Please help me with this

Answer:

2x

Step-by-step explanation:

Volume of a cylinder is:

V = πr²h

If V = 4πx³ and h = x, then:

4πx³ = πr²x

4x² = r²

r = 2x

For this case we have that by definition, the volume of a cylinder is given by:

[tex]V = \pi * r ^ 2 * h[/tex]

Where:

r: It's the radio

h: It's the height

We have as data that the volume of the cylinder is:

[tex]V = 4 \pi * x ^ 3[/tex]

They also tell us that the height is "x", then:

[tex]4 \pi * x ^ 3 = \pi * r ^ 2 * x[/tex]

We have similar terms on both sides of the equation:

[tex]4x ^ 2 = r ^ 2[/tex]

We apply root to both sides of the equation:

[tex]r = \pm \sqrt {4x ^ 2}\\r = \pm2x[/tex]

We choose the positive value. Now the radio is 2x

Answer:

[tex]2x[/tex]

Answer:

actual rate of interest is 2.08 %

Step-by-step explanation:

Given data

borrowed = $2100

time = 15 months

rate = 2.6%

to find out

interest and receive from the bank and actual rate of interest

solution

we know bank discounted the loan at 2.6%

so interest will be 2.6% of $2100

that is  = 2.6/100 × 2100

interest = $54.60

so receive from the bank  is  borrowed money - interest

receive = 2100 - 54.60

receive from the bank is $2045.40

and we can say that interest is directly proportional to time

so interest of 12 months / interest of 15 month = 12 month time/ 15 month time

here  interest of 12 months = 12/15 ×  interest of 15 month

interest of 12 months = 12/15 × 54.60

interest of 12 months is $43.68 (annually)

now we can calculate actual rate of interest

that is = interest / borrowed money × 100

actual rate of interest = 43.68/2100 × 100

actual rate of interest is 2.08 %

Final answer:

Dylan borrowed $2100 which was discounted at a 2.6% rate by the bank over 15 months, incurring $68.25 in interest, leaving him to receive $2031.75. The actual annual rate of interest calculated is approximately 2.667%.

Explanation:

The calculation of the interest on a discounted loan involves understanding the interest rate and the method by which the interest is calculated. In Dylan's case, the bank used a discount rate of 2.6% on a loan of $2100 over a period of 15 months.

To calculate discounted interest, you need to multiply the principal amount by the discount rate and then adjust for the loan term. In this case, the formula for interest (I) looks like this: I = Principal (P) × Discount Rate (r) × Time (t), where 't' is in years. Since the loan term is 15 months, we convert it to years by dividing by 12, resulting in 1.25 years.

So, the interest can be calculated as follows:
I = $2100 × 0.026 × (15/12) = $2100 × 0.026 × 1.25 = $68.25. Therefore, the interest on the loan is $68.25, rounded to the nearest penny.

The actual amount Dylan received from the bank can be found by subtracting the interest from the principal: $2100 - $68.25 = $2031.75.

To calculate the actual interest rate, we compare the amount of interest paid over the loan term to the amount received. The actual interest for 15 months is the interest Dylan would be paying, divided by the amount he received, all divided by the time in years: Actual Rate (R) = ($68.25 / $2031.75) / 1.25. After solving for R, we get an actual rate of approximately 2.667%, rounded to the nearest hundredth.

Answer:

The correct options are a,b,e and f.

Step-by-step explanation:

It is given that the nth terms of the series is defined as

[tex]T_n=3n+17[/tex]

Subtract 17 from both the sides.

[tex]T_n-17=3n[/tex]

Divide both sides by 3.

[tex]\frac{T_n-17}{3}=n[/tex]

The term Tₙ is a term of given series if n is a positive integer.

(a) The given term is 80.

[tex]n=\frac{80-17}{3}=21[/tex]

Since n is a positive integer, therefore 80 is a term of given series.

(b) The given term is 170.

[tex]n=\frac{170-17}{3}=51[/tex]

Since n is a positive integer, therefore 170 is a term of given series.

(c) The given term is 217.

[tex]n=\frac{217-17}{3}=66.67[/tex]

Since n is not a positive integer, therefore 217 is a term of given series.

(d) The given term is 312.

[tex]n=\frac{312-17}{3}=98.33[/tex]

Since n is not a positive integer, therefore 312 is a term of given series.

(e) The given term is 278.

[tex]n=\frac{278-17}{3}=87[/tex]

Since n is a positive integer, therefore 278 is a term of given series.

