which equation is equivalent to log3(x+5)=2

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].

The Geometric mean of the pair of number 6 and 10 is [tex]7.74[/tex]

The geometric mean of two number m and n is given as,

                       [tex]G.M=\sqrt{m*n}[/tex]

Geometric mean of the pair of number 6 and 10 is,

                    [tex]G.M=\sqrt{6*10} =\sqrt{60} \\\\G.M=7.74[/tex]

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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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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.

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]

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.

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.

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.

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]

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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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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Answer:

  x = 5 or x = -1

Step-by-step explanation:

x^2 – 4x = 5

Solve it by completing the square method

In completing the square method, we take the coefficient of x that is -4, divide it by 2 and then square it

-4/2 = -2  

square it (-2)^2 = 4

Now add 4 on both sides

x^2 – 4x +4 = 5+4

x^2 – 4x +4 = 9

Now factor the left hand side

(x-2)(x-2)=9

(x-2)^2 = 9

Take square root on both sides

x-2 = +-3

x-2 = 3 , so x= 5

x-2 = -3, so x= -1

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.