The total number of ways are:
462
Step-by-step explanation:When we are asked to select r items from a set of n items that the rule that is used to solve the problem is:
Method of combination.
Here the total number of bills of different values are: 7
i.e. n=7
( $1, $2, $5, $10, $20, $50, and $100 )
and there are atleast five of each type of bill.
Also, we have to choose 5 bills i.e. r=5
The repetition is allowed while choosing bills.
Hence, the formula is given by:
[tex]C(n+r-1,r)[/tex]
Hence, we get:
[tex]C(7+5-1,5)\\\\i.e.\\\\C(11,5)=\dfrac{11!}{5!\times (11-5)!}\\\\C(11,5)=\dfrac{11!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7\times 6!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2}\\\\\\C(11,5)=462[/tex]
Hence, the answer is:
462
There are 462 different possible ways to choose five bills from a set of seven types with repetition allowed.
To determine the number of different ways to choose five bills from a set of bills with values $1, $2, $5, $10, $20, $50, and $100, we can use the combinatorial concept known as "combinations with repetition."
Given:
- Bill values: [tex]\( \{1, 2, 5, 10, 20, 50, 100\} \)[/tex]
- Total number of types of bills: ( n = 7 )
- Number of bills to choose: ( r = 5 )
The formula for combinations with repetition (also known as "stars and bars") is:
\[
\binom{n + r - 1}{r}
\]
Substituting the values ( n = 7 ) and ( r = 5 ):
[tex]\[\binom{7 + 5 - 1}{5} = \binom{11}{5}\][/tex]
Now, calculate [tex]\( \binom{11}{5} \):[/tex]
[tex]\[\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!}\][/tex]
First, compute the factorials:
[tex]\[11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]\[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\][/tex]
Next, compute [tex]\( \frac{11!}{6!} \):[/tex]
[tex]\[\frac{11!}{6!} = 11 \times 10 \times 9 \times 8 \times 7\]\[= 11 \times 10 = 110\]\[110 \times 9 = 990\]\[990 \times 8 = 7920\]\[7920 \times 7 = 55440\][/tex]
Now, compute:
[tex]\[\frac{55440}{120} = 462\][/tex]
Therefore, the number of different possible ways to choose the five bills is:
[tex]\[\boxed{462}\][/tex]
Suppose a system has two modules, A and B, that function independently. Module A fails with probability 0.24 and Module B fails with probability 0.38, when the system is executed. When the system is executed, find the probability that.
(a) only Module A fails (b) only Module B fails. (c) both modules fail. (d) neither module fails. (e) What do the probabilities add up to?
Let's denote the events as follows:
- A: Module A fails
- B: Module B fails
Given probabilities:
[tex]\[ P(A) = 0.24 \][/tex]
[tex]\[ P(B) = 0.38 \][/tex]
We can use the addition rule of probability for mutually exclusive events to calculate the probabilities:
(a) Probability that only Module A fails:
[tex]\[ P(A \text{ only}) = P(A) = 0.24 \][/tex]
(b) Probability that only Module B fails:
[tex]\[ P(B \text{ only}) = P(B) = 0.38 \][/tex]
(c) Probability that both modules fail:
[tex]\[ P(\text{both}) = P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]
(d) Probability that neither module fails:
[tex]\[ P(\text{neither}) = 1 - (P(A) + P(B) - P(A \cap B)) \][/tex]
[tex]\[ = 1 - (0.24 + 0.38 - 0.0912) \][/tex]
[tex]\[ = 1 - 0.5288 \][/tex]
[tex]\[ = 0.4712 \][/tex]
(e) The probabilities should add up to 1, representing the total probability of all possible outcomes:
[tex]\[ P(A \text{ only}) + P(B \text{ only}) + P(\text{both}) + P(\text{neither}) = 0.24 + 0.38 + 0.0912 + 0.4712 = 1 \][/tex]
So, the probabilities add up to 1, as expected.
