percentage of data above that z-score using the standard normal distribution table.

z-score = (X - μ) / σ

z-score = (615 - 515) / 100

z-score = 1

Using the standard normal distribution table (or calculator), we find that approximately 84.13% of the data is below a z-score of 1. Since the distribution is symmetric, the percentage above the z-score of 1 is also approximately 100% - 84.13% = 15.87%.

Therefore, approximately 15.87% of students have an SAT math score greater than 615.

(b) We repeat the same process for a score of 715.

z-score = (715 - 515) / 100

z-score = 2

Using the standard normal distribution table (or calculator), we find that approximately 97.72% of the data is below a z-score of 2. The percentage above the z-score of 2 is approximately 100% - 97.72% = 2.28%.

Therefore, approximately 2.28% of students have an SAT math score greater than 715.

(c) We can use the empirical rule to find the percentage of students within one standard deviation of the mean and then subtract that from 100% to find the percentage between 415 and 515.

Percentage between 415 and 515 ≈ 68%

(d) We can calculate the z-score as follows:

z-score = (620 - 515) / 100

z-score = 1.05

(e) We can calculate the z-score as follows:

z-score = (405 - 515) / 100

z-score = -1.1

Remember that a negative z-score indicates a value below the mean.

Note: These calculations assume a normal distribution, and the actual percentages may vary slightly due to the discrete nature of test scores and rounding in calculations.

To know more about students:

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Answer:

Step-by-step explanation:

a) The standard deviation is usually preferred over the range because it is calculated from all of the data and will not be impacted as much as the range when they are outliers,and the standard deviation uses all of the data.

b)The IQR sometimes referred to the standard deviation when there is an outlier because the IQR is less sensitive to this features than standard deviation.

that is the IQR is not affected by an outlier,while the standard deviation is affected by an outlier.

c)The advantage of the standard deviation over the IQR is the standard deviation takes into account the values of all observation,while the IQR uses only some of the data.

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

[tex]v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}[/tex]

Where:

[tex]x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval[/tex]

a. Let's find h(3) and h(4) using the data provided by the problem:

[tex]h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f[/tex]

The average velocity over the interval [3, 4] is :

[tex]v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s[/tex]

b. Let's find h(3.5) using the data provided by the problem:

[tex]h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f[/tex]

The average velocity over the interval [3, 3.5] is :

[tex]v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s[/tex]

c. Let's find h(3.1) using the data provided by the problem:

[tex]h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f[/tex]

The average velocity over the interval [3, 3.1] is :

[tex]v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s[/tex]

d. Let's find h(3.01) using the data provided by the problem:

[tex]h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f[/tex]

The average velocity over the interval [3, 3.01] is :

[tex]v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s[/tex]

e. Let's find h(3.001) using the data provided by the problem:

[tex]h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f[/tex]

[tex]v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s[/tex]

The average velocity over a given time [3,4] is 38.19 m/sec, the average velocity over a given time [3,3.5] is 38.605 m/sec, the average velocity over a given time [3,3.1] is 38.937 m/sec and this can be determined by using the given data.

Given :

Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by [tex]\rm h(t) = 44t-0.83t^2[/tex].

a) [3,4]

At time t = 3 and 4, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(4) = 44(4)-0.83(4)^2 = 176-13.28=162.72[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{162.72-124.53}{4-3}=38.19[/tex]

b) [3,3.5]

At time t = 3 and 3.5, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.5) = 44(3.5)-0.83(3.5)^2 = 154-10.1675=143.8325[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{143.8325-124.53}{3.5-3}=38.605[/tex]

c) [3,3.1]

At time t = 3 and 3.1, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(4) = 44(3.1)-0.83(3.1)^2 = 136.4-7.9763=128.4237[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{128.4237-124.53}{3.1-3}=38.937[/tex]

d) [3,3.01]

At time t = 3 and 3.01, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.01) = 44(3.01)-0.83(3.01)^2 = 132.44-7.519883=124.920117[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{124.920117-124.53}{3.01-3}=39.0117[/tex]

e) [3,3.001]

At time t = 3 and 3.001, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.001) = 44(3.001)-0.83(3.001)^2 = 132.044-7.47498083=124.5690192[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{124.5690192-124.53}{3.001-3}=39.01917[/tex]

For more information, refer to the link given below:

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Answer:

144 observations

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.8684}{2} = 0.0668[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.0668 = 0.9332[/tex], so [tex]z = 1.5[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

[tex]\sigma = 4, M = 0.5[/tex]

We want to find n

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.5 = 1.5*\frac{4}{\sqrt{n}}[/tex]

[tex]0.5\sqrt{n} = 6[/tex]

[tex]\sqrt{n} = \frac{6}{0.5}[/tex]

[tex]\sqrt{n} = 12[/tex]

[tex]\sqrt{n}^{2} = (12)^{2}[/tex]

[tex]n = 144[/tex]