(f) The given term is 3566.

[tex]n=\frac{3566-17}{3}=1183[/tex]

Since n is a positive integer, therefore 3566 is a term of given series.

Thus, the correct options are a, b, e and f.

Answer:

hence proceeds is $2513.05 and discount is $443.47

Step-by-step explanation:

Lauren pays = $3,400

interest = 15%

time = 1 yr

[tex]P(1+\dfrac{r}{100} )^n = 3400\\P(1+\dfrac{15}{100} )^1 = 3400\\P=2956.52[/tex]

the amount to be discounted is

[tex]\dfrac{15}{100} \times 2956.52 = 443.47[/tex]

discount = $443.47

proceeds = $2956.52 -$443.47

proceeds = $2513.05

hence proceeds is $2513.05 and discount is $443.47

The discount on the loan is $510, calculated as 15% of $3,400. Lauren receives proceeds of $2,890 after the discount is subtracted from the principal amount.

To calculate the discount and proceeds given to Lauren by the bank, we use the interest rate and the principal amount. Since the bank subtracts the interest upfront (known as discounting), we need to find 15% of $3,400.

First, find the discount:
Discount = Principal x Interest Rate
Discount = $3,400 x 0.15 = $510

Next, to calculate the proceeds, subtract the discount calculated above from the principal:
Proceeds = Principal - Discount
Proceeds = $3,400 - $510 = $2,890

Therefore, the bank subtracts the discount of $510 from the original $3,400 and gives Lauren the remaining proceeds of $2,890.

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Answer:

19%

Step-by-step explanation:

First, let's notice that having 1 or more rotten oranges in the sample is the complement of having 0 rotten oranges in the sample. That means

proba (1 or more rotten oranges) + proba (0 rotten oranges) = 100%

We will focus in the case of 0 rotten oranges that's easier and then we go back to this last equation.

For 0 rotten oranges, we need that the customer picks 3 good oranges. As there are 4 rotten oranges in the crate of 60, we have [tex]60-4=56[/tex] good oranges.  

So, the customer has a 56/60 chance of getting a good orange. As he needs 3 good oranges at the same time, he has [tex]\frac{56}{60} *\frac{55}{59} *\frac{54}{60}= \frac{1386}{1711}[/tex] chance.

Therefore,  proba(0 rotten oranges) = [tex]\frac{1386}{1711}[/tex] = 81% (approximately)

Going back to the first paragraph, we have proba(1 or more rotten oranges) = 100% - proba(0 rotten oranges) = 100% - 81% = 19%

Answer:

  6 cm by 17 cm

Step-by-step explanation:

The area is the product of the dimensions; the perimeter is double the sum of the dimensions.

So, we want to find two numbers whose product is 102 and whose sum is 23.

  102 = 1·102 = 2·51 = 3·34 = 6·17

The last of these factor pairs has a sum of 23.

The dimensions are 6 cm by 17 cm.

Answer:

D.) Both (B) and (C)

Step-by-step explanation:

The slope upward depicts higher sales option (B) and (C) are causes for higher production

HOPE IT HELPS....

The supply curve slopes upward due to two primary reasons. Firstly, as a firm increases output, the cost of production typically increases, necessitating a higher price to cover the increasing costs. Also, at higher prices, it's more profitable for firms to supply more of a good. Both these factors cause an upward slope in the supply curve.

The supply curve in Economics slopes upward because it represents the relationship between the cost of production and the quantity supplied. The correct answer is option d. As a firm produces more goods (output), its cost of production often rises, typically due to diminishing returns in the short run. So, they require a higher price to cover their increasing costs, hence option (b). Further, when the price per unit of a good in a market is higher, it's more profitable for firms to supply more of that good, conforming with option (c). Both these factors together explain the upward slope of the supply curve.

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With 22 packed cheese sandwiches made, bread is the 'limiting' ingredient, and we have three slices of cheese in excess.

The number of sandwiches that can be made depends on the availability of the critical ingredients needed to complete a sandwich: 2 slices of bread and three slices of cheese. To calculate the number of sandwiches, we can determine how many sandwiches each ingredient batch could provide, then find the minimum of those two, as the 'limiting' factor will be the ingredient that finishes first.

