find the coefficient of x^6 in the binomial expression of (2x+3)^9

85% = 0.85

1-0.85 = 0.15

80 x 0.15 = 12

the price now is $12

Given:

24,000 books

Between 3 and 4 books will have a printing error in every 3000 printed books

Find: in the 24000 books, the total number of books that will have a printing error

Solution:

Based from the given, we need to know how many sets of 3000 we

have in the 24000 books so:

24000 / 3000 = 8

Now, in each set of 3000 we have between 3 and 4 errors and we have 8 sets of 3000 books:

minimum errors 3 * 8 = 24

maximum errors 4 * 8 = 32

Therefore, at this rate, the number of books that will have a printing error in the 24000

will be between 24 and 32 or 24 < E < 32.

In 24,000 books, the number of printing errors could be between 24 and 32. This is calculated based on an error rate of 3 to 4 errors per 3,000 books.

To determine the number of printing errors in 24,000 books, we need to understand the error rate. The problem states that there are between 3 and 4 books with errors per 3,000 printed books.

First, find the range of error rates per 3,000 books:

Next, scale this up to 24,000 books:

Therefore, the number of books with printing errors in 24,000 books will be between 24 and 32.

Final answer:

To find the max value of a line integral over a closed curve using Green's Theorem, consider the curl of the given vector field and apply the theorem to express the result. The maximum value of the line integral is -2y²dy, determined through vector calculus and Green's Theorem application.

Explanation:

Green's Theorem states that for a vector field f in the form given, the max value of the line integral over any positively oriented closed curve C can be found by considering the curl of f.

By applying Green's Theorem, we can find that the maximum value of the line integral is -2 y²dy.

This computation involves utilizing vector calculus and understanding how to apply Green's Theorem to find the extremum of the line integral.

Answer:

25%

Step-by-step explanation:

Just here to help cause im doing this too lol

The total amount of candy left after adding the candy received from the grandmother and the sister, and subtracting the candy eaten by the friend, is approximately 0.63 pounds.

First, we add up the amounts of candy you received. You started with 1/3 pound from your grandmother and received an additional 1/2 pound from your sister, for a total of 5/6 pound of candy. However, because your friend ate some, we subtract 1/5 pound from this total. To do this, we need to convert all fractions to have a common denominator, which is 30 in this case. Therefore, 5/6 becomes 25/30, and 1/5 becomes 6/30. Subtraction gives us (25-6)/30 = 19/30 or approximately 0.63 pounds of candy left.

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Answer:

d

Step-by-step explanation:

Answer: 108 lbs.

Step-by-step explanation:

Given : A cube is packed with decorative pebbles. If the cube has a side length of 6 inches.

Volume of cube = [tex](side)^3[/tex]

i.e. Volume of cube = [tex](6)^3=216\text{ cubic inches}[/tex]

Since , each pebble weighs on average 0.5 lb per cubic inch.

Then, the total weight of the pebbles in the cube will be

= 0.5 x Volume of cube

= [tex]0.5\times216=108\text{ lb}[/tex]

Hence,  the total weight of the pebbles in the cube =108 lbs.

The area under the graph by using the first two and then four rectangles is [tex]2.958[/tex] units square.

For reference use the below-given graph.

Given function is

[tex]f(x)=x^{2}[/tex]  when [tex]x=1[/tex] to [tex]x=2[/tex] .

The first rectangle of the first part graph goes from [tex]1.0[/tex] to [tex]1.6[/tex], so the width will be [tex]0.6[/tex] units. And the height measured from the middle point i.e. [tex]1.3[/tex] is

[tex]f(1.3)=(1.3)^{2}[/tex]

[tex]=1.69[/tex] units.

Then the area of the first rectangle is [tex]0.6\times1.69=1.014[/tex] units square.

Similarly, the second rectangle of the first part graph goes from [tex]1.6[/tex] to [tex]2.0[/tex], so the width will be [tex]0.4[/tex] units. And the height measured from the middle point i.e. [tex]1.8[/tex]  is

[tex]f(1.8)=(1.8)^{2}[/tex]

[tex]=3.24[/tex] units.

So, the area of the second rectangle is [tex]0.6\times3.24=1.944[/tex] units square.

Hence, the final area under the graph will be [tex]1.014+1.944=2.958[/tex] units square.

Further, we can do the same for another part of the graph to find the area under the graph by using four rectangles.

For example,  the first rectangle of the four has a width of [tex]0.6[/tex] units and a height of [tex]f(1.1)=(1.1)^{2}[/tex]

[tex]=1.21[/tex] units.

Therefore, the area under the graph by using the first two and then four rectangles is [tex]2.958[/tex] units square.

Know more about the area under the curve here:

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The estimated areas under the curve of the function f(x)=x^2 between x = 1 and x = 2 are 2.3125 using two rectangles and 2.3281 using four rectangles

To estimate the area under the graph of the function f(x)=x^2 between x = 1 and x = 2 using rectangles, we use the method of midpoint Riemann sums. For this question, let's use 2 rectangles and then 4 rectangles.

First, for 2 rectangles, the interval from 1 to 2 is divided into 2 equal parts: [1, 1.5] and [1.5, 2]. The midpoints of these intervals are 1.25 and 1.75. The height of each rectangle is given by the function value at these midpoints: [tex]f(1.25) = (1.25)^2 =1.5625, and f(1.75) = (1.75)^2 = 3.0625.[/tex] The total area of the rectangles is thus (0.5 * 1.5625) + (0.5 * 3.0625) = 2.3125.

