What is the future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.) Future value

Answer:

graph a

Step-by-step explanation:

The required shortest length to go from A to G is given as 8. Option C is correct.

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
The shortest distance from the figure can be given as,
= AE + EG
= 1 + 7
= 8

Thus, the required shortest length to go from A to G is given as 8. Option C is correct.

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Answer:

A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.

B) slope: (5.6-1.8)/(2-0) = 1.9

y = 1.9x + 1.8

(Line of best fit)

x = 13,

y = 1.9(13) + 1.8 = 26.5

Predicted no. of games won after 13 months of practice is 26.5

Final answer:

The y-intercept, representing initial games won and the equation for line of best fit predicting future games won, are determined from the graph data.

Explanation:

Part A:

Part B:

The equation for the line of best fit in slope-intercept form is y = 1.4x + 1.8.

To predict the number of games won after 13 months of practice, substitute x = 13 into the equation:

y = 1.4(13) + 1.8 = 19.5

So, the predicted number of games that could be won after 13 months of practice is 19.5.

Answer:

The​ P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is 4.4 which is LOW so there IS sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.

Step-by-step explanation:

Hello!

Remember:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

Let's say that the significance level of this correlation test is α: 0.05

If the p-value is the probability of obtaining

you can express it as a percentage: 4.4%Is a very low probability. The decision rule using the p-value is:

p-value < α ⇒ Reject the null hypothesis

p-value ≥ α ⇒ Do not reject the null hypothesis.

The p-value is less than the significance level, the decision is to reject the null hypothesis.

In a linear correlation analysis the statement "there is no linear correlation between the two variables" is always in the null hypothesis, so if you reject it, you can conclude that there is a linear correlation between the variables.

I hope it helps!

In statistical analysis, a P-value of 0.044 indicates there is a 4.4% chance of obtaining a linear correlation as extreme as the observed correlation coefficient. A P-value under 0.05 provides enough evidence to reject the null hypothesis of no correlation, therefore suggesting a significant correlation. Hence, there is sufficient evidence of a linear correlation between car weight and highway fuel consumption.

The P-value of 0.044 in this context represents the likelihood of obtaining a linear correlation coefficient for the data points in your dataset that is as extreme as, or more extreme than, the one you calculated, assuming there is no linear relationship between the two variables (weight and highway fuel consumption in automobiles). This probability is 4.4% - rather low. A common threshold for significance in many fields is 0.05, or 5%. If your P-value is below this threshold, we reject the null hypothesis that there is no correlation and conclude there may be a correlation. Therefore, as the P-value is 0.044, which is below the 0.05 threshold, there is sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.

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Answer:

Step-by-step explanation:

18 a)Since line DE is parallel to line BC, it means that triangle ADE is similar to triangle ABC. Therefore,

AB/AD = AC/AE = BC/DE

AC = AE + CE = 18 + 9

AC = 27

AB = AD + DB

AB = AD + 5

Therefore,

27/18 = (AD + 5)/AD

Cross multiplying, it becomes

27 × AD = 18(AD + 5)

27AD = 18AD + 90

27AD - 18AD = 90

9AD = 90

AD = 90/9

AD = 10

b) AC = AE + EC = 13 + 3

AC = 16

To find AD,

16/13 = 24/AD

16 × AD = 13 × 24

16AD = 312

AD = 312/16

AD = 19.5

DB = 24 - 19.5

DB = 4.5

Multiply the cost per hour by number of hours x and for option 1 you need to add the monthly fee:

A) option 1: C= 4x +35

Option2: C = 6.50x

B)

4x+35 < 6.50x

Subtract 4x from both sides:

35 < 2.50x

Divide both sides by 2.50 :

X < 14

You would need to rent more than 14 hours for option 1 to be cheaper.

Answer:

Perimeter = 72

Step-by-step explanation:

Mathematics Puzzle

The area of a rectangle is the product of the base and the height

A=b.h

We know the base and the height are two consecutive prime numbers, not much of useful information so far.

We also know the area is composed only of the two smallest prime digits. Those digits are 2 and 3. If the number is reversed and it's not changed, then we only have two possible values for the area: 232 and 323.

We only need to find two consecutive prime numbers which product is one of the above. The number 232 has no prime factors: 232 = 8*29.

The number 323 is the product of 17 and 19, two consecutive prime numbers, thus the dimensions of the rectangle are 17 and 19.

The perimeter of that rectangle is 2*17+2*19= 72

Answer:

72 units

Step-by-step explanation:

72 units

Answer:

( 5 x 10 ) cm

Step-by-step explanation:

Given:

- The dimensions of the bar are:

                                  ( 2 x 5 x 10 ) cm

Find:

Which two faces with dimensions ( _x _ ) should be connected to get smallest possible resistance.

Solution:

- The electrical resistance R of any material with density ρ and corresponding dimensions is expressed as:

                                R = ρ*L / A

- Where, A: cross sectional Area

              L: The length in between the two faces.

- We need to minimize the electrical resistance of the bar. For that the Area must be maximized and Length should be minimized.

-                               A_max & L_min ---- > R_min

                               (5*10) & ( 2) ------> R_min

Hence, the electrical resistance is minimized by connecting the face with following dimensions ( 5 x 10 ) cm

Answer:

Percentage of scores that fall between 70 and 80 = 24.34%

Step-by-step explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N([tex]\mu,\sigma^{2}[/tex])

The z probability is given by;

           Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)    where, [tex]\mu[/tex] = 75  and  [tex]\sigma[/tex] = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{80-75}{16}[/tex] ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{70-75}{16}[/tex] ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

                                              = 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%

Answer:

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Step-by-step explanation:

Total plants = 11

Domestic plants = 7

Outside the US plants = 4

Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total no. of trials

           x = no. of successful trials

           p = probability of success

           q = probability of failure

Here we have n=4, p=4/11 and q=7/11

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰

          = 1 - 0.16399

P(X≥1) = 0.836

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Answer:

The probability that the next customer will arrive in 3 minutes or less is 0.45.

Step-by-step explanation:

Let N (t) be a Poisson process with arrival rate λ. If X is the time of the next arrival then,

[tex]P(X>t)=e^{-\lambda t}[/tex]

Given:

[tex]\lambda=\frac{12}{60}[/tex]

t = 3 minutes

Compute the probability that the next customer will arrive in 3 minutes or less as follows:

P (X ≤ 3) = 1 - P (X > 3)

              [tex]=1-e^{-\frac{12}{60}\times3}\\=1-e^{-0.6}\\=1-0.55\\=0.45[/tex]

Thus, the probability that the next customer will arrive in 3 minutes or less is 0.45.

The probability that the next customer will arrive in 3 minutes or less is; 0.4512

This is a Poisson distribution problem with the formula;

P(X > t) = e^(-λt)

Where;

λ is arrival rate

t is arrival time

We are given;

λ = 12/60

t = 3 minutes

P (X ≤ 3) = 1 - (P(X > 3))

Thus;

P (X ≤ 3) = 1 - e^((12/60) × 3)

P (X ≤ 3) = 1 - 0.5488

P (X ≤ 3) = 0.4512

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Answer:

x = 6 units.

Step-by-step explanation:

By Geometric mean property:

[tex]x = \sqrt{3 \times 12} = \sqrt{36} = 6 \\ \hspace{20 pt} \huge \orange{ \boxed{ \therefore \: x = 6}}[/tex]

Hence, x = 6 units.

Answer:

First part

[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]

And for this case we can use the z score formula given by:

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

And using this formula we got:

[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]

And we can use the normal standard table or excel and we got:

[tex]P(Z<-0.6) = 0.274[/tex]

Second part

For the other part of the question we want to find the following probability:

[tex] P(-1.715 <X< 7.12)[/tex]

And using the score we got:

[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]

And we can find this probability with this difference:

[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/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 data of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(3.4,3.1)[/tex]  

Where [tex]\mu=3.4[/tex] and [tex]\sigma=3.1[/tex]

First part

And for this case we want this probability:

[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]

And for this case we can use the z score formula given by:

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

And using this formula we got:

[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]

And we can use the normal standard table or excel and we got:

[tex]P(Z<-0.6) = 0.274[/tex]

Second part

For the other part of the question we want to find the following probability:

[tex] P(-1.715 <X< 7.12)[/tex]

And using the score we got:

[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]

