given the equation W=CE2/2, solve for C

Solution:

The given differential equation is,

[tex]y=C_{1}e^{4 x}+C_{2}e^{-x}[/tex]------(A)

Differentiating once,with respect to x,

[tex]y'=4C_{1}e^{4 x}-C_{2}e^{-x}[/tex]-------(1)

Differentiating again with respect to x,

[tex]y"=16C_{1}e^{4 x}+C_{2}e^{-x}[/tex]-------(2)

Equation (1) + Equation (2)

y' +y" [tex]=20 C_{1}e^{4 x}[/tex]

[tex]C_{1}=\frac{y'+y"}{20e^{4 x}}[/tex]

4 ×Equation (1) - Equation (2)

4 y'- y"[tex]=-5 C_{2}e^{-x}[/tex]

[tex]C_{2}=\frac{4y'-y"}{-5e^{-x}}[/tex]

Substituting the value of [tex]C_{1},C_{2}[/tex] in A,we get

[tex]y=\frac{y'+y"}{20}+\frac{4 y'-y"}{-5}\\\\ 20 y=y'+y"-16 y'+4 y"\\\\ 20 y=-15 y'+5y"\\\\ 4 y+3 y'-y"=0[/tex]

As, y(0)=1 , and y'(0)=2, gives

[tex]C_{1}+C_{2}=1\\\\ 4C_{1}-C_{2}=2[/tex]

gives , [tex]5C_{1}=3\\\\ C_{1}=\frac{3}{5}\\\\ 5 C_{2}=2\\\\ C_{2}=\frac{2}{5}[/tex]

So, member of the family that is a solution of the initial-value problem, [tex]y=C_{1}e^{4 x}+C_{2}e^{-x}[/tex] is

[tex]5 y=3 e^{4 x}+2 e^{-x}[/tex]

The inequality -4(x+3)<-2-2x simplifies to x > -5. Therefore, the number line representing this solution should shade the values greater than -5, thus, highlighting that x is any value greater than -5.

Let's solve the inequality -4(x+3)<-2-2x step by step:

As a result, the number line that represents this solution is one where the values greater than -5 are shaded. This means the value of x is anything greater than -5. Any number line visualization should start at -5 and include everything to the right of that point.

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Answer:

To write 0.080 as a fraction you have to write 0.080 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

0.080 = 0.080/1 = 0.8/10 = 8/100

So your final answer is:  0.080 as a fraction equals 8/100

Step-by-step explanation:

The decimal number 0.080 as a fraction is A = 2/25

Given data:

The decimal number is represented as 0.080

Step 1: Count the number of decimal places in the given decimal. So, there are three decimal places.

Step 2: Write the decimal as the numerator of the fraction. The decimal 0.080 becomes the numerator 80.

Step 3: Determine the denominator of the fraction. Since there are three decimal places, the denominator should be 10 raised to the power of the number of decimal places. So, the denominator is 10³ = 1000.

Putting it all together, 0.080 as a fraction is 80/1000.

Dividing both the numerator and denominator by 40:

80 ÷ 40 / 1000 ÷ 40 = 2/25

Hence, 0.080 as a simplified fraction is 2/25.

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Answer:

He paid $ 21.6 for the haircut.

Step-by-step explanation:

Given,

The amount charged by hair stylist = $ 18,

The percentage of tip = 20 %,

So, the amount of the tip = 20 % of 18

= 0.2 × 18

= $ 3.6,

Hence, the total amount paid by him for haircut = The amount charged by hair stylist + Tip

= $ 18 + $ 3.6

= $ 21.6

Number of cups of Strawberries are there for 3 pies =2

Now we apply unitary method here to find how many strawberries are need for 1 pie.

Number of cups of Strawberries required for 1 pie = 2/3

Answer: So Grace would require 2/3 cups of Strawberries for making one pie.

Final answer:

To represent the decimal numbers from 0 to 999 in binary code, we require 10 bits. This is determined by finding the binary equivalent of the largest number, 999, which necessitates at least 10 powers of 2 (from [tex]2^0 \ to \ 2^9[/tex]).