a. The probability that only Module A fails is 0.1488
b. The probability that only Module B fails is 0.2888
c. The probability that both modules fail is 0.0912
d. The probability that neither module fails is 0.4712
e. The probabilities add up to is 1
Let's denote the events as follows:
Event A: Module A fails
Event B: Module B fails
Given probabilities:
P(A) = 0.24
P(B) = 0.38
(a) To find the probability that only Module A fails:
[tex]\[ P(A \cap \neg B) = P(A) \times P(\neg B) \][/tex]
To calculate [tex]\( P(\neg B) \)[/tex], we subtract the probability of B from 1:
[tex]\[ P(\neg B) = 1 - P(B) = 1 - 0.38 = 0.62 \][/tex]
[tex]\[ P(A \cap \neg B) = 0.24 \times 0.62 = 0.1488 \][/tex]
(b) Similarly, to find the probability that only Module B fails:
[tex]\[ P(\neg A \cap B) = P(\neg A) \times P(B) \][/tex]
[tex]\[ P(\neg A) = 1 - P(A) = 1 - 0.24 = 0.76 \][/tex]
[tex]\[ P(\neg A \cap B) = 0.76 \times 0.38 = 0.2888 \][/tex]
(c) To find the probability that both modules fail:
[tex]\[ P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]
(d) To find the probability that neither module fails:
[tex]\[ P(\neg A \cap \neg B) = P(\neg A) \times P(\neg B) \][/tex]
[tex]\[ P(\neg A \cap \neg B) = 0.76 \times 0.62 = 0.4712 \][/tex]
(e) The probabilities add up to :
[tex]\[ P(A \cap \neg B) + P(\neg A \cap B) + P(A \cap B) + P(\neg A \cap \neg B) = 0.1488 + 0.2888 + 0.0912 + 0.4712 = 1 \][/tex]
Devise the exponential growth function that fits the given data, then answer to accompanying question. Be sure to identify the references point (t = 0) and The current population of a town is 70,000 and is growing exponentially. If the population to be 75,000 in 10 years, then what will be the populations 20 years What is the reference point (t = 0)? the initial population 70,000 the current year the population in 10 years, 75,000 What are the units of time? percent people decades years Write the exponential growth function. Round any numerical values to three decimal places as needed. y(t) = What is the population 20 years from now?
Answer:
In 20 years, the population will be about 80.3 thousand people
Step-by-step explanation:
If our first time is 0 and the population that goes along with that time is 70,000, we have a coordinate point where x is the time (0), and y is the population at that time (70). Our next time is 10 years later, when the population is 75,000. The coordinate point for that set of data is (10, 75). Now we will use those 2 points in the standard form of an exponential equation to write the model for this particular situation.
Exponential equations are of the form
[tex]y=a(b)^x[/tex]
where x and y are the coordinates from our points, one at a time; a is the initial value, and b is the growth rate. Filling in an equation with the first set of data:
[tex]70=a(b)^0[/tex]
Anything raised to the power of 0 = 1, so b to the power of 0 = 1 and we simply have that a = 70.
Now we use that value of a along with the x and y from the next coordinate pair to solve for b:
[tex]75=70(b)^{10}[/tex]
Begin by dividing both sides by 70 to get
[tex]1.071428571=b^{10}[/tex]
Undo the power of 10 on the right by taking the 10th root of both sides:
[tex](1.071428571)^{\frac{1}{10}}=(b^{10})^{\frac{1}{10}}[/tex]
On the right side we simply have b now, and on the left we have
1.006923142=b
Now we have a and b to write the model for this situation:
[tex]y=70(1.006923142)^x[/tex]
We need to find y, the population, in x = 20 years:
[tex]y=70(1.006923142)^{20}[/tex]
Raise the parenthesis to the 20th power giving you
y = 70(1.147959784) and
y = 80.3 thousand people
Final answer:
The exponential growth function that fits the given data is y(t) = a * (1 + r)^t. Using this function, we can find the population 20 years from now.
Explanation:
The exponential growth function that fits the given data is:
y(t) = a * (1 + r)^t
where:
a represents the initial population (70,000)r represents the growth rate per yeart represents the time in yearsTo find the growth rate per year, we can use the formula: r = (P/P0)^(1/t) - 1
Given that the population is projected to be 75,000 in 10 years, we can substitute these values into the formula to find the growth rate:
r = (75,000/70,000)¹/¹⁰ - 1 ≈ 0.035
The exponential growth function becomes:
y(t) = 70,000 * (1 + 0.035)^t
To find the population 20 years from now, we can substitute t = 20 into the exponential growth function:
y(20) = 70,000 * (1 + 0.035)²⁰ ≈ 95,212
The mean and the standard deviation of a sampled population are, respectively, 113.9 and 32.1. Find the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64. Round your answers to one decimal place.
Answer: The mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 is
[tex]113.9\text{ and }4.0[/tex] respectively.