Next, for 4 rectangles, the interval from 1 to 2 is divided into 4 equal parts: [1, 1.25], [1.25, 1.5], [1.5, 1.75], [1.75, 2]. The midpoints of these intervals are 1.125, 1.375, 1.625, 1.875. The height of each rectangle is given by the function value at these midpoints: [tex]f(1.125) = (1.125)^2 = 1.26562, f(1.375) = (1.375)^2 = 1.8906, f(1.625) = (1.625)^2 = 2.6406[/tex], and f(1.875) = (1.875)^2 = 3.5156. The total area of the rectangles is thus [tex](0.25 * 1.26562) + (0.25 * 1.8906) + (0.25 * 2.6406) + (0.25 * 3.5156) = 2.3281.[/tex]

These are the estimated areas under the curve for 2 rectangles and 4 rectangles respectively. And as you can see, the more rectangles we use, the closer we get to the actual area under the curve.

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Answer:

An employee’s rate of pay is $36 per hour, with time and a half for all hours worked in excess of 40 during a week. The employee worked 48 hours during the week. The amount of the employee’s gross pay for the week is:

Step-by-step explanation:

Answer:  The required number of ways is 184756.

Step-by-step explanation:  Given that a test consists of 20 problems and students are told to answer any 10 of these questions.

We are to find the number of different ways in which the students choose 10 questions.

We know that

the number of ways in which r things can be chosen from n different things is given by

[tex]N=^nC_r.[/tex]

Therefore, the number of ways in which students chose 10 questions from 20 different questions is given by

[tex]N\\\\=^{20}C_r\\\\\\=\dfrac{20!}{10!(20-10)!}\\\\\\=\dfrac{20\times19\times18\times17\times16\times15\times14\times13\times12\times11\times10!}{10!\times 10\times9\times8\times7\times6\times5\times4\times3\times2\times1}\\\\\\=184756.[/tex]

Thus, the required number of ways is 184756.

Final answer:

Harris ate 12/k - 1 slices of pizza after a 12-slice pizza was divided evenly among k friends and he ate one less than he was given.

Explanation:

To find out how many slices of pizza Harris ate, we initially need to determine how many slices each person would get if the 12-slice pizza is divided evenly among k friends.

Each friend would get 12/k slices.

Since Harris ate 1 fewer slice than he was given, we subtract 1 from the number of slices he was supposed to get.

Therefore, the expression for the number of slices Harris ate is 12/k - 1.

1.
The midpoint MPQ of PQ is given by  (a + c / 2, b + d / 2)

2.
Let the x coordinates of the vertices of P_1 be : 

x1, x2, x3,…x33

the x coordinates of P_2 be :

z1, x2, x3,…z33

and the x coordinates of P_3 be:


w1, w2, w3,…w33

3.
We are given with: 

X1 + x2 + x3… + x33 = 99

We also want to find the value of w1 + w2 + w3… + w33.

4.

Now, based from the midpoint formula:

Z1 = (x1 + x2) / 2

Z2 = (x2 + x3) / 2

Z3 = (x3 + x4) / 2

Z33 = (x33 + x1) / 2

and 

W1 = (z1 + z1) / 2


W2 = (z2 + z3) / 2

W3 = (z3 + z4) / 2

W13 = (z33 + z1) / 2

.
.

5.

W1 + w1 + w3… + w33 = (z1 + z1) / 2 +  (z2 + z3) / 2 + (z33 + z1) / 2 = 2 (z1 + z2 + z3… + z33) / 2

Z1 + z1 + z3… + z33 = (x1 + x2) / 2 + (x2 + x3) / 2 + (x33 + x1) / 2

2 (x1 + x2 + x3… + x33) / 2 = (x1 + x2 + x3… + x33 = 99


Answer: 99

In the process of calculating probabilities of events on a pair of dice, we found P(F), the probability of rolling a total of two, to be 1/36. P(E), the probability of rolling an even total, to be 1/2. However, when determining P(F|E), the probability of rolling a two given we've rolled an even total, the probability changes to 1/18 due to the reduced sample space.

The concepts involved in this question are related to probability, specifically the principles governing dice rolls. In this particular scenario, the events are rolling two dice and getting a total of two (Event F), and rolling an even total on the dice (Event E).

In this specific scenario, event F (the total is two) can only occur in one way - when both dice show 1. Since there are 36 potential outcomes when two dice are rolled (6 possibilities for the first die and 6 for the second), the probability of event F, P(F), is 1/36.

Event E (an even total) can occur in 18 ways (2,4,6 for the first die and 1,3,5 for the second or 1,3,5 for the first die and 2,4,6 for the second). So, P(E) = 18/36 = 1/2. However, when considering P(F|E) (the probability of event F given that event E has occurred), you need to adjust your consideration of 'total possibilities' based only on event E. Since P(E) = 1/2, your total possibilities now become 18. From these 18, only one will result in a total of two. Therefore, P(F|E) = 1/18.

Of course, there's different perspectives to consider how adding the condition that E had occurred would change the probability of event F. Essentially, by narrowing down the potential outcomes to only those that involve event E, you're working with a reduced sample space. This in turn affects the likelihood of event F occurring, hence the alteration in probability.

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