And we can find this probability with this difference:

[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/tex]

Answer:

P ( X > 5) = 0.0374

Step-by-step explanation:

Given:

n = 12

p = 0.23

Using Binomial distribution formula,

X ~ Binomial ( n = 12, p = 0.23)

[tex]=\frac{n!}{(n-x)! x!}. p^{x} q^{n-x}[/tex]

Substitute for n = 12, p = 0.23, q = 1-0.23 for  x = 6,7,8,9,10,11 and 12

P (X > 5)  = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)

P ( X > 5 ) = 0.0285 + 0.007299 + 0.00136 + 0.000181 + 0.0000162 + 1E-6 + 1E-6

P ( X > 5) = 0.0374

Final answer:

The student's question pertains to calculating the probability of selecting two individuals from a sorority consisting of full members and pledges. The calculation involves using combinatorial formulas to determine the likelihood of different types of selections.

Explanation:

The student is asking about finding the probability of selecting two persons at random from a group consisting of full members and pledges in a sorority. With 38 members in total, 28 full members and 10 pledges, we need to calculate the probability of different pairs of members being selected. This is a combinatorial probability question.

In such problems, if there is no specific requirement about who needs to be picked (like how many full members or pledges), the total number of ways to select two members from the group without regard to order is calculated using the combination formula C(n, k) = n! / [k!(n-k)!], where 'n' is the total number of items, and 'k' is the number of items to choose.

For instance, the probability of selecting two full members can be calculated by finding the number of ways to choose two full members from the 28 available, divided by the number of ways to choose any two members from the entire group of 38.

Use simple unitary method:

∵ 1.5 cm on map = 25 miles in reality

∴ x cm on map = 190 miles in reality

[tex]\frac{1.5}{x} = \frac{25}{190}\\ x = 11.4 cm[/tex]

Thus, the two cities are 11.4 cm apart on the map

Final answer:

To determine the map distance between two cities that are 190 miles apart using the scale 1.5 cm = 25 miles, divide the actual distance by the scale ratio (16.67 miles/cm), resulting in 11.4 cm.

Explanation:

For the map with scale 1.5 cm = 25 miles, we first calculate how many miles one centimeter represents by dividing the miles by the centimeters in the scale:

25 miles / 1.5 cm = 16.67 miles/cm

Next, we find the map distance for 190 miles by dividing the actual distance by the distance one centimeter represents:

190 miles / 16.67 miles/cm = 11.4 cm

So, the two cities are 11.4 cm apart on the map.

Answer:

The confidence interval is 27.5 hg less than mu less than 29.1 hg

(A) Yes, because the confidence interval limits are not similar.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17

The confidence intervals for the two experiments have overlapping intervals, suggesting that the results are not very different.

The first experiment resulted in a 95% confidence interval of 3.070-3.164 g for the population mean weight of newborn girls. The second experiment had a 95% confidence interval of 3.035-3.127 g. Although the two confidence intervals are not identical, the mean for each experiment is within the confidence interval of the other experiment. This suggests that the results are not very different and there is an appreciable overlap between the two intervals. Therefore, the answer is C. No, because each confidence interval contains the mean of the other confidence interval.

We can estimate the mean, variance, and standard deviation by using class midpoints and the number of observations. First, find the midpoint of each age group. Then, to find the mean, multiply each midpoint by the number of shoplifters in that age group, sum those products, and divide by the total number of observations. To find the variance, subtract the mean from each midpoint, square the result, multiply by the number of shoplifters in that group, sum those products, and divide by the total number of observations minus one. The standard deviation is the square root of the variance.

To estimate the mean age, sample variance, and sample standard deviation for the shoplifters from a random sample of 895 incidents we need to follow a few steps. Here's how to do it:

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Answer: 0.06 ft long

Step-by-step explanation:

Porcelli's board = 2% feet long; to convert 2% into fraction, divide by 100= 2/100 = 0.02 ft

Smith's board, from the question is 3 times porcelli's board = 3 x 0.02= 0.06 ft

I hope this helps.

In damped harmonic motion, we calculate damping coefficient γ by comparing the periods of damped and undamped motion. For the given situation where the quasi-period is 90% greater than the undamped period, the damping coefficient is approximately 0.7416.

The subject of this question involves Damped Harmonic Motion, a concept in Physics, related to vibrations and waves. The equation given, u'' + γu' + u = 0, describes the motion where γ denotes the damping coefficient. Here, we have to calculate this damping coefficient when the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.

To solve this, we must use the relationship between damped and undamped periods. The quasi-period T' of a damped harmonic motion relates to the undamped period T as: T' = T/(sqrt(1 - (γ/2)^2)). Now, given that T' = 1.9T, we can but these two equations together:

1.9 = 1/(sqrt(1 - (γ/2)^2))

Solving this for γ, we get γ ≈ 0.7416. Hence, the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is approximately 0.7416.

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The value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the undamped motion is the one that satisfies γ=2*ω*0.9, where ω is the natural frequency of oscillation.

The given equation is for a damped harmonic oscillator, a physical system that oscillates under both a restoring force and a damping force proportional to the velocity of the system. The damping coefficient γ determines the behavior of the system and in this case, we need to find the value of γ such that the quasi period of the damped motion is 90% greater than the period of the undamped motion.

The period of the undamped motion, T₀, is calculated by the formula T₀=2π/sqrt(ω), where ω is the natural frequency of oscillation. The quasi period of the damped motion, Td, is increased by a factor of 1+η (in this case, 1.9 as the increase is 90%) and calculated by the formula Td=T₀(1+η) = T₀*1.9.

The damping ratio η is determined by the damping coefficient γ as η=γ/2ω. Therefore, by combining these expressions and rearranging the terms, we extract γ from these formulas as γ=2ω*η => γ=2*ω*(0.9). Thus, the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is the one which satisfies γ=2*ω*0.9.

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Answer: the third option is correct

Step-by-step explanation:

The first matrix is a 2 × 3 matrix while the second matrix is a 3 × 2 matrix. To get the product of both matrices, we would multiply each term in each row by the terms in the corresponding column and add.

1) row 1, column 1

1×2 + 3×3 + 1×4 = 2 + 9 + 4 = 15

2) row 1, column 2

1×-2 + 3×5 + 1×1 = - 2 + 15 + 1 = 14

3) row 2, column 1

-2×-2 + 1×3 + 0×4 = - 4 + 3 + 0 = - 1

4) row 2, column 2

- 2×-2 + 1×5 + 0×1 = 4 + 5 = 9

The solution becomes

15 14

- 1 9

Answer:

a) Discrete

b) Continuous

c) Continuous

d) Discrete

e) Continous

Step-by-step explanation:

Continuous:

Real numbers, can be integer, decimal, etc.

Discrete:

Only integer(countable values). So can be 0,1,2...

(a) number of traffic fatalities per year in the state of Florida

You cannot have half of a traffic fatality, for example. So this is discrete

(b) distance a golf ball travels after being hit with a driver

The ball can travel 10.25m, for example, which is a decimal number. So this is continuous.

(c) time required to drive from home to college on any given day

You can take 10.5 minutes, for example, which is a decimal value. So this is continuous.

(d) number of ships in Pearl Harbor on any given day

There is no half ship, for example. So this is discrete.

(e) your weight before breakfast each morning continuous discrete

You can weigh 80.4kg, for example, which is a decimal number. So this is continuous.

a) Discrete

b) Continuous

c) Continuous

d) Discrete

e) Continuous

Discrete data is the numerical type of data which includes whole, concrete numbers that has specific and fixed data values. therefore, they can be determined by counting.

example: Number of students in the class, ect.,

Continuous type of data includes complex numbers or the varying data values that are measured over a specific time interval.

example: Height of students in a school, etc.,

(a) Number of traffic fatalities per year in the state of Florida.

This is a Discrete data, as it is a countable data and will be always a whole be number. Number of traffic fatalities per year will be 75, 150, 200, etc.

(b) distance a golf ball travels after being hit with a driver.

This is a Discrete data, as the data can be measures and will be not always be a whole number. distance a golf ball travels after being hit can be 1.25 meters,  10.9 meters, 21.1 meters, etc.

(c) time required to drive from home to college on any given day.

This is a Discrete data, as the data can be measures and will be not always be a whole number. time required to drive from home to college can be 5minutes 30 seconds, 15 minutes 26 seconds.

(d) number of ships in Pearl Harbor on any given day.

This is a Discrete data, as it is a countable data and will be always a whole number. number of ships in Pearl Harbor on any given day will be always 1, 10, 8, etc.

(e) your weight before breakfast each morning.

This is a Discrete data, as the data can be measures and will be not always be a whole number. weight can be 44.56 kgs., 52.3 kgs, etc.

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Answer:

The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.

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{1-0.99}{2} = 0.005[/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]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find 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.

[tex]M = 2.575*\frac{62}{\sqrt{38}} = 25.90[/tex]

The lower end of the interval is the mean subtracted by M. So it is 500 - 25.90 = 474.10 milligrams.

The upper end of the interval is the mean added to M. So it is 500 + 25.90 = 525.90 milligrams

The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.

Answer:

Option A

Step-by-step explanation:

The 99% confidence interval is (472.69, 527.31). We are 99% confident that the true population mean of sodium loss for tennis players will be between 472.69 milligrams per pound and 527.31 milligrams per pound.

Answer:

For the velocity vector, we have

V(t) = dR

dt = (1, 2t, 3t

2

).

For the acceleration vector, we get

A =

dV

dt = (0, 2, 6t).

The velocity vector at t = 1 is

V(1) = (1, 2, 3).

The speed at t = 1 is

kV(1)k =

p

1

2 + 22 + 32 =

√

14.

Answer:

1/3

Step-by-step explanation:

60-90 is 30 numbers, right? So it is 30/90, or 1/3

Answer:

rm(list=ls(all=TRUE))

set.seed(12345)

N=c(10,100,500)

Rate=0.2

B=1000

MN=SE=rep()

for(i in 1:length(N))

{

n=N[i]

X=rexp(n,rate=Rate)

EST=1/mean(X)

ESTh=rep()

for(j in 1:B)

{

Xh=rexp(n,rate=EST)

ESTh[j]=1/mean(Xh)

}

MN[i]=mean(ESTh)

SE[i]=sd(ESTh)

}

cbind(N,Rate,MN,SE)

Answer:

7.51% probability that none of the 9 employees will say their company is loyal to them.

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they think that their company is loyal to them, or they do not think this. The probability of an employee thinking that their company is loyal to them is independent of other employees. So we use the binomial probability distribution to solve this question.

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.

25% of employees said their company is loyal to them.

This means that [tex]p = 0.25[/tex]

9 employees are selected randomly

This means that [tex]n = 9[/tex]

What is the probability that none of the 9 employees will say their company is loyal to them?

This is P(X = 0).

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

[tex]P(X = 0) = C_{9,0}.(0.25)^{0}.(0.75)^{9} = 0.0751[/tex]

7.51% probability that none of the 9 employees will say their company is loyal to them.

Using the normal distribution, it is found that:

a) P(X < 230) = 0.734.

b) P(180 < X < 245) = 0.6885.

c) P( X >190) = 0.734.

d) X = 151.76.

e) X = 188.56.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

In this problem:

Item a:

This probability is the p-value of Z when X = 230, hence:

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

[tex]Z = \frac{230 - 210}{32}[/tex]

[tex]Z = 0.625[/tex]

[tex]Z = 0.625[/tex] has a p-value of 0.734.

Hence:

P(X < 230) = 0.734.

Item b:

This probability is the p-value of Z when X = 245 subtracted by the p-value of Z when X = 180, hence:

X = 245

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

[tex]Z = \frac{245 - 210}{32}[/tex]

[tex]Z = 1.09[/tex]

[tex]Z = 1.09[/tex] has a p-value of 0.8621.

X = 180

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

[tex]Z = \frac{180 - 210}{32}[/tex]

[tex]Z = -0.94[/tex]

[tex]Z = -0.94[/tex] has a p-value of 0.1736.

0.8621 - 0.1736 = 0.6885.

Then:

P(180 < X < 245) = 0.6885.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 190, hence:

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

[tex]Z = \frac{190 - 210}{32}[/tex]

[tex]Z = -0.625[/tex]

[tex]Z = -0.625[/tex] has a p-value of 0.266.

1 - 0.266 = 0.734.

Hence:

P( X >190) = 0.734.

