Answer:
[tex]C^{102}_{100}=5151[/tex]
Step-by-step explanation:
Let's imagine the following situation, if we want to distribute 100 coins between three pirates we could represent this situation with a line arrangement. For example if we had7 coins and 3 pirates one possible distribution of coins would be given by CC|CCCC|C, the C's represent coins and the bars the boundaries between two pirates, for the particular line arrangement shown, we have that pirate A has 2 coins, B has 4 coins and C has a single coin. Another possible arrangement is,
|CCC|CCCC, where pirate A has no coin, pirate B has 3 coins and C has 7 coins. If we take notice of the fact that the arrangement representing a distribution is composed of 9 elements, that is 7 C's and 2 | (bars), then a way to make an arrangement would be to fill 9 empty boxes with our available coins and bars in all the possible ways. This means that if we first choose to fill 7 out of 9 boxes with coins then the number of possible combinations is [tex]C^{9}_7=\frac{9!}{7!(9-7)!}36[/tex]. In general if we want to distribute n elements in k boxes, where the boxes can either be filled with any number of elements (including 0 number of elements), we have that the number of possible distributions will be [tex]C^{n+k-1}_{n}=\frac{(n+k-1)!}{n!(k-1)!}[/[tex], where we used the fact that we need k-q bars to represent k boxes. Thus pirate A can choose from [tex]C^{102}_{100}=5151[/tex] possible divisions.
Bonus:
If every pirate wants to have the maximum number of coins possible without being executed, here's how pirate A has to divide the coins in order to keep the largest amount of coins.
We have to think backwards to figure this out. Imagine pirate A was executed and there are only two remaining players. Pirate B should propose to keep all the coins, pirate C could oppose but pirate B's vote would break the vote and keep all the loot. Pirate A, B and C are all aware of this, so pirate A should propose to keep 99 coins and give the remaining gold piece to pirate C, Pirate B will of course oppose the division, but pirate C should accept because if not he would get no coins. Thus the division would be.
A: 99 coins
B: 0 coins
C: 1 coins
You buy g gallons of gasoline at $3.05 per gallon and pay $36.60. Write an equation to find the number of gallons purchased. Then find the number of gallons of gasoline that you purchased.
Answer:
The equation to find the number of gallons purchased is:
[tex]C(n) = 3.05n[/tex]
You purchased 12 gallons of gasoline
Step-by-step explanation:
This problem can be modeled by the following first order function
[tex]C(n) = P_{G}n[/tex]
Where C(n) is the cost in function to the number of gallons, P is the price of the gallon and n is the number of gallons
The problem states that the price of gasoline is $3.05 per gallon, so P = 3.05
The equation to find the number of gallons purchased is:
[tex]C(n) = 3.05n[/tex]
If you pay $36.60, you have C = 36.60, and want to find n, so:
[tex]36.60 = 3.05n[/tex]
[tex]n = \frac{36.60}{3.05}[/tex]
[tex]n = 12[/tex]
You purchased 12 gallons of gasoline
Limit of a line: Which statements are true and which are false? Justify your answer.
Please help me out. I have a test and I'm completely lost. I have the basic idea of limits, but I'm not sure if I'm doing this correctly.
The important thing to note here is that a limit is defined when the value of f(x) as it approaches x is the same when approaching from both the left and the right sides (unless a specific side from which to approach has been defined in the question, eg. lim (x -> 10⁻) f(x) or lim (x -> 10⁺) f(x) ).
1. The first line states that the limit of f(x) as x approaches 10 is 0. This means that the value of f(x) as x approaches 10 from both the left and the right sides must be 0.
Looking at the graph, we can see that this is indeed the case, and may write this as:
lim (x -> 10⁻) f(x) = lim (x -> 10⁺) f(x) = 0
Thus, the statement is True.
2. The second line states that the limit of f(x) as x approaches -2 from the right side is 3.
Looking at the graph, whilst f(x) approaches 3 as x approaches -2 from the left side, f(x) approaches 7 as x approaches -2 from the right side. We can write this as:
lim (x -> -2⁺) f(x) = 7
Thus, since lim (x -> -2⁺) f(x) ≠ 3, the statement is False.
3. The third line states that the limit of f(x) as x approaches -8 is the value of f(x) at x = -8 (ie. the value of f(x) for which the graph exists at x = -8).
Looking at the graph, we can see that the graph of f(x) approaches -6 as x approaches -8 from both the left and right sides. We can write this as:
lim (x -> -8⁻) f(x) = lim (x -> -8⁺) f(x) = -6
Now, this is where a knowledge of what open and closed circles represent on a graph is crucial. A closed circle means that a point exists on a graph, whereas an open circle means that there is a break in the graph at that point.
If we look at the graph, we can see that the closed circle at x = -8 is actually the value f(x) = -3. We can write this as:
f(-8) = -3
Moving from the line before this, where we defined the limit as -6, we can thus write:
lim (x -> -8) f(x) ≠ -3
Therefor, lim (x -> -8) f(x) ≠ f(-8)
Thus, the statement is False.
4. The fourth line states that the limit of f(x) as x approaches 6 is 5. Thus, for this to be true, f(x) must approach 5 as x approaches 6 from both the left and right sides.
Looking at the graph, we can see that as x approaches 6 from the left side, f(x) approaches 2; whilst as x approaches 6 from the right side, f(x) approaches 5. We can write this as:
lim (x -> 6⁻) f(x) = 2
lim (x -> 6⁺) f(x) = 5
These limits are not the same:
lim (x -> 6⁻) ≠ lim (x -> 6⁺), therefor a limit does not exist at x = 6.