Step-by-step explanation:
Given : The mean of sampled population : [tex]\mu = 113.9[/tex]
The standard deviation of sampled population : [tex]\sigma = 32.1[/tex]
We know that the mean and standard deviation of the sampling distribution of the sample mean for samples of size n is given by :_
[tex]\mu_s=\mu\\\\\sigma_s=\dfrac{\sigma}{\sqrt{n}}[/tex]
Now, the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 will be :-
[tex]\mu_s=113.9\\\\\sigma_s=\dfrac{32.1}{\sqrt{64}}=4.0125\approx4.0[/tex]
The mean of the sample is 113.9, and the standard deviation of the sample is 4.0
The given parameters are:
[tex]\mu = 113.9[/tex] --- the population mean
[tex]\sigma = 32.1[/tex] --- the population standard deviation
[tex]n = 64[/tex] --- the sample size
The sample mean is calculated as:
[tex]\bar x = \mu[/tex]
So, we have:
[tex]\bar x = 113.9[/tex]
The sample standard deviation is calculated as:
[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]
This gives
[tex]\sigma_x = \frac{32.1}{\sqrt {64}}[/tex]
[tex]\sigma_x \approx 4.0[/tex]
Hence, the mean of the sample is 113.9, and the standard deviation of the sample is 4.0
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identify the type of data that would be used to describe a response (A. Quantitative discrete B. Quantitative continuous, or C. Qualitative), and give an example of the data. (a) Number of tickets sold to a concert (b) Percent of body fat (c) Favorite baseball team (d) Time in line to buy groceries (e) Number of students enrolled at Evergreen Valley College (f) Most-watched television show (g) Brand of toothpaste (h) Distance to the closest movie theatre (i) Age of executives in Fortune 500 companies (j) Number of competing computer spreadsheet software packages.
Answer:
a) Quantitative discrete
b) Quantitative continuous
c) Qualitative
d) Quantitative continuous
e) Quantitative discrete
f) Qualitative
g) Qualitative
h) Quantitative continuous
i) Quantitative continuous
j) Quantitative discrete
Step-by-step explanation:
Okay, so an easy way to remember the difference between qualitative data and quantitative data:
Quantitative is measurable, it measures "QUANTITY."
Qualitative is categorical, meaning it has numeric value.
Now what about the difference between quantitative discrete and quantitative continuous.
Quantitative discrete there is NO IN BETWEEN. It counts things as whole numbers. The example would be your problem:
Number of tickets sold to a concert. Tickets are sold WHOLE, because you can't sell half a ticket.
Quantitative continuous has numbers in between, specifically, there could be decimals or fractions in between. Like in our problem:
Distance to the closest movie theater.
It can be 10.5m, 1.6 km. It does not necessarily have to be EXACTLY 1 Km away or 10 m away. So this is a continuous.
The type of data (Quantitative discrete, Quantitative continuous, or Qualitative) depends on whether a response involves counting, measuring, or categorizing. Examples were given to illustrate these different types of data, from counting tickets (Quantitative discrete) to measuring time (Quantitative continuous), to choosing a preferred product (Qualitative).
Explanation:The type of data used to describe a response differs depending on the nature of the response. Let's identify them:
(a) Number of tickets sold to a concert - Quantitative discrete (because we count the number of tickets)(b) Percent of body fat - Quantitative continuous (since it involves a measurement that can fall anywhere within a certain range)(c) Favorite baseball team - Qualitative (this is categorical data)(d) Time in line to buy groceries - Quantitative continuous (it can be measured precisely)(e) Number of students enrolled at Evergreen Valley College - Quantitative discrete (because students can be counted)(f) Most-watched television show - Qualitative (it's a category, not a numerical value)(g) Brand of toothpaste - Qualitative (it's categorical data)(h) Distance to the closest movie theatre - Quantitative continuous (since it's a measurable distance)(i) Age of executives in Fortune 500 companies - Quantitative continuous (age can fall anywhere within a certain range)(j) Number of competing computer spreadsheet software packages - Quantitative discrete (since we count the number of software packages).Learn more about Data Types here:https://brainly.com/question/37915128
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PLEASE GIVE AN EXPLANATION WITH YOUR ANSWER! The table below shows the change in the value of shares over the last three years. Calculate the percentage change in shares from the start of 2013 to the end of 2015.
First, lets convert them into multipliers:
The multiplier for a:
25% increase = 1.25
40% decrease = 0.6
40% increase = 1.4
Now to work out the overall percentage change, we just times all of the multipliers together, and convert it back to a percentage:
1.25 x 0.6 x 1.4 = 1.05
So the overall multiplier is 1.05
And a multiplier of 1.05 = a 5% increase.
That means that the percentage change is + 5%
_________________________________________
Answer:
The percentage change in shares from the start of 2013 to the end of 2015 is:
+ 5%
_______________________________________
Note: if you haven't been taught multipliers - then ask and I'll try my best to explain!
Solve the equation and check your answer. 0.95 t plus 0.05 left parenthesis 100 minus t right parenthesis equals 0.49 left parenthesis 100 right parenthesis
The value of t is:
t=48.8889
Step-by-step explanation:We are asked to solve the linear equation in terms of variable t.
The equation is given by:
[tex]0.95t+0.05(100-t)=0.49(100)[/tex]
Firstly we will solve the parentheses term in the left and right hand side of the equality as follows:
[tex]0.95t+0.05\times 100-0.05t=49[/tex]
Now we combine the like terms on the left side of equality by:
[tex]0.95t-0.05t+5=49\\\\i.e.\\\\0.90t+5=49[/tex]
Now we subtract both side of the equation by 5 to get:
[tex]0.90t=44[/tex]
Now on dividing both side of the equation by 0.90 we get:
[tex]t=\dfrac{44}{0.90}\\\\i.e.\\\\t=48.8889[/tex]
Hence, the value of t is: 48.8889
After simplifying and rearranging the given equation, the solution for the variable 't' comes out to be approximately 48.89. The solution was verified by substituting 't' back into the original equation.
