Mathematics College

Please help, thank you!

Two numbers total 63 and have a difference of 11. Find the two numbers.

Answers

Answer 1

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

Related Questions

100pts: What is the remainder when 3^128 is divided by 17?

Answers

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

In an experimental study on friendliness and tipping, every alternate customer to whom the waiter is extra friendly toward are referred to as the...A. Control group
B. Experimental group
C. Nonexperimental group
D. Dependent group

Answers

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.

Defects in a product occur at random according to a Poisson distribution with parameter ???? = 0.04. What is the probability that a product has one or more defects? If the manufacturing process of the product is improved and the occurrence rate of defects is cut in half to ???? = 0.02. What effect does this have on the probability that the product has one or more defects?

Answers

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!

The game of European roulette involves spinning a wheel with 37 slots: 18 red, 18 black, and 1 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money.(a) Suppose you play roulette and bet $3 on a single round. What is the expected value and standard deviation of your total winnings?(b) Suppose you bet $1 in three dierent rounds. What is the expected value and standard deviation of your total winnings?(c) How do your answers to parts (a) and (b) compare? What does this say about the riskiness of the two games?

Answers

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.

Final answer:

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.

Explanation:

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.

Learn more about European roulette here:

https://brainly.com/question/35806988

#SPJ2

The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a sample of size 100, resulting in the following frequencies:
Number of particles: 0, 1, 2, 3, 4, , 5, 6, 7
Frequency: 1, 2, 3, 12, 11, 15, 18, 10
Number of particles: 8, 9, 10, 11, 12, 13, 14
Frequency: 12, 4, 5, 3, 1, 2, 1
(a.) What proportion of the sampled wafers had at least one particle? At least five particles?
(b.) What proportion of the sampled wafers had between five and ten particles, inclusive? Strictly between five and ten particles?
(c.) Draw a histogram using relative frequency on the vertical axis. How would you describe the shape of the histogram.

Answers

Final answer:

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.

Explanation:

(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.

Learn more about Proportions and Histogram here:

https://brainly.com/question/27004795

#SPJ12

An experiment was performed upon rats to investigate the effect of ingesting Alar (a chemical sprayed on apple trees to keep fruit from dropping before ripe) upon subsequent cancer rates.
The following variables were measured:
gender (0=female, 1=male); weight (g); dose of Alar (nil, low, high); and number of tumors.
Which of the following is FALSE?

A) Gender is categorical; dose is ordinal
B) Gender is discrete; weight is continuous
C) Number of tumors is categorical
D) Dose is discrete
E) Weight is continuous

Answers

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.

Final answer:

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.

Explanation:

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.

Learn more about Variable Types here:

https://brainly.com/question/30463740

#SPJ3

Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes. What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?

Answers

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

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.

Find out more on Z score at: https://brainly.com/question/25638875

Express the negations of each of these statements so that all negation symbols immediately precede predicates. a. ∃z∀y∀xT (x, y, z) b. ∃x∃yP (x, y) ∧ ∀x∀yQ(x, y) c. ∃x∃y(Q(x, y) ↔ Q(y, x)) d. ∀y∃x∃z(T (x, y, z) ∨ Q(x, y))

Answers

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.

For the statement ∃z∀y∀xT(x, y, z), the negation would be ∀z∃y∃x¬T(x, y, z), which states that there is no z for which every y and every x make T(x, y, z) true.The negation of ∃x∃yP(x, y) ∧ ∀x∀yQ(x, y) is ∀x∀y¬P(x, y) ∨ ∃x∃y¬Q(x, y), meaning there are no such x and y that P(x, y) is true or there exists some x and y for which Q(x, y) is not true.For ∃x∃y(Q(x, y) ↔ Q(y, x)), the negation would be ∀x∀y¬(Q(x, y) ↔ Q(y, x)), indicating that for all x and y, it is not the case that Q(x, y) if and only if Q(y, x).The negation of the statement ∀y∃x∃z(T(x, y, z) ∨ Q(x, y)) is ∃y∀x∀z(¬T(x, y, z) ∧ ¬Q(x, y)), stating there exists a y such that for all x and z, neither T(x, y, z) nor Q(x, y) are true.

According to a survey of adults, 64 percent have money in a bank savings account. If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.

Answers

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].

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

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

The perimeter of the window of the camper shell is 130 in. Find the length of one of the shorter sides of the window.
in

Answers

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.

Final answer:

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.

Explanation:

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.

Learn more about Perimeter of a Rectangle here:

https://brainly.com/question/29595517

#SPJ12

As a result of discharges from local dry cleaner, dinitrotoluene concentration in the groundwater is 8 mg/L. RfD for dinitrotoluene is 2.0 x 10-3 mg/kg-day. The average 70 Kg person drinks 2L/day water. The hazard ratio is most nearly:

Answers

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.

In a study of environmental lead exposure and IQ, the data was collected from 148 children in Boston, Massachusetts. Their IQ scores at age of 10 approximately follow a normal distribution with mean of 115.9 and standard deviation of 14.2. Suppose one child had an IQ of 74. The researchers would like to know whether an IQ of 74 is an outlier or not.

Calculate the lower fence for the IQ data, which is the lower limit value that the IQ score can be without being considered an outlier. Keep a precision level of two decimal places for the lower fence.