Explanation:

To determine how many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code, we need to find the binary equivalent of the largest number in that range, which is 999. In binary, this would require the highest power of 2 that is less than or equal to 999. The largest power of 2 less than 999 is 512 (29), and we continue to add powers of 2 to represent the number:

Adding these up, we see that we need at least 10 bits to represent the number 999 in binary, because adding powers of 2 from 20 to 29 will give us the required range. Therefore, we need 10 bits to represent any number from 0 to 999 in binary.

The standard deviation of the customer's book purchases is 0.72.

The standard deviation of a discrete probability distribution can be calculated using the following formula:

σ = √(p[tex](x - \mu)^2[/tex])

where:

σ is the standard deviation

p(x) is the probability of the event x

μ is the mean of the distribution

In this case, the probabilities of the customer purchasing 0, 1, or 2 books are 0.2, 0.3, and 0.5, respectively. So, the mean of the distribution is:

μ = (0 * 0.2) + (1 * 0.3) + (2 * 0.5) = 1.2

The standard deviation is then:

σ = √([tex](0 - 1.2)^2 (0.3)^2[/tex] + [tex](1 - 1.2)^2 (0.3)^2[/tex] + [tex](2 - 1.2)^2 (0.5)^2)[/tex] = 0.72

So, the standard deviation of the customer's book purchases is 0.72.

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The mean of the values in the stem-and-leaf plot is 42.

To find the mean of the values in the stem-and-leaf plot, we first need to interpret the plot correctly. Here's the stem-and-leaf plot description:

Stem | Leaves

 1  | 5, 8

 2  |

 3  |

 4  | 6

 5  | 0, 0, 0, 0, 7

 6  |

 7  |

 8  |

 9  |

Let's list the individual values from the plot:

- From stem 1: Leaves are 5 and 8

- From stem 4: Leaf is 6

- From stem 5: Leaves are 0, 0, 0, 0, 7

Now, we calculate the mean (average) of these values.

Step-by-step calculation:

1. List of values:

  - Values from stem 1: 15, 18

  - Values from stem 4: 46

  - Values from stem 5: 50, 50, 50, 57

2. Count the number of values:

  - There are 2 values from stem 1, 1 value from stem 4, and 5 values from stem 5.

  - Total count = 2 + 1 + 5 = 8

3. Calculate the sum of all values:

  - Sum = 15 + 18 + 46 + 50 + 50 + 50 + 50 + 57

  - Sum = 336

4. Calculate the mean:

  - Mean = Sum / Count

  - Mean = 336 / 8

  - Mean = 42

Therefore, the mean of the values in the stem-and-leaf plot is [tex]{42} \).[/tex]

Final answer:

To find the cost of small and large prints, we create two equations based on Andy's and Alex's orders and solve for the two variables. The small 4x6 inch prints cost $0.50 each, and the large 8x10 inch prints cost $1.25 each.

Explanation:

To solve how much small and large sized prints cost, we can set up a system of linear equations based on the information given:

Let s represent the cost of a small 4x6 inch print, and let l represent the cost of a large 8x10 inch print. We can then form the equations:

By dividing the first equation by 2, we get:

Subtracting the second equation from this result gives:

With the cost of a large print found, we can plug this back into the second equation to find the cost of a small print:

Therefore, the cost of a small 4x6 inch print is $0.50 and a large 8x10 inch print is $1.25.

If you have an item that costs 10 dollars and a sales tax of 7% you could do it this way

10 * 7/100 = 0.70 dollars.

The total cost is 10.70 dollars. or 10 dollars and 70 cents.

The best way to do the total however is to multiply the cost by 1.07. You will get the number the customer has to pay.

10 * 1.07 = 10.70

So for any price (p) it would be 1.07*p

Answer:

1.07p

Step-by-step explanation:

The tax  equals the price times the tax rate

We know the price p and the tax rate 7%

tax = p*7%

Changing this to decimal form

tax = .07p

The  final price of the item is the sales tax plus the price

final price = tax + price

                   .07p +p

                  1.07p

a. Probability defective hard drive from F2 ≈ 0.3226.

b. Probability defective hard drive from F4 ≈ 0.7742.

c. Probability non-defective hard drive ≈ 0.9785.