Thus, this statement is False.
Hope this helps, but if anything is unclear, please feel free to leave a comment below :)
Line&lies in the xy-plane. The x-intercept of line K is -4, and line k passes through the midpoint of the line segment whose endpoints are (2, 9) and (2, 0). What is the slope of line k? Give your answer as a fraction
Answer: [tex]\dfrac{3}{4}[/tex]
Step-by-step explanation:
Given : Line k lies in the xy-plane.
The x-intercept of line K is -4. i.e. line k is intersecting the x-axis at (-4,0).
We know that the mid point of a line passing through any two points (a,b) and (c,d) is [tex](\dfrac{a+c}{2},\dfrac{b+d}{2})[/tex].
Then, the midpoint of the line segment whose endpoints are (2, 9) and (2, 0) will be :-
[tex](\dfrac{2+2}{2},\dfrac{9+0}{2})=(2,4.5)[/tex]
The slope of line passing through (p,q) and (r,s) will be :-
[tex]m=\dfrac{s-q}{r-p}[/tex]
Then, the slope of line passing through (-4,0) and (2,4.5) will be :-
[tex]m=\dfrac{4.5-0}{2-(-4)}=\dfrac{4.5}{2+4}=\dfrac{4.5}{6}\\\\=\dfrac{45}{60}=\dfrac{3}{4}[/tex]
Hence, the slope of line k [tex]=\dfrac{3}{4}[/tex]
Workers at paper company count the number of boxes of paper in a warehouse each month. In january, there were 160341 boxes of paper. In February, there were 32698 boxes of paper. How does the digit 6 in February compare to the digit 6 in january?
Answer:
The digit 6 in February is worth 1/100 of its value in January.
Step-by-step explanation:
The value of a given digit in a number can be found by setting the other digits to zero.
In February, the digit 6 has a value of 00600 = 600.
In January, the digit 6 has a value of 060000 = 60,000.
The ratio of the value in February to the value in January is ...
600/60000 = 1/100
The digit 6 in February is one hundredth of the digit 6 in January.
Two sides of a rectangle are 4cm in length. The other two sides are 6cm in length. What is the perimeter of the rectangle? Include the abbreviation for millimeter as the units.
Answer: 200 mm
Step-by-step explanation:
The perimeter of rectangle is given by :-
[tex]P=2(l+w)[/tex], where l is length and w is width of the rectangle.
Given : Two sides of a rectangle are 4 cm in length. The other two sides are 6 cm in length.
The perimeter of the rectangle will be :_
[tex]P=2(4+6)=2(10)=20\ cm[/tex]
We know that 1 cm = 10 mm
Therefore, perimeter of the rectangle = [tex]20\times10=200\ mm[/tex]
Eric wants to estimate the percentage of elementary school children who have a social media account. He surveys 450 elementary school children and finds that 280 have a social media account. Identify the values needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval.
The confidence interval for population mean is given by :-
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where [tex]\hat{p}[/tex] is the sample proportion, n is the sample size , [tex]z_{\alpha/2}[/tex] is the critical z-value.
The values needed to calculate a confidence interval at the 99% confidence level are :
Given : Significance level : [tex]\alpha:1-0.99=0.01[/tex]
Sample size : n=450
Critical value : [tex]z_{\alpha/2}=2.576[/tex]
Sample proportion: [tex]\hat{p}=\dfrac{280}{450}\approx0.62[/tex]
Now, the 99% confidence level will be :
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.62\pm(2.576)\sqrt{\dfrac{0.62(1-0.62)}{450}}\\\\\approx0.62\pm0.023\\\\=(0.62-0.023,\ 0.62+0.023)=(0.597,\ 0.643)[/tex]
Hence, the 99% confidence interval is [tex](0.597,\ 0.643)[/tex]
To calculate a 99% confidence interval for the true proportion, find the values needed and apply the formula. In this case, the confidence interval is between 0.5601 and 0.6811.
1. Sample proportion [tex](\( \hat{p} \))[/tex]:
[tex]\[ \hat{p} = \frac{x}{n} = \frac{280}{450} \approx 0.6222 \][/tex]
2. Margin of error E :
For a 99% confidence level, z = 2.576.
[tex]\[ E = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]\[ E = 2.576 \times \sqrt{\frac{0.6222 \times (1-0.6222)}{450}} \]\[ E = 2.576 \times \sqrt{\frac{0.6222 \times 0.3778}{450}} \]\[ E = 2.576 \times \sqrt{\frac{0.235052}{450}} \]\[ E \approx 2.576 \times \sqrt{0.00052233} \]\[ E \approx 2.576 \times 0.022854 \]\[ E \approx 0.058893 \][/tex]
3. Confidence interval:
[tex]\[ \text{Confidence interval} = \hat{p} \pm E \]\[ \text{Confidence interval} = 0.6222 \pm 0.0589 \][/tex]
Therefore, the confidence interval for the percentage of elementary school children who have a social media account at the 99% confidence level is approximately (0.5633, 0.6811) .
Construct a truth table for the logical operator NAND.
Answer:
The output for NAND gate is false if all the inputs are true. it is true if either A or B is False.
Step-by-step explanation:
Step1
NAND operator:
NAND is a logic or Boolean expression and is a combination of NOT and AND Gate.
NAND gate is built using various junction diodes and transistors.
