A book has a front and a back and 400 pages. The front and back cover are each 0.9mm thick measured to 1 decimal place. Each page is 0.13 mm thick measured to 2 decimal places. Calculate the minimum thickness of the book.

Lucy will be paid $1260 in interest in the first 6 years.

To calculate the interest Lucy will be paid in the first 6 years, we will use the simple interest formula:

Interest = Principal x Rate x Time

Given:

Plugging in the values, we get:

Interest = $7000 x 0.03 x 6 = $1260

Therefore, Lucy will be paid $1260 in interest over the first 6 years.

https://brainly.com/question/22621039

#SPJ2

Answer:

Your answer is B=4

Step-by-step explanation:

hope it helps once again

The explicit formula for the given sequence is a_n = 3 + 4(n-1).

The given sequence is (3, 7, 11, 15, 19, 23, 27, ...).

To find the explicit formula for this sequence, we can observe that each term is obtained by adding 4 to the previous term. So, the formula can be written as:

an = 3 + 4(n-1)

where n represents the position of the term in the sequence.

x = square feet

total cost = 500 + 600x

The expected number of bad apples is [tex]\( \frac{2}{10} \times 3 = 0.6 \) bad apples.[/tex]

The expected number of bad apples when grabbing 3 apples from a pool of 10, of which 2 are bad, can be calculated using probability.

To calculate this, follow these steps:

1. Calculate the probability of picking a bad apple:

  The probability of picking a bad apple is the ratio of the number of bad apples to the total number of apples. In this case, it's [tex]\( \frac{2}{10} = 0.2 \) or 20%.[/tex]

2. Multiply the probability by the number of apples picked:  

  Multiply the probability of picking a bad apple by the number of apples you're picking.  

 [tex]\( 0.2 \times 3 = 0.6 \)[/tex]

When you pick 3 apples randomly from a pool of 10, you're essentially conducting a sampling experiment. The probability of picking a bad apple on any given pick is[tex]\( \frac{2}{10} \),[/tex]  which simplifies to 0.2 or 20%.

Since you're picking 3 apples, you multiply the probability of picking a bad apple by 3 to find the expected number of bad apples. This is because in expectation, you would expect that proportion of bad apples in your sample. So, [tex]\( 0.2 \times 3 = 0.6 \)[/tex], which means you would expect to pick approximately 0.6 bad apples when picking 3 from the pool of 10, assuming the picking is random.

So, the final answer is that you would expect about 0.6 bad apples when picking 3 apples from a pool of 10, of which 2 are bad.

complete question

There are 10 apples in the fridge and 2 of them are bad. if you grab 3 apples, what is the expected number of bad apples

To find the number of quarters and nickels, we can set up a system of equations using the total number of coins and the total value of the coins. Solving this system of equations, we find that Kevin and Randy have 27 quarters and 14 nickels.

To solve this problem, we can set up a system of equations. Let's use the variables q (number of quarters) and n (number of nickels). We know that there are a total of 41 coins, so we can write the equation q + n = 41. We also know that the total value of the coins is $7.45, so we can write the equation 0.25q + 0.05n = 7.45.

Now we can solve this system of equations using substitution:

Therefore, Kevin and Randy have 27 quarters and 14 nickels in their jar.

see picture for the graph

equation: y = 1.50x

Answer:

Step-by-step explanation:

took the test (k12) and its halfway to 50 if you see in the picture

The rate of change of the area of the circle when the radius is 39 centimeters is 1976π cm²/min. This uses the concept of related rates in calculus and the area formula A = πr².

The problem deals with the concept of related rates in calculus. In this problem, we're looking at how the rate of change of the radius of a circle impacts the rate of change of the area of the circle. The formula for the area of a circle is A = πr².

Differentiating both sides with respect to time(t) gives dA/dt = 2πr(dr/dt). In this case, dr/dt (the rate of change of the radius) is given as 8 cm/min. To find dA/dt (the rate of change of the area) when r = 39 cm, we substitute these values into the differentiated equation: dA/dt = 2π(39cm)(8 cm/min) = 1976π cm²/min

So, the rate of change of the area when r = 39 centimeters is 1976π cm²/min. As the radius increases, the area of the circle increases at a rate directly proportional to the radius.

https://brainly.com/question/33509029

#SPJ11

Final answer:

The radius of a circle with a circumference of 201 centimeters is approximately 32 centimeters, found by dividing the circumference by 2 times pi (π), using 3.14 as the value of pi. The correct answer is A) 32 centimeters.

Explanation:

The question asks us to find the radius of a circle when its circumference is given to be 201 centimeters. We use the circumference formula of a circle, which is C = 2πr, where C stands for circumference, π (pi) is a constant approximately equal to 3.14, and r represents the radius of the circle.

To find the radius, we rearrange this formula to solve for r:

Divide both sides of the equation by 2π to isolate r.

Plug in the given circumference value, C = 201 cm, and the approximate value of pi, π = 3.14, into the rearranged formula.

So the calculation would be r = C / (2π) = 201 / (2 * 3.14) = 201 / 6.28. When we compute this, we get r ≈ 32 centimeters to the nearest whole number.

Therefore, the correct answer is A) 32 centimeters.

a) [tex]\[ \theta = 67.9^\circ, 112.1^\circ \][/tex], b) [tex]\[ \theta = 133.7^\circ, 226.3^\circ \][/tex], c) [tex]\[ \theta = 122.7^\circ, 302.7^\circ \][/tex], d) [tex]\[ \theta = 35.8^\circ, 215.8^\circ \][/tex], e) [tex]\[ \theta = 45.5^\circ, 314.5^\circ \][/tex] and f) [tex]\[ \theta = 154.4^\circ, 334.4^\circ[/tex].

Let's solve each part:

(a) [tex]\(\sin \theta = 0.9263\)[/tex]

1. Find the reference angle using [tex]\(\theta = \sin^{-1}(0.9263)\)[/tex]:

[tex]\[ \theta \approx 67.9^\circ \][/tex]

2. Since [tex]\(\sin \theta\)[/tex] is positive, the angles are in the first and second quadrants:

[tex]\[ \theta_1 \approx 67.9^\circ \][/tex]

[tex]\[ \theta_2 \approx 180^\circ - 67.9^\circ = 112.1^\circ \][/tex]

So, the solutions are:

[tex]\[ \theta = 67.9^\circ, 112.1^\circ \][/tex]

(b) [tex]\(\cos \theta = -0.6909\)[/tex]

1. Find the reference angle using [tex]\(\theta = \cos^{-1}(-0.6909)\)[/tex]:

[tex]\[ \theta \approx 133.7^\circ \][/tex]

2. Since [tex]\(\cos \theta\)[/tex] is negative, the angles are in the second and third quadrants:

[tex]\[ \theta_1 \approx 133.7^\circ \][/tex]

[tex]\[ \theta_2 \approx 360^\circ - 133.7^\circ = 226.3^\circ \][/tex]

So, the solutions are:

[tex]\[ \theta = 133.7^\circ, 226.3^\circ \][/tex]

(c) [tex]\(\tan \theta = -1.5416\)[/tex]

1. Find the reference angle using [tex]\(\theta = \tan^{-1}(-1.5416)\)[/tex]:

[tex]\[ \theta \approx -57.3^\circ \][/tex]

2. Adjust the reference angle to fall within the interval [tex]\([0^\circ, 360^\circ)\)[/tex]:

[tex]\[ \theta_1 = 360^\circ - 57.3^\circ = 302.7^\circ \][/tex]

3. Since [tex]\(\tan \theta\)[/tex] is negative, the angles are in the second and fourth quadrants:

[tex]\[ \theta_2 = 180^\circ + (-57.3^\circ) = 122.7^\circ \][/tex]

So, the solutions are:

[tex]\[ \theta = 122.7^\circ, 302.7^\circ \][/tex]

(d) [tex]\(\cot \theta = 1.3952\)[/tex]

1. Find the reference angle using [tex]\(\theta = \cot^{-1}(1.3952)\)[/tex]:

[tex]\[ \theta \approx 35.8^\circ \][/tex]

2. Since [tex]\(\cot \theta\)[/tex] is positive, the angles are in the first and third quadrants:

[tex]\[ \theta_1 \approx 35.8^\circ \][/tex]

[tex]\[ \theta_2 \approx 180^\circ + 35.8^\circ = 215.8^\circ \][/tex]

So, the solutions are:

[tex]\[ \theta = 35.8^\circ, 215.8^\circ \][/tex]

(e) [tex]\(\sec \theta = 1.4293\)[/tex]

1. Convert to [tex]\(\cos \theta\)[/tex]:

[tex]\[ \cos \theta = \frac{1}{1.4293} \approx 0.6996 \][/tex]

2. Find the reference angle using [tex]\(\theta = \cos^{-1}(0.6996)\)[/tex]:

[tex]\[ \theta \approx 45.5^\circ \][/tex]