For bread, we have 44 pieces/ 2 pieces per sandwich = 22 sandwiches. With cheese, we have 69 pieces/ 3 pieces per sandwich = 23 sandwiches. Therefore, we can fully make a maximum of 22 sandwiches because we would run out of bread first. This means the bread is the 'limiting' ingredient, while cheese is in excess. With 22 sandwiches made, we would use three slices of cheese per sandwich for 66 slices used, leaving three slices of cheese remaining. The quantity in excess is three slices of cheese.

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There are 0 slices of bread in excess, and there are 3 slices of cheese in excess.

Let's start by calculating the maximum number of sandwiches that can be made using 44 slices of bread:

Each sandwich requires 2 slices of bread.

So, the maximum number of sandwiches that can be made using 44 slices of bread is 44/2 = 22 sandwiches.

Now, let's calculate the maximum number of sandwiches that can be made using 69 slices of cheese:

Each sandwich requires 3 slices of cheese.

So, the maximum number of sandwiches that can be made using 69 slices of cheese is 69/3 = 23 sandwiches.

Now, we compare the results:

With 44 slices of bread, we can make 22 sandwiches.

With 69 slices of cheese, we can make 23 sandwiches.

Since we can only make 22 sandwiches due to the limitation of the bread, the bread is the limiting ingredient.

The quantity of the ingredient in excess can be found by subtracting the number of sandwiches that can be made using the limiting ingredient from the total number of slices of that ingredient.

For the bread:

Quantity of bread in excess = Total slices of bread - (Number of sandwiches × Slices of bread per sandwich)

Quantity of bread in excess = 44 - (22 *2)

= 44 - 44

= 0

For the cheese:

Quantity of cheese in excess = Total slices of cheese - (Number of sandwiches × Slices of cheese per sandwich)

Quantity of cheese in excess = 69 - (22 *s 3)

= 69 - 66

= 3

So, there are 0 slices of bread in excess, and there are 3 slices of cheese in excess.

Answer:

The mean absolute deviation is $26.38

Step-by-step explanation:

$22+$16+$35+$19+$34+$20+$26= $211

$211 ÷ 8 (backpacks) = $26.375= $26.38

Step-by-step explanation:

There's a lot of information here, so first things first, let's get organized.

Let's start by assigning variable names to each group.  We want the variables to be short but easy to understand.

For example, let's say the number of female seniors on the dean's list is FSD (F for female, S for senior, and D for dean's list).

FSD = 31

Sticking to this naming scheme:

FD = 62

MSD = 45

FS = 87

MS = 96

F = 275

MD = 88

M = 227

Now we can begin.

a) We want to know how many seniors there are.  So all we have to do is add up all the variables with an S in them.

FSD + MSD + FS + MS

31 + 45 + 87 + 96

259

b) We want to know how many women there are.  So add up all the variables with an F in them.

FSD + FD + FS + F

31 + 62 + 87 + 275

455

c) We want to know how many are on the dean's list.  So add up all the variables with a D in them.

FSD + FD + MSD + MD

31 + 62 + 45 + 88

226

d) Now we want to know how many are seniors AND on the dean's list.  So add up all the variables that have both an S and a D.

FSD + MSD

31 + 45

76

e) We want to know how many female seniors there are, so add up all the variables with both an F and an S.

FSD + FS

31 + 87

118

f) We want to know how many women were on the dean's list, so add up all the variables with both an F and a D.

FSD + FD

31 + 62

93

g) Finally, we want to know how many students there are total.  So add up all the variables.

FSD + FD + MSD + FS + MS + F + MD + M

62 + 45 + 87 + 96 + 275 + 88 + 227

880

Using arithmetic operations based on the provided data, we deduced the total number of seniors, women, students on the dean's list, and overall students at the college, among other specific categorizations.

To solve these problems, we will first interpret the given data, then perform simple arithmetic operations based on the information provided to answer each part of the question.

Answer:

dy/dt = 7y / (t − 1000)

Step-by-step explanation:

Change in mass of salt = mass of salt going in − mass of salt going out

dy/dt = 0 − (C kg/L × 7 L/min)

where C is the concentration of salt in the tank.

The concentration is mass divided by volume:

C = y / V

The volume in the tank as a function of time is:

V = 1000 + 6t − 7t

V = 1000 − t

Therefore:

C = y / (1000 − t)

Substituting:

dy/dt = -7y / (1000 − t)

dy/dt = 7y / (t − 1000)

If we wanted, we could separate the variables and integrate.  But the problem only asks that we find the differential equation, so here's the answer.

The differential equation for the change in the mass of salt over time in the tank is expressed as dy/dt = -(y/(1000-1*t))*7. This is based on the concentration of salt in the water and the rates at which water enters and leaves the tank.