Step2
The output of NAND can be 0 or 1. Here, LOW (0) and HIGH (1). 0 or 1 is commonly named as FALSE or TRUE respectively.
It is basically the complement of AND gate. If all its inputs are True or 1, then the output is False or 0 else if either A or B is 0 or 1 then it will give the output as TRUE.
Step3
Here, A and B are 2 inputs used to make the truth table for NAND.
In 1st row, when A and B are both 0, the output for NAND is 1.
Likewise, in 2st row and 3rd row, when either A or B is 0 or 1, the output for NAND is 1 that is TRUE.
But in last row, when both inputs A and B are true, then the output for NAND gate is FALSE or 0.
The diagram and truth table for NAND is shown below.
Final answer:
A truth table for the NAND logical operator lists all possible binary input combinations of variables A and B, applies the AND operation, and then negates the output to present the NAND result. The resulting truth table will show that a NAND gate outputs true for all cases except when both inputs are true.
Explanation:
Constructing a truth table for the logical operator NAND is a way to visualize how this operator works with different input values.
The NAND operator is essentially the negation of the AND operator, meaning that it outputs the opposite of what an AND gate would. Here is how you construct a truth table for NAND:
List all possible input combinations of the variables A and B in binary (00, 01, 10, 11).Under the AND column, apply the AND operation to each pair of inputs (true only if both A and B are true).Finally, under the NAND column, negate the output of the AND operation (true if the AND output is false).The resulting truth table for the NAND operation is as follows:
A B A AND B A NAND B
false false false true
false true false true
true false false true
true true true false
An urn contains 2 red marbles and 3 blue marbles. 1. One person takes two marbles at random from the urn and does not replace them. a) State the general ways in which the person could get a red marble and a blue marble b) State the number of ways this can occur. c) What is the probability the person gets a red and a blue marble? P(R & B) =
Given : An urn contains 2 red marbles and 3 blue marbles.
Total marbles = 2+3=5
a) The general ways in which the person could get a red marble and a blue marble are :
1) He draws red marble first and then second marble as blue.
2) He draws blue marble first marble and then second marble as red.
b) The number of ways to get one red and one blue marble is given by :-
[tex]^2C_1\times^3C_1=2\times3=6[/tex] (i)
c) Number of ways to get 2 marbles from 5 is given by :-
[tex]^5C_2=\dfrac{5!}{2!(5-2)!}=\dfrac{5\times4\times3!}{3!\times2}=10[/tex] (ii)
Now, The probability the person gets a red and a blue marble will be :-
[tex]P(R\ \&B)=\dfrac{6}{10}=0.6[/tex] [Divide (i) by (ii)]
Hence, the probability the person gets a red and a blue marble= 0.6
Consider the following equation: f(x) = 64-7x+ 4x^2
a) Take the deriivative and write out the equation, y= df/dx. Call this equation 1.
b)Let equation 2 be, y= f(x)/ x
c)Find the intersection of equation 1 and equation 2
Answer:
Considering the equation f(x) = 64-7x+ 4x^2 this are the answers
a) Equation 1:
[tex]y=8x-7[/tex]
b) Equation 2:
[tex]y=\frac{64}{x}-7+4x[/tex]
c) Intersection of Equation 1 and Equation 2:
The lines intersects in the points:
P(x,y)=(4,25) and P(x,y)=(-4,-39)
Step-by-step explanation:
a) Drivate f(x) to find Equation 1:
y=f'(x)
y=0-7+8x
y=8x-7
b) Equation 2 is f(x)/x
y=f(x)/x
[tex]y=\frac{64-7x+4x^2}{x}[/tex]
[tex]y=\frac{64}{x}-7+4x[/tex]
c) The intersection between the two equations is the only point that they have in common, this means that the points (x,y) satisfies both equations
To find it lets write the equations side by side
[tex]\left \{ {{y=8x-7} \atop {y=\frac{64}{x}-7+4x}} \right.[/tex]
Given the fact that the y points are the same for both equations, you can replace the equation 1 into equation 2, this means, instead of write y, write equation 2:
[tex]8x-7=\frac{64}{x}-7+4x[/tex]
now you can solve this for x
[tex]8x-4x-7+7=\frac{64}{x}\\4x=\frac{64}{x}\\4x^2=64[/tex]
[tex]x^2=\frac{64}{4}\\x=\sqrt{16}\\ x=±4[/tex]
With the values of x, you can find the values of y by putting it into equation 1:
y=8*(+4)-7 and y=8*(-4)-7
y=25 and y=-39
Finally, the points where these two equations intersect are P(x,y)=(4,25) and P(x,y)=(-4,-39).
Write a linear cost function for the situation. Identify all variables used. A parking garage charges 4 dollars plus 65 cents per half-hour. A linear cost function for the situation is C(x)=L (Use integers or decimals for any numbers in the expression)
Answer:
Step-by-step explanation:
Given that the cost function is linear.
Fixed charges = 4 dollars
Variable charges perhalf hour = 65 cents = 0.65 dollars
Suppose a man leaves for x hours say
then we have he would be charged 4 dollars besides 0.65 for 2x half hours.
Hence
[tex]C(x) = 5+4(2x)\\C(x) = 8x+5[/tex] where x is the number of hours.
The variable charge is 1.3 dollars per hour and the fixed cost is 4 then the linear cost equation is C(x) = 1.3x + 4.
What is the linear system?It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.
Given
A parking garage charges 4 dollars plus 65 cents per half-hour.
Let C(x) be the linear cost.
The fixed charge is 4 dollars
The variable charge per half-hour is 0.65 dollars.
Suppose the men leave for x half-hours for one hour the x will be 2x.
Then the equation will be
[tex]\rm C(x) = 0.65*2x + 4\\\\C(x) = 1.3x + 4[/tex]
Thus, the equation is [tex]\rm C(x) = 1.3x + 4[/tex].
More about the linear system link is given below.
https://brainly.com/question/20379472
Find a general solution of y" – 4y = 0.
Answer:
y = [tex]C_{1}e^{2x} + C_{2}e^{-2x}[/tex]
Step-by-step explanation:
We are given the differential equation: y'' - 4y = 0
We have to find the general solution.
The auxiliary equation for the above differential equation can be written as:
m² - 4 = 0
We solve for m.
⇒m² = 4
⇒m = ±2
⇒[tex]m_{1}[/tex] = +2 and [tex]m_{2}[/tex] = -2
Thus, we have two distinct roots or we have two distinct values of m.
Thus, the general solution will be of the form:
y = [tex]C_{1}e^{m_{1}x} + C_{2}e^{m_{2}x}[/tex]
y = [tex]C_{1}e^{2x} + C_{2}e^{-2x}[/tex]
Drug A has a concentration of 475 mg/10 mL. How many grams are in 100 mL of Drug A? (Round to the nearest tenth if applicable).
Answer: There are 4.8 grams in 100 mL of Drug A
Step-by-step explanation:
In order to determinate how many grams are in 100 mL of Drug A you can multiply and divide the expression of the concentration by 10 (To obtain 100 mL in the denominator)
Notice that the value remains unaltered.
(475 mg/10 mL )(10/10) = 4750 mg/ 100 mL
But the question is how many grams are in 100 mL, so you have to convert the value from mg to g.
The prefix m (mili) is equivalent to 0,001 so you can use 0.001 instead of the prefix
4750(0.001) g/ 100 mL
4.75 g/ 100 mL
The rounded result is 4.8 g/ 100 mL
A survey of 1,116 tourists visiting Orlando was taken. Of those surveyed:
280 tourists had visited the Magic Kingdom
292 tourists had visited LEGOLAND
94 tourists had visited both the Magic Kingdom and LEGOLAND
97 tourists had visited both the Magic Kingdom and Universal Studios
80 tourists had visited both LEGOLAND and Universal Studios
38 tourists had visited all three theme parks
91 tourists did not visit any of these theme parks
How many tourists only visited the Universal Studios (of these three)?
Your Answer:
Answer:
Step-by-step explanation:
We can solve this using a Venn Diagram,
First we write down the 91 tourists that didn't visit any of the theme parks.
Then we focus on the 38 tourists that visited all three theme parks and write that down in the intersection of the 3 circles.
80 tourists visited both LEGOLAND and Universal Studios but we already know that 38 visited the three of them, so 80-38= 42 tourists visited only those two parks.
97 tourists visited both Magic Kingdom and Universal Studios, but again, we know that 38 visited the three of them, so 97-38 = 59 tourists visited only those two parks.
94 tourists visited both Magic Kingdom and LEGOLAND, but since 38 visited the three of them, we subtract 94-38 = 56 tourists visited those two parks.
292 tourists visited LEGOLAND so we'll focus on that circle 292 - (56+38+42) = 156 tourists visited only LEGOLAND
280 tourists visited Magic Kingdom so we focus now on the Magic Kingdom circle and we have 290 - (56+38+59) = 127 only visited Magic Kingdom.
Now we will sum up all the quantities we have and subtract them from the total amount of tourists (1116) to find out how many people visit Universal Studios
1116 - (91+156+42+38+56+59+127) = 547
The n × n identity matrix is the matrix with diagonal entries are all 1’s and the rest are all 0’s. Show that, for any n×n matrix A, we have AI=IA=A.
Express the matrix [tex]I[/tex] like [tex](\delta_{ij})_{n\times n}[/tex], where [tex]\delta_{ij}=1[/tex] if [tex]i=j[/tex] and [tex]\delta_{ij}=0[/tex] if [tex]i\neq j[/tex]. ([tex]\delta_{ij}[/tex] is known as kronecker's delta)
In the same form we express the matrix [tex]A=(a_{ij})_{n\times n}[/tex].
The firs index indicate the row and the second the column.
By the multiplication of matrices we have [tex]AI=(c_{ij})_{n\times n}[/tex], where
[tex]c_{ij}=\sum_{k=1}^n a_{ik}\delta _{kj} = a_{ij}[/tex]
because only [tex]\delta_{jj}[/tex] is non-zero in the last sum.
therfore we have [tex]AI=(c_{ij})_{n\times n}=(a_{ij})_{n\times n}=A[/tex].
In the same manner we have
[tex]IA=(d_{ij})_{n\times n}[/tex], where
[tex]d_{ij}=\sum_{k=1}^n \delta _{ik}a_{kj} = a_{ij}[/tex]
And so, [tex]IA=A[/tex]
Matrix multiplication with the identity matrix leaves the original matrix unchanged. This is analogous to multiplying a number by 1, which leaves the number unchanged. Thus, for any square matrix A, AI = IA = A.
Explanation:The n×n identity matrix, denoted by I, is a special square matrix that has ones on its main diagonal and zeros elsewhere. It's called the 'identity' matrix because multiplication with it leaves a matrix unchanged, similar to how multiplying a number by 1 leaves it unchanged.
Let's prove that for any n×n matrix A, we have AI=IA=A. As A is an n×n matrix, suppose it has elements aij. Because of the definition of matrix multiplication, element at ith row and jth column of the product of two matrices (say A and I) is given by summation of the products of corresponding elements in ith row of first matrix and jth column of the second matrix.
In the case of AI, for any element in the result, we get it by summation of product of corresponding elements in ith row of A and ith column of I. Now, since except for the ith position, rest of the elements in ith column of I are zero, we only get aii in the summation. Hence, AI = A. Similar argument can be made for IA. Therefore, AI=IA=A.
Learn more about Identity Matrix Multiplication here:https://brainly.com/question/13595132
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Find a power series solution of the differential equation y" + 4xy = 0 about the ordinary point x = 0.
Answer:
[tex]y=c_0(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(2*3)(5*6)...(3k-1)(3k)} x^{3k}])+c_1(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(3*4)(6*7)...(3k)(3k+1)} x^{3k+1}])[/tex]
Step-by-step explanation:
[tex]y"+4xy=0[/tex]
Assume that the problem has a solution of the form:
[tex]y=f(x)=\sum_0^{\infty}(c_nx^n)[/tex]
then calculate the derivative:
[tex]y'=\sum_1^{\infty}(nc_nx^{n-1})[/tex]
and the second derivative:
[tex]y"=\sum_2^{\infty}(n(n-1)c_nx^{n-2})[/tex]
In these sums the subindex indicates the first non cero term.
Making the substitution in the original equation:
[tex]\sum_2^{\infty}(n(n-1)c_nx^{n-2})+4x\sum_0^{\infty}(c_nx^{n})=0\\[/tex]
Now look at the subindex of the sum: they do not match; you need to make them match:
[tex]\sum_0^{\infty}((n+2)(n+1)c_{n+2}x^{n})+4\sum_0^{\infty}(c_nx^{n+1})=0\\[/tex]
Now they match and they could be joined together on the same sum symbol, however the x term doesn't have the same exponent (after multiplying the x from the original formula with the one inside the sum). Therefore you need to expand the second derivative term and displace the other one (as we did in the previous step) to match them both:
[tex]2c_2+\sum_1^{\infty}((n+2)(n+1)c_{n+2}x^{n})+4\sum_1^{\infty}(c_{n-1}x^{n})=0[/tex]
Now you can join them together:
[tex]2c_2+\sum_1^{\infty}[(n+2)(n+1)c_{n+2}x^{n}+4(c_{n-1}x^{n})]=0[/tex]
[tex]2c_2+\sum_1^{\infty}x^n=0[/tex]
If the solution is valid for all x, then all terms must be independantly 0. therefore:
[tex]a_2=0\\(n+2)(n+1)c_{n+2}+4(c_{n-1})=0\\c_{n+2}=\frac{-4(c_{n-1})}{(n+2)(n+1)}[/tex]
Now you need to use that recurrent function to calculate the coefficients:
[tex]n=1 \rightarrow c_3=\frac{-4c_0}{3*2}=-\frac{2}{3} c_0\\n=2 \rightarrow c_4=\frac{-4c_1}{4*3}=-\frac{1}{3} c_1\\n=3 \rightarrow c_5=\frac{-4c_2}{5*4}=-\frac{1}{5} c_2=0\\n=4 \rightarrow c_6=\frac{-4c_3}{6*5}=-\frac{2}{15} c_3\\n=5 \rightarrow c_7=\frac{-4c_4}{7*6}=-\frac{2}{21} c_4\\n=6 \rightarrow c_8=\frac{-4c_5}{8*7}=0\\n=7 \rightarrow c_9=\frac{-4c_6}{9*8}=-\frac{1}{18} c_6[/tex]
you found that for each n divisible by 3, therefore for each [tex]c_{3n-1}[/tex] the coefficient is 0:
[tex]c_2=c_5=c_8....=c_{3n-1}=0[/tex]
now look at the terms [tex]a_3,a_6,a_9[/tex], they are all recurrent to c_0, therefore you can wirte a rule for them:
[tex]c_{3n} = \frac{(-4)^nc_0}{(2*3)(5*6)...(3n-1)(3n)}[/tex]
Finally, look at the terms [tex]a_4,a_7[/tex], they recurr to c_1, therefore:
[tex]c_{3n+1} = \frac{(-4)^nc_1}{(3*4)(6*7)...(3n)(3n+1)}[/tex]
So you have two constants that are independant, c_0 and c_1.
Therefore the solution must be writen in terms of these two arbitrary constants:
[tex]y=c_0(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(2*3)(5*6)...(3k-1)(3k)} x^{3k}])+c_1(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(3*4)(6*7)...(3k)(3k+1)} x^{3k+1}])[/tex]
Assume the random variable X is normally distributed with meanmu equals 50μ=50and standard deviationsigma equals 7σ=7.Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded.Upper P left parenthesis 34 less than Upper X less than 63 right parenthesisP(34
Answer: 0.9575465
Step-by-step explanation:
Let the random variable X is normally distributed with mean [tex]\mu=50[/tex] and standard deviation[tex]\sigma=7[/tex] .
Using the formula , [tex]z=\dfrac{x-\mu}{\sigma}[/tex] , we have the z-value for x= 34
[tex]z=\dfrac{34-50}{7}\approx-2.29[/tex]
For x= 63
[tex]z=\dfrac{63-50}{7}\approx1.86[/tex]
P-value : P(34<x<63)=P(-2.29<z<1.86)
[tex]=P(z<1.86)-P(z<-2.29)\\\\=0.9685572-(1-P(z<2.29))\\\\1=0.9685572-(1-0.9889893)\\\\=0.9575465[/tex]
Hence, the required probability = 0.9575465
Use the method of your choice to determine the probability below. Being dealt three sixes off the top of a standard deck of well-shuffled cards. The probability is . (Type an integer or a simplified fraction.)
Answer:
1/5525
Step-by-step explanation:
We now that a standard deck has 52 different cards. Also we know that a standard deck has four different suits, i.e., Spades, Hearts, Diamonds and Clubs. We can find the following cards for each suit: Ace, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen and King.
Now, the probability of getting any of these cards off the top of a standard deck of well-shuffled cards is 1/52. As we have 4 different sixes, we have that the probability of getting a six is 4/52. When we get a six, in the deck only remains 3 sixes and 51 cards, so, the probability of getting another six later is 3/51. When we get the second six, in the deck only remains 2 sixes and 50 cards, so, the probability of getting the third six is 2/50. As we have independet events, we should have that the probability of getting 3 sixes off the top of a standard deck of well-shuffled cards is
(4/52)(3/51)(2/50)=
24/132600=
12/66300=
6/33150=
3/16575=
1/5525
The probability of being dealt three sixes off the top of a standard deck of well-shuffled cards is approximately 1/5513 or 0.00018 when rounded to five significant digits.
Explanation:The subject of this question is probability in mathematics. In a standard deck of 52 playing cards, there are four sixes: one each of hearts, diamonds, clubs, and spades. When looking at the probability of being dealt three sixes off the top of a well-shuffled deck, looking at drawing one card at a time in succession grants us the solution.
For the first card, the probability of drawing a six is 4/52. If you draw a six, there are now three sixes left in a 51-card deck. So, the probability of drawing a six on the second draw is 3/51. Using the same logic, the probability of drawing a six on the third draw is 2/50.
The probability of these three events happening in succession is the product of their individual probabilities, which is calculated as follows: (4/52) * (3/51) * (2/50) = 24/132600 = 0.00018 when rounded to 5 significant digits, or simplified, this is approximately 1/5513.
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Is 57.3 a whole number
Answer:
No, whole numbers are numbers that are whole, which would mean that they are not a decimal or a fraction. This is because a fraction or a decimal is a portion of a number, which would mean that the decimal or fraction is not complete/ not whole!
If one 20-mL ampul contains 0.5 g of aminophylline, how many milliliters should be administered to provide a 25-mg dose of aminophylline?
Answer:
1mL should be administered to provide a 25-mg dose of aminophylline.
Step-by-step explanation:
The problem states that one 20-mL ampul contains 0.5 g of aminophylline, and asks how many milliliters should be administered to provide a 25-mg dose of aminophylline.
The first step is converting 0.5g to mg.
Each g has 1000mg, so:
1g - 1000mg
0.5g - xmg
x = 1000*0.5
x = 500mg.
Now we have that one 20-mL ampul contains 500mg of aminophylline. How many milliliters should be administered to provide a 25-mg dose of aminophylline?
As the dose increases, so does the quantity of aminophylline. It means that we have a direct rule of a three, there is a cross multiplication. So:
20mL - 500mg
x mL - 25mg
500x = 500
[tex]x = \frac{500}{500}[/tex]
x = 1 mL
1mL should be administered to provide a 25-mg dose of aminophylline.
Final answer:
1 mL of the aminophylline solution should be administered to provide a 25-mg dose. This calculation was made by first converting the total amount of aminophylline to milligrams and then establishing the amount per milliliter to solve for the necessary volume.
Explanation:
To determine how many milliliters of aminophylline should be administered to provide a 25-mg dose, we need to use the following information:
1 ampul = 20 mL contains 0.5 g of aminophyllineDesired dose = 25 mg of aminophyllineFirst, we need to convert 0.5 g of aminophylline to milligrams:
0.5 g × 1000 mg/g = 500 mg
Now that we have the total amount of aminophylline in milligrams, we can determine the amount per milliliter:
500 mg / 20 mL = 25 mg/mL
To find the volume that contains 25 mg, we set up a proportion:
(25 mg of aminophylline) / (X mL) = (25 mg/mL)
By solving for X, we find:
X = 1 mL
So, 1 mL of the aminophylline solution should be administered to provide a 25-mg dose.
translate phrases algebraic expressions, half a number
Answer: It is 1\2x and the next one is 1\2x+10 then the last one is 1\2x+10=22
Step-by-step explanation:For sure they are right
The algebraic expression of half a number is expressed as: x/2
How to write algebraic expressions?Algebraic expressions are defined as the idea of expressing numbers using letters or alphabets without specifying their actual values.
Now, we can represent a number in general by x.
Halving a number means dividing it by two, so the expression that shows the half of a number is:
x/2, since you divide a number by two.
Thus, we conclude that the algebraic expression of half a number is x/2
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4 1/6 divided by 1 1/3
Answer:
[tex]\frac{25}{8}[/tex]
Step-by-step explanation:
A fraction is a part of a whole .
A proper fraction is a fraction whose numerator is less than denominator .
For example [tex]\frac{2}{3}\,,\,\frac{3}{4}[/tex]
An improper fraction is a fraction whose numerator is greater than denominator .
For example [tex]\frac{5}{4}\,,\,\frac{8}{7}[/tex]
A mixed fraction is made up of whole number and a proper fraction .
Given : [tex]4\frac{1}{6}\,,1\frac{1}{3}[/tex]
Solution :
[tex]4\frac{1}{6}=\frac{4\times 6+1}{6}=\frac{25}{6}\\1\frac{1}{3}=\frac{3\times 1+1}{3}=\frac{4}{3}[/tex]
We need to divide [tex]4\frac{1}{6}[/tex] by [tex]1\frac{1}{3}[/tex] .
[tex]4\frac{1}{6}\div 1\frac{1}{3}=\frac{25}{6}\div\frac{4}{3} \\\Rightarrow 4\frac{1}{6}\div 1\frac{1}{3}= \frac{25}{6}\times \frac{3}{4}=\frac{25}{8}[/tex]
You arrive in London with $400. How many British pounds can you purchase with your US dollars if the conversion is $1.414 per British pound? Your goal is to prepare a recipe for a gathering of 21 people. If the original recipe serves 6 people and requires 5 cups of flour, how much flour should you use?
Answer:
a) You can buy 282.88 British pounds.
b) you should use 17.5 cups of flour.
Step-by-step explanation:
a) You arrive in London with $400. How many British pounds can you purchase with your US dollars if the conversion is $1.414 per British pound?
We can solve this problem by using a rule of three
1.414 dollars --- 1 British pound
400 dollars --- x British pounds
We solve for x,
x = 400/1.414 = 282.88
We can buy 282.88 British pounds.
b) Your goal is to prepare a recipe for a gathering of 21 people. If the original recipe serves 6 people and requires 5 cups of flour, how much flour should you use?
Again, we can use a rule of three.
6 people --- 5 cups of flour
21 people --- x cups of flour
We solve for x:
x= (21)(5)/6 = 105/6=17.5
We should use 17.5 cups of flour.
Use Polya's four-step problem solving strategy and the problem-solving procedures presented in this lesson to solve the following exercise. A shirt and a tie together cost $57. The shirt costs $21 more than the tie. What is the cost of the shirt?
Answer:
$ 30
Step-by-step explanation:
Let x be the cost of tie ( in dollars ),
∵ The shirt costs $21 more than the tie,
So, the cost of shirt = x + 21
Thus, the cost of a tie and a shirt = x + x + 21 = 2x + 21,
According to the question,
2x + 21 = 57
2x = 57 - 21
2x = 36
x = 18
Hence, the cost of the shirt = 18 + 12 = $ 30
The problem is solved using Polya's four-step problem-solving process. By setting up two equations based on the problem, we determine that the tie costs $18 and the shirt costs $39 which is confirmed by substitution into the original problem statement.
Explanation:According to Polya's four-step problem solving strategy, we first need to understand the problem. We know that a shirt and tie cost $57 together, and the shirt costs $21 more than the tie.
Step two of Polya's strategy involves devising a plan. In this case, we can write two expressions to represent the cost of the two items: S (the cost of the shirt) and T (the cost of the tie). The total cost (S+T) equals $57. We also know that S (the cost of the shirt) is T (the cost of the tie) plus $21.
For step three, we carry out the plan. We can substitute S from the second expression into the first expression: (T + $21) + T = $57. Solving for T gives us T = $18. Hence, the tie costs $18.
Finally, for step four, we review our answer. We can plug T back into the second expression to get S = T + $21 = $18 + $21 = $39. So, the shirt costs $39, which sounds reasonable given that the shirt is supposed to be more expensive than the tie.
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Vector A⃗ has magnitude 5.00 and is at an angle of 36.9∘ south of east. Vector B⃗ has magnitude 6.40 and is at an angle of 20.0∘ west of north. Choose the positive x-direction to the east and the positive y-direction to the north. Find the components of A⃗
Answer:
(x, y) = (4, -3)
Step-by-step explanation:
Relative to straight east with angles measured CCW, the vector is ...
5∠-36.9° = 5(cos(-36.9°), sin(-36.9°)) = 5(0.8, -0.6) = (4, -3)
The components of vector A that has a magnitude of 5.00 and is at a 36.9 degrees angle south of east, are 3.96 towards east (x-direction), and -3.00 towards south (y-direction).
Explanation:The components of a vector can be calculated by using trigonometry. Where the x-component or the horizontal component can be found using the cosine of the angle and the y-component or the vertical component can be found using the sine of the angle. However, there are different coordinate systems and conventions. In this case, positive x-direction is to the east and positive y-direction is to the north, and the angle is measured from the positive x-axis (east) but the vector points towards the negative y-axis (south). So you need to apply a negative to the sine of the angle.
For the vector A⃗:
A_x = A⃗ cos(θ) = 5.00 cos(36.9) = 3.96 (east) A_y = -A⃗ sin(θ) = -5.00 sin(36.9) = -3.00 (south)
Therefore, the components of vector A⃗ are 3.96 in x-direction (east) and -3.00 in y-direction (south).
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• 10 17 (10 complete) Find an equation of the line in the form ax + byc whose x-interceptis 4 and y-intercept is 2 where a, b, and care integers with no factor common to all three and a 20 The equation of the line is IN (Type an equation)
Answer:
y+0.5x-2=0
Step-by-step explanation:
Given,
X-intercept of line = 4
So, the line intersect the x axis at (4,0)
Y-intercept of the line = 2
So, the line intersect the y axis at (0,2)
So, the slope of the line can be given by
[tex]m\ =\ \dfrac{y_2-y_1}{x_2-x_1}[/tex]
[tex]=\ \dfrac{2-0}{0-4}[/tex]
= -0.5
Hence, the equation of line can be given by
y=mx+c
where, m=slope of the line
c= y-intercept of line
So, after putting the value of m and c in above equation, the equation of line will be
y = -0.5x + 2
=> y+0.5x-2=0
So, the equation of line will be y+0.5x-2=0.