3. Since [tex]\(\cos \theta\)[/tex] is positive, the angles are in the first and fourth quadrants:

[tex]\[ \theta_1 \approx 45.5^\circ \][/tex]

[tex]\[ \theta_2 \approx 360^\circ - 45.5^\circ = 314.5^\circ \][/tex]

So, the solutions are:

[tex]\[ \theta = 45.5^\circ, 314.5^\circ \][/tex]

(f) [tex]\(\csc \theta = -2.3174\)[/tex]

1. Convert to [tex]\(\sin \theta\)[/tex]:

[tex]\[ \sin \theta = \frac{1}{-2.3174} \approx -0.4316 \][/tex]

2. Find the reference angle using [tex]\(\theta = \sin^{-1}(-0.4316)\)[/tex]:

[tex]\[ \theta \approx -25.6^\circ \][/tex]

3. Adjust the reference angle to fall within the interval [tex]\([0^\circ, 360^\circ)\)[/tex]:

[tex]\[ \theta_1 = 360^\circ - 25.6^\circ = 334.4^\circ \][/tex]

4. Since [tex]\(\sin \theta\)[/tex] is negative, the angles are in the third and fourth quadrants:

[tex]\[ \theta_2 = 180^\circ + (-25.6^\circ) = 154.4^\circ \][/tex]

So, the solutions are:

[tex]\[ \theta = 154.4^\circ, 334.4^\circ[/tex]

The complete question is:

Approximate, to the nearest 0.1°, all angles θ in the interval [0°, 360°) that satisfy the equation. (Enter your answers as a comma-separated list.)

(a) sin θ = 0.9263

θ = °

(b) cos θ = −0.6909

θ = °

(c) tan θ = −1.5416

θ = °

(d) cot θ = 1.3952

θ = °

(e) sec θ = 1.4293

θ = °

(f) csc θ = −2.3174

θ = °

There are 6 sides of the die, let us say that side 1 is S1, side 2 is S2 and so on.

Assigning a value of 3 to S1, so the probabilities are:

                                          Probability

S1 = 3                                    3/8

S2 = 1                                    1/8

S3 = 1                                    1/8

S4 = 1                                    1/8

S5 = 1                                    1/8

S6 = 1                                    1/8

total = 8                               

The combinations that the sum of the two rolls will be 4 are:

S1 and S3

S3 and S1

S2 and S2

So the total probability is:

P = (3/8) * (1/8) + (1/8) * (3/8) + (1/8) * (1/8)

P = 0.1094 = 10.94%

So there is about 0.1094 or 10.94% probability that the sum will be 4.

Answer: [tex]\frac{1}{4}[/tex]

Step-by-step explanation: If this is from a quiz from Buzz (Acceleration Education) Then this is the answer.

Hope This Helps You!

The probability that the stock closed at less than the mean for that year is approximately 0.5, assuming a normal distribution.

1. In a normal distribution, the mean divides the distribution into two equal halves, with 50% of the data falling below the mean and 50% above it.

2. Therefore, the probability that a randomly chosen data point (in this case, the closing price of  stock) is below the mean is 0.5 or 50%.

Now, let's dive into the step-by-step calculation to illustrate this:

1. Identify the Mean and Standard Deviation: While you mentioned not to calculate the mean, it's important to understand that for a normal distribution, the mean is at the center and separates the distribution into equal probabilities on either side. However, we do need to know the standard deviation, which measures the spread of data points around the mean. Let's assume the mean closing price for stock over the past year is $X, and the standard deviation is $Y.

2. Use the Standard Normal Distribution: Since we're dealing with a normal distribution, we can use the standard normal distribution to find probabilities. This distribution has a mean of 0 and a standard deviation of 1.

3. Standardize the Data: To work with the standard normal distribution, we need to standardize the data point we're interested in. Let's denote the closing price of  stock on the chosen day as $Z. We standardize this value using the formula:

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

  Here, ( Z ) is the standardized value.

4. Find the Probability: Once we have the standardized value ( Z ), we can look it up in a standard normal distribution table or use software/tools to find the probability that a randomly chosen value from a standard normal distribution is less than ( Z ). This probability will be the same as the probability that the stock closed at less than the mean for that year.

5. Conclusion: Based on the properties of a normal distribution, the probability that a randomly chosen data point (stock closing price) is below the mean is always 0.5 or 50%. This holds true regardless of the specific mean or standard deviation values for  stock.

complete question

And 1. if a person bought 1 share of  stock within the last year, what is the probability that the stock on that day closed at less than the mean for that year? hint: you do not want to calculate the mean to answer this one. the probability would be the same for any normal distribution.