To set up the differential equation for the student's question, we must take into account the rate at which the water (and hence the salt solution) is entering and leaving the tank. Let's denote the amount of salt in the tank after t minutes as y (kg). The rate of water entering the tank is 6 L/min of pure water, so no additional salt is added. The water leaving the tank, which has a concentration of salt, is 7 L/min.

The concentration of the salt at any time t is given by the mass of the salt y divided by the volume of the solution in the tank. Since the solution is leaving the tank at 7 L/min, the rate at which salt leaves the tank is the concentration times the outflow rate, which is (y/(1000-1×t))×7 kg/min. The negative sign represents the loss of salt from the tank.

The differential equation accounting for this change in mass of salt over time would be: dy/dt = -(y/(1000-1×t))×7. Note that this equation is valid until the tank is empty, at which point a different model would be needed as there would be no more solution left to leave the tank.

Answer:

68600 will there be at​ midnight ( approx )

Step-by-step explanation:

Let P shows the population of the bacteria,

Since, the number of bacteria in a culture grew at a rate proportional to its size,

[tex]\implies \frac{dP}{dt}\propto P[/tex]

[tex]\frac{dP}{dt}=kP[/tex]

Where, k is the constant of proportionality,

[tex]\frac{dP}{P}=kdt[/tex]

[tex]\int \frac{dP}{P}=\int kdt[/tex]

[tex]ln P=kt + C_1[/tex]

[tex]P=e^{kt+C_1}[/tex]

[tex]P=e^{kt}.e^{C_1}=C e^{kt}[/tex]

Now, let the population of bacteria is estimated from 10:00 AM,

So, at t = 0, P = 4,000 ( given )

[tex]4000 = Ce^{0}[/tex]

[tex]\implies C=4000[/tex]

Now, at noon there are 4,600 bacterias,

That is, at t = 2, P = 4600

[tex]4600=Ce^{2k}[/tex]

[tex]4600 = 4000 e^{2k}[/tex]

[tex]\implies e^{2k}=\frac{4600}{4000}=1.15[/tex]

[tex]2k=ln(1.5)\implies k=\frac{ln(1.5)}{2}=0.202732554054\approx 0.203[/tex]

Hence, the equation that represents the population of bacteria after t hours,

[tex]P=4000 e^{0.203t} [/tex]

Therefore, the population of the bacteria at midnight ( after 14 hours ),

[tex]P=4000 e^{0.203\times 14}=4000 e^{2.842}= 68600.1252903\approx 68600[/tex]

To find the probability that more than 1 student will have their automobile stolen during the current semester, we can use a Poisson distribution. The average number of automobile thefts per semester is 7. By calculating 1 minus the probability of 0 or 1 thefts, we find that the probability of more than 1 theft is approximately 0.9991.

To find the probability that more than 1 student will have their automobile stolen during the current semester, we can use a Poisson distribution. The Poisson distribution models the number of events that occur in a fixed interval of time or space. In this case, the average number of automobile thefts per semester is given as 7. We want to find the probability that more than 1 student will have their automobile stolen, so we need to calculate 1 minus the probability that 0 or 1 student will have their automobile stolen.

The probability mass function of a Poisson distribution is given by P(X=k) = e^{-λ} * (λ^k) / k! where λ is the average number of events. In this case, λ = 7.

So, the probability that more than 1 student will have their automobile stolen is P(X>1) = 1 - (P(X=0) + P(X=1)).

To calculate P(X=0) and P(X=1), we can substitute k=0 and k=1 into the probability mass function and sum them up.

P(X=0) = e^{-7} * (7^0) / 0! = e^{-7}

P(X=1) = e^{-7} * (7^1) / 1! = 7e^{-7}

Therefore, P(X>1) = 1 - (e^{-7} + 7e^{-7}).

Rounding this answer to four decimal places, we get P(X>1) ≈ 0.9991.

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[tex]y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}[/tex]

Divide both sides by [tex]\dfrac13y^{-2}(x)[/tex]:

[tex]3y^2y'-y^3=xe^x\ln x[/tex]

Substitute [tex]v(x)=y(x)^3[/tex], so that [tex]v'(x)=3y(x)^2y'(x)[/tex].

[tex]v'-v=xe^x\ln x[/tex]

Multiply both sides by [tex]e^{-x}[/tex]:

[tex]e^{-x}v'-e^{-x}v=x\ln x[/tex]

The left side can be condensed into the derivative of a product.

[tex](e^{-x}v)'=x\ln x[/tex]

Integrate both sides to get

[tex]e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C[/tex]

Solve for [tex]v(x)[/tex]:

[tex]v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]

Solve for [tex]y(x)[/tex]:

[tex]y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]

[tex]\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}[/tex]

Answer: [tex] 495[/tex]

Step-by-step explanation:

We know that if the population is normally distributed with mean [tex]\mu[/tex]  and standard deviation [tex]\sigma[/tex], then the sampling distribution of the sample mean , [tex]\overline{x}[/tex] is also normally distribution with :-

Mean : [tex]\mu_{\overlien{x}}=\mu[/tex]

Given : The scores on the Critical Reading part of the SAT exam in a recent year were roughly Normal with mean [tex]\mu= 495[/tex]

Then , the mean of the average scores you get will be close to [tex] 495[/tex]

Answer: z= 1.51

Step-by-step explanation:

Test statistic for proportion is given by :-

[tex]z=\dfrac{p-P}{\sqrt{\dfrac{PQ}{n}}}[/tex]

Where n is sample size ,p is the sample proportion , P Is the population proportion and Q =1 - P.

Given : P=57% = 0.57

Q= 1- P = 1-0.57=0.43

n = 25

[tex]p=\dfrac{18}{25}=0.72[/tex]

Test statistic for proportion will be :-

[tex]z=\dfrac{0.72-0.57}{\sqrt{\dfrac{0.57\times0.43}{25}}}\approx1.51[/tex]

We should report the value of test statistic z= 1.51

The mean/expected value is

[tex]E[X]=\displaystyle\sum_x x\,P(X=x)=0P(X=0)+1P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)[/tex]

[tex]\implies E[X]=\boxed{1.987}[/tex]

d. The standard deviation is the square root of the variance, which itself is

[tex]V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]

We have

[tex]E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=0^2P(X=0)+1^2P(X=1)+2^2P(X=2)+3^2P(X=3)+4^2P(X=4)[/tex]

[tex]\implies E[X^2]=5.577[/tex]

Then the variance is

[tex]V[X]=5.577-1.987^2\approx1.629[/tex]

and so the standard deviation is

[tex]\sqrt{V[X]}\approx\boxed{1.276}[/tex]

e. We know this immediately from the table:

[tex]P(X=3)=\boxed{0.289}[/tex]

f. A parent can be involved in either 3 or 4 activities, but not simultaneously 3 and 4 activities (i.e. these events are disjoint), so

[tex]P(X=3\text{ or }X=4)=P(X=3)+P(X=4)=\boxed{0.39}[/tex]

Applying statistical concepts, we have that:

c) The mean is of 1.987 activities.

d) The standard deviation is of 1.67 activities.

e) The probability is 0.289.

f) The probability is 0.39.

The probability distribution is given by:

[tex]P(X = 0) = 0.216[/tex]

[tex]P(X = 1) = 0.072[/tex]

[tex]P(X = 2) = 0.322[/tex]

[tex]P(X = 3) = 0.289[/tex]

[tex]P(X = 4) = 0.101[/tex]

Item c:

The mean is given by the sum of the multiplications of each outcome by it's probability, thus:

[tex]E(X) = 0.216(0) + 0.072(1) + 0.322(2) + 0.289(3) + 0.101(4) = 1.987[/tex]

The mean is of 1.987 activities.

Item d:

The standard deviation is given by the square root of the sum of the squares of each outcome subtracted by the mean, multiplied by it's probability, thus:

[tex]\sqrt{V(X)} = \sqrt{0.216(0-1.987)^2 + 0.072(1-1.987)^2) + 0.322(2-1.987)^2) + ...}[/tex]

[tex]\sqrt{V(X)} = 1.67[/tex]

The standard deviation is of 1.67 activities.

Item e:

This probability is P(X = 3), thus 0.289.

Item f:

This probability is:

[tex]p = P(X = 3) + P(X = 4) = 0.289 + 0.101 = 0.39[/tex]

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Answer:

The equation to find the number of fish after t years where y is the number of fish is:

y(t) = 160×3^(t) ( t <= 2.93 assuming that the maximum number of fish is 4000 )

Therefore when y(t) = 2000 2000 = 160×3^(t)

3^(t) = 25/2 log 3 (3^(t)) = log 3 (25/2) t = 2.29 years.

Using the logistic equation, we have that:

a)

The equation is:

[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

b)

It will take 2.68 years for the population to increase to 2000.

The logistic equation is:

[tex]P(t) = \frac{K}{1 + Ae^{-kt}}[/tex]

With:

[tex]A = \frac{K - P(0)}{P(0)}[/tex]

The parameters are:

In this problem:

Then:

[tex]A = \frac{4000 - 160}{160} = 24[/tex]

Thus:

[tex]P(t) = \frac{4000}{1 + 24e^{-kt}}[/tex]

Item a:

Tripled during the first year, thus [tex]P(1) = 3P(0) = 3(160) = 480[/tex].

This is used to find k.

[tex]480 = \frac{4000}{1 + 24e^{-k}}[/tex]

[tex]480 + 11520e^{-k} = 4000[/tex]

[tex]e^{-k} = \frac{3520}{11520}[/tex]

[tex]\ln{e^{-k}} = \ln{\frac{3520}{11520}}[/tex]

[tex]k = -\ln{\frac{3520}{11520}}[/tex]

[tex]k = 1.1856[/tex]

Thus, the equation is:

[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

Item b:

This is t for which P(t) = 2000, thus:

[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

[tex]2000 = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

[tex]\frac{1}{1 + 24e^{-1.1856t}} = 0.5[/tex]

[tex]0.5 + 12e^{-1.1856t} = 1[/tex]

[tex]e^{-1.1856t} = \frac{1}{24}[/tex]

[tex]\ln{e^{-1.1856t}} = \ln{\frac{1}{24}}[/tex]

[tex]t = -\frac{\ln{\frac{1}{24}}}{1.1856t}[/tex]

[tex]t = 2.68[/tex]

It will take 2.68 years for the population to increase to 2000.

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The five-number summary for the given data set of strontium-90 levels in baby teeth is 124 (minimum), 131.5 (Q1), 139 (median), 149.5 (Q3), and 160 (maximum).

The five-number summary consists of the minimum, first quartile (Q1), median, third quartile (Q3), and maximum. Here's the calculation using the given data, already sorted in ascending order:

Determine the minimum and maximum values.

- The minimum value is the first element of the sorted data set: 124.

- The maximum value is the last element of the sorted data set: 160.

Calculate the median (Q2).

- Since there are 20 numbers, the median is the average of the two middle values (10th and 11th):

[tex]\[ \frac{138 + 140}{2} = 139. \][/tex]

Calculate the first quartile (Q1).  

- The first quartile is the median of the lower half of the data (the first 10 numbers):

[tex]\[ \frac{131 + 132}{2} = 131.5. \][/tex]

Calculate the third quartile (Q3).

- The third quartile is the median of the upper half of the data (the last 10 numbers):

[tex]\[ \frac{149 + 150}{2} = 149.5. \][/tex]

Thus, the five-number summary for this data set is 124, 131.5, 139, 149.5, 160.

The complete question is : Listed below are amounts of strontium-90 (in millibecquerels, or mBq) in a simple random sample of baby teeth obtained from residents in a region born after 1979. Use the given data to construct a boxplot and identify the 5-number summary.

124 125 128 131 132 134 134 135 137 138

140 140 142 144 146 149 150 150 154 160

The 5 – number summary is _, _,_,_ and _ all in mBq. (use ascending order. Type integers or decimals do not round

Answer:

[tex]14878.04878miles/hours^2[/tex]

Step-by-step explanation:

Let's find a solution by understanding the following:

The acceleration rate is defined as the change of velocity within a time interval, which can be written as:

[tex]A=(Vf-Vi)/T[/tex] where:

A=acceleration rate

Vf=final velocity

Vi=initial velocity

T=time required for passing from Vi to Vf.

Using the problem's data we have:

Vf=65miles/hour

Vi=6miles/hour

T=14.8seconds

Using the acceleration rate equation we have:

[tex]A=(65miles/hour - 6miles/hour)/14.8seconds[/tex], but look that velocities use 'hours' unit while 'T' uses 'seconds'.

So we need to transform 14.8seconds into Xhours, as follows:

[tex]X=(14.8seconds)*(1hours/60minutes)*(1minute/60seconds)[/tex]

[tex]X=0.0041hours[/tex]

Using X=0.0041hours in the previous equation instead of 14.8seconds we  have:

[tex]A=(65miles/hour - 6miles/hour)/0.0041hours[/tex]

[tex]A=(61miles/hour)/0.0041hours[/tex]

[tex]A=(61miles)/(hour*0.0041hours)[/tex]

[tex]A=61miles/0.0041hours^2[/tex]

[tex]A=14878.04878miles/hours^2[/tex]

In conclusion, the acceleration rate is [tex]14878.04878miles/hours^2[/tex]

Answer:

D. None of the Above

Step-by-step explanation:

Answer:

15%

Step-by-step explanation:

Answer:

$ 1.96

Step-by-step explanation:

Number of spots with outcome of $1 = 9

Number of spots with outcome of $2 = 18

Number of spots with outcome of $10 = 1

Total number of spots = 28

Probability that ball will land on $1 = [tex]\frac{9}{28}[/tex]

Probability that ball will land on $2 = [tex]\frac{18}{28}[/tex]

Probability that ball will land on $10 = [tex]\frac{1}{28}[/tex]

The amount that player should expect to win on average in equal to expected value of the game. Expected value is calculated as the summation of product of probabilities with their respective outcomes.

i.e. for this case:

Expected Value will be:

[tex](1 \times \frac{9}{28})+(2 \times \frac{18}{28})+(10 \times \frac{1}{28})\\\\ =1.96[/tex]

This means, on average the player should expect to win $ 1.96

Probability is the ratio of the favorable event to the total number of events. The average amount is $1.96.

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen.

A new game is being introduced at the Hard Rock Cafe.

A ball is spun around a wheel until it comes to rest in one of many spots. Whatever is listed in that spot will be the player's winnings.

If the wheel following spots

Number of spots labeled $1 = 9

Number of spots labeled $2 = 18

Number of spots labeled $10 = 1

The total number of spots will be

Total spots = 28

The probability of the ball landing on $1 will be

[tex]\rm P(\$1) = \dfrac{9}{28}[/tex]

The probability of the ball landing on $2 will be

[tex]\rm P(\$2) = \dfrac{18}{28}[/tex]

The probability of the ball landing on $10 will be

[tex]\rm P(\$10) = \dfrac{1}{28}[/tex]

The amount that player should expect to win on average is equal to the expected value of the game will be

[tex]\rm Average\ amount = 1*\dfrac{9}{28} + 2*\dfrac{18}{28} + 10*\dfrac{1}{28}\\\\\\Average \ amount = 1.96[/tex]

More about the probability link is given below.

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Answer:

  x = {kπ, arccos(1/5) +2kπ, 2kπ -arccos(1/5)}

Step-by-step explanation:

The double-angle trig identity for sine is useful:

  5(2sin(θ)cos(θ)) -2sin(θ) = 0

  2sin(θ)(5cos(θ) -1) = 0

This has solutions that make the factors zero:

  θ = arcsin(0) = kπ

and ...

  cos(θ) = 1/5

  θ = arccos(1/5) +2kπ . . . . or . . . . 2kπ -arccos(1/5)

_____

Some numerical values are shown on the graph attached. values for multiples of pi are ...

  {..., -12.566, -9.425, -6.283, -3.142, 0, 3.142, 6.283, 9.425, 12.566, ...}

We can express the recurrence,

[tex]\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}[/tex]

in matrix form as

[tex]\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}[/tex]

By substitution,

[tex]\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}[/tex]

and continuing in this way we would find that

[tex]\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]

Diagonalizing the coefficient matrix gives us

[tex]\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}[/tex]

which makes taking the [tex](n-2)[/tex]-th power trivial:

[tex]\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}[/tex]

[tex]\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}[/tex]

So we have

[tex]\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]

and in particular,

[tex]a_n=\dfrac{29\left(2(-5)^{n-1}+5\cdot2^{n-1}\right)-47\left(-(-5)^{n-1}+2^{n-1}\right)}7[/tex]

[tex]a_n=\dfrac{105(-5)^{n-1}+98\cdot2^{n-1}}7[/tex]

[tex]a_n=15(-5)^{n-1}+14\cdot2^{n-1}[/tex]

[tex]\boxed{a_n=-3(-5)^n+7\cdot2^n}[/tex]

Answer:

0.1971 ( approx )

Step-by-step explanation:

Let X represents the event of weighing more than 20 pounds,

Since, the binomial distribution formula is,

[tex]P(x)=^nC_r p^r q^{n-r}[/tex]

Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Given,

The probability of weighing more than 20 pounds, p = 25% = 0.25,

⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

[tex]P(X<3) = P(X=0)+P(X=1)+P(X=2)[/tex]

[tex]=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}[/tex]

[tex]=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}[/tex]

[tex]=0.1971110499[/tex]

[tex]\approx 0.1971[/tex]

Answer:

Step-by-step explanation:

What is a regular tessellation?