To solve this problem, we can use Bayes' theorem, which states:

[tex]\[ P(A|B) = \frac{P(B|A) \times P(A)}{P(B)} \][/tex]

Where:

- [tex]\( P(A|B) \)[/tex] is the probability of event A happening given that event B has occurred.

- [tex]\( P(B|A) \)[/tex] is the probability of event B happening given that event A has occurred.

- [tex]\( P(A) \)[/tex] and [tex]\( P(B) \)[/tex] are the probabilities of events A and B, respectively.

Let's solve each part of the problem:

a. If a defective HDC hard drive is picked at random, what is the probability that it was produced at F2?

Let:

- A be the event that the hard drive is defective.

- B be the event that the hard drive was produced at F2.

We need to find [tex]\( P(B|A) \)[/tex], the probability that the hard drive was produced at F2 given that it is defective.

[tex]$\begin{aligned} & P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ & P(A \cap B)=P(A \mid B) \times P(B)=0.02 \times 0.25=0.005 \\ & P(A)=P(A \cap F 1)+P(A \cap F 2)+P(A \cap F 3)+P(A \cap F 4) \\ & P(A)=0.015 \times 0.20+0.02 \times 0.25+0.01 \times 0.15+0.03 \times 0.40=0.0155 \\ & P(B \mid A)=\frac{0.005}{0.0155} \approx 0.3226\end{aligned}$[/tex]

b. If a defective HDC hard drive is picked at random, what is the probability that it was produced at F4?

We need to find [tex]\( P(F4|A) \)[/tex], the probability that the hard drive was produced at F4 given that it is defective.

[tex]\[ P(F4|A) = \frac{P(A \cap F4)}{P(A)} \][/tex]

[tex]\[ P(A \cap F4) = P(A|F4) \times P(F4) = 0.03 \times 0.40 = 0.012 \][/tex]

[tex]\[ P(F4|A) = \frac{0.012}{0.0155} \approx 0.7742 \][/tex]

c. If an HDC hard drive is picked at random, what is the probability that it is non-defective?

Let P(non-defective) = P(non-defective at F1) + P(non-defective at F2) + P(non-defective at F3) + P(non-defective at F4)

P(non-defective) = [tex](1 - 0.015) \times 0.20 + (1 - 0.02) \times 0.25 + (1 - 0.01) \times 0.15 + (1 - 0.03) \times 0.40[/tex]

P(non-defective) = [tex]0.985 \times 0.20 + 0.98 \times 0.25 + 0.99 \times 0.15 + 0.97 \times 0.40[/tex]

P(non-defective) = 0.197 + 0.245 + 0.1485 + 0.388 = 0.9785

So, the probability that an HDC hard drive picked at random is non-defective is approximately [tex]\( 0.9785 \)[/tex].

There are 86 lilies he did plant in 12th row of the garden.

Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence.

Here, The number of lilies plants in first, second third,..., and last row are respectively.

20,26,32,........

the number of rows of lilies plants is 12.

The sequence 20,26,32,........ is an A.P. with first term a =20, common difference d = 6 and n =12

formula for nth term.

∴aₙ = a+(n−1)d

aₙ=20+ (12-1).6

aₙ = 20 + 11 X 6

aₙ = 86

Thus, there are 86 lilies he did plant in 12th row of the garden.

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Answer:

Option B

Step-by-step explanation:

The given expression is ( 2x + 5)²

We have to form a statement to describe the expression given.

(2x + 5)² square of sum of 2 times x and 5

Option B is the correct answer.

Answer:

6

Step-by-step explanation:

To find the number of different routes the delivery driver can take to make the deliveries at 5 out of 6 locations, we can use the concept of permutations.

To find the number of different routes the delivery driver can take, we can use the concept of permutations. Since the delivery driver has to make deliveries at 5 out of the remaining 6 locations, we need to calculate the number of ways to arrange these locations.

We can use the formula for permutations to find the number of different routes:

nPr = n! / (n - r)!

Using this formula, we can calculate the number of different routes as:

6P5 = 6! / (6 - 5)! = 6! / 1! = 6 × 5 × 4 × 3 × 2 × 1 / 1 = 720

Therefore, there are 720 different routes the delivery driver can take to make the deliveries.