Answers

Answer:

a) Lower inner fence = 77.6168 = 77.62 to 2 d.p

Lower outer fence = 48.9044 = 48.90 to 2 d.p

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159

Step-by-step explanation:

Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Q₁ = 25th percentile = lower quartile

IQR = Inter quartile Range = Q₃ - Q₁

Q₃ = 75th percentile = upper quartile

To calculate Q₁ for a normal distribution with only mean and standard deviation known,

We need the standardized score whose probability is 0.25 P(z) = 0.25

From the normal distribution table

z = (± 0.674)

z = (x - xbar)/σ

x = the value in the data we're interested in,

xbar = mean = 115.9

σ = standard deviation = 14.2

Lower quartile corresponds to (z = - 0.674)

- 0.674 = (x - 115.9)/14.2

Q₁ = X = 106.3292

The upper quartile, Q₃ corresponds to z = (+0.674)

Q₃ = 125.4708

IQR = 125.4708 - 106.3292 = 19.1416

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168

Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)

We standardize 74 by obtaining its z-score

z = (x - xbar)/σ

z = (74 - 115.9)/14.2 = - 2.95

P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)

Final answer:

The lower fence for the IQ data, which determines whether an IQ score is an outlier, is calculated as the mean minus two times the standard deviation. In this case, the lower fence is 87.5, which makes an IQ score of 74 an outlier as it falls significantly below this threshold.

Explanation:

To determine if an IQ score is an outlier, we often use the interquartile range (IQR) and calculate the fences. However, since the data is approximately normally distributed and we have the mean and standard deviation, we can also consider an IQ score to be an outlier if it falls more than two standard deviations from the mean. In this question, we do not have the IQR, so we'll use standard deviations to calculate the outlier threshold.

The mean IQ score is 115.9 and the standard deviation is 14.2. An outlier is typically defined as a value that is more than two or three standard deviations away from the mean. These thresholds are sometimes called the outer fences in statistical outlier detection. Using two standard deviations, we can calculate the lower limit as follows:

Lower Limit = Mean - 2 × Standard Deviation
Lower Limit = 115.9 - 2(14.2)
Lower Limit = 115.9 - 28.4
Lower Limit = 87.5

Therefore, an IQ score of 74 is considerably lower than the lower limit of 87.5, suggesting that it could indeed be considered an outlier.

Solve, graph, and give interval notation for the inequality:

4(3x − 4) < 32 AND 2x + 1 ≤ 8x + 25

Answers

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)

Final answer:

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.

Explanation:

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, ∞).

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

Answers

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

Find the direction cosines and direction angles of the given vector. (Round the direction angles to two decimal places.) a = 5, 9, 3 cos(α) = cos(β) = cos(γ) = α = ° β = ° γ = °

Answers

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)

Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a week, what's the probability that he gets greater than 9 hits?

Answers

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]

You would like to make a nutritious meal of eggs, mixed vegetables and brown rice. The meal should provide at least 35 g of carbohydrates, at least 30 g of protein, and no more than 45 g of fat. One serving of eggs contains 2 g of carbohydrates, 18 g of protein, and 12 g of fat. A serving of mixed vegetables contains 14 g of carbohydrates, 15 g of protein, and 8 g of fat. A serving of rice contains 40 g of carbohydrates, 6 g of protein, and 1 g of fat. A serving of eggs costs $3.75, a serving of mixed vegetables costs $3.50, and a serving of rice costs $2. It is possible to order a partial serving, e.g. 0.75 servings of rice. Formulate a linear optimization model that could be used to determine the number of servings of eggs, mixed vegetables, and rice for your meal that meets the nutrition requirements at minimal cost.

Answers

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

Suppose that vehicles taking a particular freeway exit can turnright (R), turn left (L), or go straight (S). Consider observingthe direction for each of three successive vehicles.
a. List all outcomes in the event A that all three vehicles go inthe same direction.
b. List all outcomes in the event B that all three vehicles takedifferent directions.
c. list all outcomes in the event C that exactly two of the threevehicles turn right.
d. List all outcomes in the event C that exactly two of the threevehicles turn right.
e. List outcomes in D', C U D, and C D

Answers

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]

Exercise 1.28. We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color? (a) Assume that the balls are sampled without replacement. (b) Assume that the balls are sampled with replacement. (c) When is the answer to part (b) larger than the answer to part (a)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?

Answers

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.

Final answer:

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.

Explanation:

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:

(a) When the balls are drawn without replacement, the probability that the two balls drawn have the same color is the sum of the probability of drawing two green balls and the probability of drawing two yellow balls. The probability of drawing two green balls is (m/(m+n)) * ((m-1)/(m+n-1)). Similarly, the probability of drawing two yellow balls is (n/(m+n)) * ((n-1)/(m+n-1)). The sum of these two probabilities gives the required probability. (b) When the balls are drawn with replacement, the same logic applies; however, since the balls are replaced, the denominator term doesn't decrease for the second draw. Thus, the probability of drawing two green balls is (m/(m+n)) * (m/(m+n)), and the probability of two yellow balls is (n/(m+n)) * (n/(m+n)). (c) The answer to part (b) becomes larger than the answer to part (a) when m and n are small numbers. This is because, when m and n are small, the probability of drawing a similarly colored ball in the second draw becomes more significant if the ball is replaced after the first draw, compared to if it is not replaced, causing the probability with replacement to be higher.

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).

Learn more about Probability here:

https://brainly.com/question/22962752

#SPJ3

Use the concept thaty = c, −[infinity] < x < [infinity],is a constant function if and only ify' = 0to determine whether the given differential equation possesses constant solutions.9xy' + 5y = 10

Answers

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!

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 h(t)=44t−0.83t2h(t)=44t-0.83t2. Find the average velocity over the given time intervals.

a. [3, 4]:
b. [3, 3.5]:
c. [3, 3.1]:
d. [3, 3.01]:
e. [3, 3.001]

Answers

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: