Find the general solution to the following differential equations:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x^2

The predicted total packing cost for 25,000 orders is $150,800

To predict the total packing cost for 25,000 orders,  to use the information provided and apply regression analysis. Let's assume we have a linear regression model with the following variables:

X: Number of orders

Y: Packing cost

Based on the given information, the following data:

X (Number of orders) = 25,000

Total weight of orders = 40,000 pounds

Number of fragile items = 4,000

Now, let's assume a regression equation in the form: Y = b0 + b1 × X + b2 ×Weight + b3 × Fragile

Where:

b0 is the regression intercept (rounded to the nearest whole dollar)

b1, b2, and b3 are coefficients (rounded to two decimal places or nearest cent)

Weight is the total weight of the orders (40,000 pounds)

Fragile is the number of fragile items (4,000)

Since the exact regression equation and coefficients, let's assume some hypothetical values:

b0 (intercept) = $50 (rounded)

b1 (coefficient for number of orders) = $2.75 (rounded to two decimal places or nearest cent)

b2 (coefficient for weight) = $0.05 (rounded to two decimal places or nearest cent)

b3 (coefficient for fragile items) = $20 (rounded to two decimal places or nearest cent)

calculate the predicted packing cost for 25,000 orders:

Y = b0 + b1 × X + b2 × Weight + b3 × Fragile

Y = 50 + 2.75 × 25,000 + 0.05 × 40,000 + 20 × 4,000

Y = 50 + 68,750 + 2,000 + 80,000

Y = 150,800

Keep in mind that the actual values of the regression intercept and coefficients might be different, but this is a hypothetical calculation based on the information provided.

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The number of tables that will be required to seat all students present at the cafeteria is 100.

By applying simple logic, the answer to this question can be obtained.

First, let us state all the information given in the question.

No. of students in the whole group = 800

Amount of students that each table can accommodate is 8 students.

So, the number of tables required can be defined as:

No. of Tables = (Total no. of students)/(No. of students for each table)

This means,

N = 800/8

N = 100 tables.

So, with the availability of a minimum of 100 tables in the cafeteria, all the students can be comfortably seated.

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After 7 hours, the mass of yeast will be approximately 9.718 grams. After 2 days (48 hours), the mass of yeast will be approximately 128.041 grams.

To calculate the mass of yeast after a certain time using exponential growth, we can use the formula:

[tex]M = M_0 * e^{(rt)}[/tex]

Where:

M is the final mass

M0 is the initial mass

e is the base of the natural logarithm (approximately 2.71828)

r is the growth rate (expressed as a decimal)

t is the time in hours

Let's calculate the mass of yeast after 7 hours:

M = 3.7 (initial mass)

r = 13% per hour

= 0.13

t = 7 hours

[tex]M = 3.7 * e^{(0.13 * 7)}[/tex]

Using a calculator, we can find that [tex]e^{(0.13 * 7)[/tex] is approximately 2.628.

M ≈ 3.7 * 2.628

≈ 9.718 grams

Now, let's calculate the mass of yeast after 2 days (48 hours):

M = 3.7 (initial mass)

r = 13% per hour

= 0.13

t = 48 hours

[tex]M = 3.7 * e^{(0.13 * 48)][/tex]

Using a calculator, we can find that [tex]e^{(0.13 * 48)}[/tex] is approximately 34.630.

M ≈ 3.7 * 34.630

≈ 128.041 grams

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a) After 7 hours, the mass will be approximately 7.8272.

b) After 2 days, the mass will be approximately 69.1614.

The growth of the yeast culture is exponential at a rate of 13% per hour.