A regular tessellation is a pattern made by repeating a regular polygon. In simpler words regular tessellations are made up entirely of congruent regular polygons all meeting vertex to vertex.

How many regular tessellation are possible?

There are only 3 regular tessellation.

1. Triangle

2. Square

3. Hexagon

Why aren't there infinitely many regular tessellations?

Not more than 3 regular tessellations are possible because the sums of the interior angles are either greater than or less than 360 degrees....

Final answer:

A regular tessellation is a repeating pattern of a regular polygon that fills a plane without gaps or overlaps, with only three possible: equilateral triangles, squares, and regular hexagons. There aren't infinitely many because a tessellation requires the angles at a vertex to sum to 360 degrees, a condition only satisfied by these three shapes.

Explanation:

A regular tessellation is a pattern made by repeating a regular polygon to fill a plane without any gaps or overlaps. There are exactly three regular tessellations possible, which are constituted by:

There aren't infinitely many regular tessellations because for a regular polygon to tessellate, the sum of the angles at a vertex where polygons meet must be exactly 360 degrees.

This criterion is satisfied only by triangles (each angle is 60 degrees), squares (each angle is 90 degrees), and hexagons (each angle is 120 degrees).

Polygons with more sides have larger interior angles, such that the sum exceeds 360 degrees, preventing them from tessellating regularly.

Using the hypergeometric distribution, the required probability is 0.0129.

Given that, the town had 7 employees over 50 years of age and 15 under 50.

Let, n₁=7 and n₂=15

Total employees in the town = 7+15  = 22

We have to find the probability that exactly 1 dismissed employee was over 50.

That means the remaining 8 dismissed employees are under 50.

Let the number of dismissed employees under 50 is denoted by x and above 50 denoted by y.

Then, using the hypergeometric distribution, the required probability can be calculated as:

[tex]P(x=1, y=8) = \dfrac{\binom{7}{1}+\binom{15}{8}}{\binom{22}{9}}[/tex]

[tex]P(x=1, y=8) = \dfrac{7+6435}{ 497420}[/tex]

[tex]P(x=1, y=8) = \dfrac{6442}{ 497420}[/tex]

[tex]P(x=1, y=8) = 0.0129[/tex]

Hence, the probability that exactly 1 employee was over 50 is 0.0129.

Learn more about hypergeometric distribution here:

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Answer:

Simple interest=4500

maturity value= 10500

Step-by-step explanation:

Given: Principal P = 6000

Rate % R = 5%

time T = 15 years

we know that simple interest [tex]SI=\frac{PRT}{100}[/tex]

⇒[tex]SI=\frac{6000\times5\times15}{100}[/tex]

on calculating we get SI = 4500

and maturity value = Principal amount + Simple interest

maturity value = 6000+4500 = 10500

hence the final answers are

Simple interest= 4500

and maturity value= 10500

hope this helps!!

The number of different possible 7-number selections from a pool of 48, where order does not matter, can be calculated using the mathematical concept of combinations. The formula to calculate the total number of combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of options and k is the number of options selected.

In the LOTTO game you described, you must select 7 numbers from a pool of 48, and the order of the numbers does not matter. This is a problem of combinations in mathematics. The formula for combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of options, k is the number of options selected, and '!' denotes a factorial. In this case, n=48 (the numbers 1 through 48) and k=7 (the seven numbers you select).

By plugging these values into the formula, we can calculate the total number of different selections possible: C(48, 7) = 48! / [7!(48-7)!]. This calculation would give us the total number of combinations of 7 numbers that can be selected from a pool of 48, which represents all the different possible LOTTO selections. It should be remembered that factorials such as 48! or 7! represent a product of an integer and all the integers below it, down to 1.

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There are 85,900,584 different selections possible in the LOTTO game.

To calculate the number of different selections possible in the LOTTO game, we need to focus on the concept of permutations.

With 48 numbers to choose from and 7 numbers to be selected, the number of permutations can be calculated using the formula P(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected.

In this case, n = 48 and r = 7.

Using the formula, we have P(48, 7) = 48! / (48-7)! = (48 * 47 * 46 * 45 * 44 * 43 * 42) / (7 * 6 * 5 * 4 * 3 * 2 * 1)

= 85,900,584 different selections possible.

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