Item d:

This is X = c when Z has a p-value of 0.0344, hence X when Z = -1.82.

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

[tex]-1.82 = \frac{X - 210}{32}[/tex]

[tex]X - 210 = -1.82(32)[/tex]

[tex]X = 151.76[/tex]

Item e:

This is X when Z has a p-value of 1 - 0.7486 = 0.2514, hence X when Z = -0.67.

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

[tex]-0.67 = \frac{X - 210}{32}[/tex]

[tex]X - 210 = -0.67(32)[/tex]

[tex]X = 188.56[/tex]

More can be learned about the normal distribution at https://brainly.com/question/24663213

The question involves calculating probabilities and specific values (c) for a given normal distribution with mean 210 and standard deviation 32. The probabilities for certain ranges and tail ends are computed using the normal cumulative distribution function and its inverse.

The question pertains to finding probabilities and specific values associated with a normal distribution X which is denoted as N(210, 32), meaning it has a mean (μ) of 210 and a standard deviation (σ) of 32.

Answer:

5/10 & 1/2: anything that simplifies to 1/2 or .5

Final answer:

The expression 7/10-2/10 simplifies to 5/10, which can be further reduced to 1/2 by dividing both the numerator and denominator by their greatest common divisor, 5.

Explanation:

The expression 7/10-2/10 involves the subtraction of two fractions with the same denominator. When subtracting fractions with the same denominator, you simply subtract the numerators and keep the denominator the same. Thus, the calculation would be:

Numerator: 7 - 2 = 5

Denominator: 10 (remains the same)

Therefore, the expression 7/10-2/10 is equivalent to 5/10. This fraction can be further simplified by dividing the numerator and the denominator by their greatest common divisor, which in this case is 5. This gives us:

(5 ÷ 5) / (10 ÷ 5) = 1/2

So, 7/10-2/10 simplifies to 1/2, which is the equivalent fraction.

Answer:

[tex] P(X >2) [/tex]

And we can calculate this with the complement rule like this:

[tex] P(X>2) = 1-P(X<2)[/tex]

And using the cdf we got:

[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

[tex]P(X=x)=\lambda e^{-\lambda x}, x>0[/tex]

And 0 for other case. Let X the random variable of interest:

[tex]X \sim Exp(\lambda=\frac{1}{2.725})[/tex]

Solution to the problem

We want to calculate this probability:

[tex] P(X >2) [/tex]

And we can calculate this with the complement rule like this:

[tex] P(X>2) = 1-P(X<2)[/tex]

And using the cdf we got:

[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]

Answer:

2522520

Step-by-step explanation:

Number of ways logs can be classified = 14C5* 9C4*5C3

= 2002*126*10

= 2522520

Number of ways to select 5 good = 14C5, out of remaining 9, number of ways to select acceptable log = 9C4, out of remaining 5, number of ways to select prime log = 5C3 and remaking two unacceptable in 2C2 ways

Final answer:

To find how many ways 14 logs can be classified into categories to match the usual counts, use the multinomial coefficient formula based on the given category counts, resulting in a single calculations. to be 54.6.

Explanation:

The question asks in how many ways 14 logs can be classified into four categories (Prime, Good, Acceptable, Not Acceptable) if we know the usual counts for three of these categories are 3 Prime, 5 Good, 4 Acceptable, and the rest are Not Acceptable. This is a combinatorial problem that can be solved using combinations.

First, we distribute the logs into the Prime, Good, and Acceptable categories as given. This leaves us with 14 - (3 + 5 + 4) = 2 logs to be classified as Not Acceptable. Hence, all of the logs are accounted for with the specified counts.

The number of ways to classify the logs can thus be calculated as the number of ways to choose 3 out of 14 for Prime, then 5 out of the remaining 11 for Good, then 4 out of the remaining 6 for Acceptable. The remaining 2 are automatically classified as Not Acceptable. However, since we're simply fulfilling given counts, and the categories are distinct without overlap, we actually approach this as a partition of 14 objects into parts of fixed sizes, which is a straightforward calculation given by the multinomial coefficient:

The formula for the calculation is: 14! / (3! × 5! × 4! × 2!)

Which simplifies to the total number of ways these logs can be classified according to the specified counts

= 54.6