To find the mass present after a certain time, we can use the formula for exponential growth:

Final mass = Initial mass × [tex](1 + growth ~rate)^{(number~ of~ hours)}[/tex]

a) After 7 hours:

Final mass = 3.7 ×[tex](1 + 0.13)^7[/tex]

To calculate this, we can plug in the values into a calculator or use the exponent rules:

Final mass = 3.7 × [tex](1.13)^{7}[/tex] ≈ 7.8272

Therefore, the mass present after 7 hours will be approximately 7.8272.

b) After 2 days:

Since there are 24 hours in a day, 2 days will be equivalent to 2 × 24 = 48 hours.

Final mass = 3.7 × [tex](1 + 0.13)^{48}[/tex]

Again, we can use a calculator or simplify using the exponent rules:

Final mass = 3.7 ×[tex](1.13)^{48}[/tex] ≈ 69.1614

Therefore, the mass present after 2 days will be approximately 69.1614.

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(a) To find the probability that all 10 trustworthy individuals pass the polygraph test,

we can use the binomial probability formula:

P(X = 10) = C(10, 10) * (0.15)^10 * (1 - 0.15)^(10 - 10)

Calculating the values:

C(10, 10) = 1 (since choosing all 10 out of 10 is only one possibility)

(0.15)^10 ≈ 0.0000000778

(1 - 0.15)^(10 - 10) = 1 (anything raised to the power of 0 is 1)

P(X = 10) ≈ 1 * 0.0000000778 * 1 ≈ 0.0000000778

The probability that all 10 trustworthy individuals pass the polygraph test is approximately 0.0000000778.

(b) To find the probability that more than 2 trustworthy individuals fail the test, we need to calculate the probability of exactly 0, 1, and 2 individuals failing and subtract it from 1 (to find the complementary probability).

P(more than 2 fail, even though all are trustworthy) = 1 - P(X = 0) - P(X = 1) - P(X = 2)

Using the binomial probability formula:

P(X = 0) = C(10, 0) * (0.15)^0 * (1 - 0.15)^(10 - 0)

P(X = 1) = C(10, 1) * (0.15)^1 * (1 - 0.15)^(10 - 1)

P(X = 2) = C(10, 2) * (0.15)^2 * (1 - 0.15)^(10 - 2)

Calculating the values:

C(10, 0) = 1

C(10, 1) = 10

C(10, 2) = 45

(0.15)^0 = 1

(0.15)^1 = 0.15

(0.15)^2 ≈ 0.0225

(1 - 0.15)^(10 - 0) = 0.85^10 ≈ 0.1967

(1 - 0.15)^(10 - 1) = 0.85^9 ≈ 0.2209

(1 - 0.15)^(10 - 2) = 0.85^8 ≈ 0.2476

P(more than 2 fail, even though all are trustworthy) = 1 - 1 * 0.1967 - 10 * 0.15 * 0.2209 - 45 * 0.0225 * 0.2476 ≈ 0.0004

The probability that more than 2 trustworthy individuals fail the polygraph test, even though all are trustworthy, is approximately 0.0004.

(c) The mean (expected value) of a binomial distribution is given by μ = np, where n is the number of trials (400 agents tested) and p is the probability of success (the probability of failing for a trustworthy agent, which is 0.15).

Mean = μ = np = 400 * 0.15 = 60

The standard deviation of a binomial distribution is given by σ = sqrt(np(1-p)).

Standard deviation = σ = sqrt(400 * 0.15 * (1 - 0.15)) ≈ 4

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The Options (a) and (c) apply to the question, i.e. the digit 2 increases in value from 2 ones to 2 hundred, and, the digit 3 shifts 2 places to the left, from the tens place to the thousands place.

32.4×10²=32.4×100=3240

Hence, digit 2 moves from one's place to a hundred's. (a) satisfied

And similarly, digit 3 moves from ten's place to thousand's place. Now, 1000=10³=10²×10.

Hence, it shifts 2 places to the left.

Therefore, (c) is satisfied.

As for (b), where the statement: Each place is multiplied by 1,000; the statement does not hold true since each digit is shifted 2 places, which indicates multiplied by 10²=100, not 1000.

Hence (a) and (c) applies to our question.

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