Consider strings of length 10 which contain only letters from the set {A, E, I, O, U} and digits from {1, 3, 5, 7, 9}. Suppose repitition of letters is not allowed.
How many different strings are there?
How many different strings are there if the letters, i.e. A, E, I, O, U and the digits, i.e. 1,3, 5, 7, 9 must alternate?
How many different strings are there if all five letters must be adjacent in each string?
Answer:
a) There are 3,628,800 different strings.
b) There are 28,800 different strings if the letters ad digits must alternate.
c)There are 86,400 different string if all five letters must be adjacent in each string.
Step-by-step explanation:
There are 10 digits.
Our strings have the following format:
C1 - C2 - C3 - C4 - C5 - C6 - C7 - C8 - C9 - C10
repitition of letters is not allowed.
a) How many different strings are there?
C1 can be any of the 10, C2 can be 9, C3 can be 8, ...
So in total there are:
[tex]T = 10*9*8*7*6*5*4*3*2*1 = 3,628,800[/tex]
There are 3,628,800 different strings.
b) How many different strings are there if the letters, i.e. A, E, I, O, U and the digits, i.e. 1,3, 5, 7, 9 must alternate?
There are the following possiblities:
5(l) - 5(d) - 4(l) - (4d) - ...
Or
5(d) - 5(l) - 4(d) - 4(l) - ...
So:
[tex]T = 2*(5*5*4*4*3*3*2*2*1*1) = 28,800[/tex]
There are 28,800 different strings if the letters ad digits must alternate.
c) How many different strings are there if all five letters must be adjacent in each string?
L - L - L - L - L - D - D - D - D - D
D - L - L - L - L - L - D - D - D - D
D - D - L - L - L - L - L - D - D - D
D - D - D - L - L - L - L - L - D - D
D - D - D - D - L - L - L - L - L - D
D - D - D - D - D - L - L - L - L - L
There are [tex]T = 6*(5*5*4*4*3*3*2*2*1*1) = 86,400[/tex]
There are 86,400 different string if all five letters must be adjacent in each string.
A commercial development project requires annual outlays of $65,000 for 10 years. Net cash inflows beginning in year 11 are expected to be $170,000 per year for 20 years. If the developer requires a rate of return of 16% , compute the net present value of the project.
Answer:
Net Present Value = - $99,360
Step-by-step explanation:
As provided,
Cash outlay = $65,000 each year for 10 years
Since the first outlay will be immediately, the cumulative discounting factor for cash outlay will be @ 16% = 1 for year 0 + 4.606 for 9 years = 5.606
Therefore, cumulative present value of total cash outlay = $65,000 [tex]\times[/tex] 5.606 = $364,390
Cash inflows beginning in year 11 = $170,000 for another continuous 20 years.
these cash flow will occur in the beginning of year 11 and end of year 10
Discounting factor will be [tex]\frac{1}{(1+0.16)^1^0}[/tex] = 0.2267
For, consecutive 20 years = 1.559
Therefore, value of inflows = $170,000 [tex]\times[/tex] 1.559 = $265,030
Net Present Value = Present Value of Cash Inflows - Present Value of Cash Outflows = $265,030 - $364,390 = - $99,360
U fill containers with an average of 12 ounces of oil with
astanderd deviation of .25 ounces. take a random sample of 40
cans,what is the probability that the sample mean,X is grater
then12.05.
Answer: 0.1038
Step-by-step explanation:
We assume that oil in each container is filled will normal distribution.
Given : Population mean : [tex]\mu=12[/tex]
Standard deviation: [tex]\sigma=0.25[/tex]
Sample size : [tex]n=40[/tex]
Let x be the random variable that denotes the amount of oil filled in container.
z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 12.05
[tex]z=\dfrac{12.05-12}{\dfrac{0.25}{\sqrt{40}}}=1.26491106407\approx1.26[/tex]
Now by using the standard normal table for z, we have the probability that the sample mean,X is greater then 12.05:-
[tex]P(z>1.26)=1-0.8961653=0.1038347\approx0.1038[/tex]
Hence, the probability that the sample mean,X is greater then 12.05 = 0.1038
5. Answer the question below
Answer:
144
Step-by-step explanation:
when you square a negative number, it would become positive.
4*-3 = -12^2 = 144
A ream of paper contains 500 sheets of paper. Norm has 373 sheets of paper left from a ream. Express the portion of a ream Norm has as a fraction and as a decimal.
Answer: In fraction : [tex]\dfrac{373}{500}[/tex]
In Decimal : 0.746
Step-by-step explanation:
Given : A ream of paper contains 500 sheets of paper.
Norm has 373 sheets of paper left from a ream.
Then, the fraction of ream Norm has will be :-
[tex]\dfrac{\text{Number of sheets left from ream }}{\text{Total sheets in a ream }}\\\\=\dfrac{373}{500}[/tex]
To convert in decimal we divide 373 by 500, we get 0.746.
Hence, The portion of a ream Norm has as = [tex]\dfrac{373}{500}[/tex] or 0.746