Explanation:The equation given is: 0.95t + 0.05(100 - t) = 0.49(100). Start by simplifying the left side of the equation, which gives us: 0.95t + 5 - 0.05t = 49. Combine the t terms to get 0.9t + 5 = 49. Rearranging for t gives us 0.9t = 44. Dividing both sides by 0.9 yields t = 48.89. Checking our answer, we can substitute t back into the original equation: 0.95(48.89) + 0.05(100 - 48.89) = 49; simplifying this, we get 46.45 + 2.56 = 49, which checks out.
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Question 6 3 pts Joshua's diastolic blood pressure reading had been 80. After several weeks of medication, his reading is now 63. By what percent did he reduce his blood pressure? Round to two decimals.
Answer: There is 21.25% of reduction in blood pressure.
Step-by-step explanation:
Since we have given that
Diastolic blood pressure reading previously = 80
Diastolic blood pressure reading now = 63
Decrement is given by
[tex]80-63\\\\=17[/tex]
So, percentage he reduced his blood pressure is given by
[tex]\dfrac{17}{80}\times 100\\\\=\dfrac{1700}{80}\\\\=21.25\%[/tex]
Hence, there is 21.25% of reduction in blood pressure.
Rewrite the Product in Exponential Form : a. a. a. b. b. b. b
Answer:
a^3b^4
Step-by-step explanation:
You have 3 a's, so that would be
[tex]a^{3}[/tex]
and you have 4 b's so that would be
[tex]b^4[/tex]
so putting it together gives you
[tex]a^3b^4[/tex]
What values of c and d make the equation true?
Answer:
the equation is true only if c=6 and d=2.
Step-by-step explanation:
We have the following expression:
[tex]\sqrt[3]{162x^{c}y^{5}} = 3x^{2}y\sqrt[3]{6y^{d}}[/tex]
Elevating to the power of three:
[tex]162x^{c}y^{5}=27x^{6}y^{3}(6y^{d})[/tex]
Simplifying:
→ [tex]162x^{c}y^{5}=162x^{6}y^{3}y^{d}[/tex]
→ [tex]x^{c}y^{5}=x^{6}y^{3}y^{d}[/tex]
→ [tex]x^{c}y^{5}=x^{6}y^{d+3}[/tex]
By comparing the two expression, we can say that:
[tex]c=6[/tex]
[tex]d+3 = 5[/tex] → [tex]d=2[/tex]
Therefore, the equation is true only if c=6 and d=2.
Answer:
c = 6 d = 2
on edge
Factor the higher degree polynomial
5y^4 + 11y^2 + 2
[tex]5y^4 + 11y^2 + 2=\\5y^4+10y^2+y^2+2=\\5y^2(y^2+2)+1(y^2+2)=\\(5y^2+1)(y^2+2)[/tex]
For this case we must factor the following expression:
[tex]5y ^ 4 + 11y ^ 2 + 2[/tex]
We rewrite[tex]y^ 4[/tex] as[tex](y^ 2) ^ 2[/tex]:
[tex]5 (y ^ 2) ^ 2 + 11y ^ 2 + 2[/tex]
We make a change of variable:
[tex]u = y ^ 2[/tex]
So, we have:
[tex]5u ^ 2 + 11u + 2[/tex]
we rewrite the term of the medium as a sum of two terms whose product is 5 * 2 = 10 and whose sum is 11. Then:[tex]5u ^ 2 + (1 + 10) u + 2\\5u ^ 2 + u + 10u + 2[/tex]
We factor the highest common denominator of each group:
[tex]u (5u + 1) +2 (5u + 1)[/tex]
We factor [tex](5u + 1):[/tex]
[tex](5u + 1) (u + 2)[/tex]
Returning the change:
[tex](5y ^ 2 + 1) (y ^ 2 + 2)[/tex]
ANswer:
[tex](5y ^ 2 + 1) (y ^ 2 + 2)[/tex]
Use natural logarithms to solve the equation. e2x = 1.4 Round to the nearest thousandth.
Answer:
x = 0.16823611831
Step-by-step explanation:
ln (natural log), when multiplied by e, cancels it out.
Therefore ln*e2x = ln*1.4
2x = 0.33647223662
x = 0.16823611831
Answer: [tex]x=0.168[/tex]
Step-by-step explanation:
To solve the equation [tex]e^{2x}=1.4[/tex] you need to apply natural logarithm to both sides of the equation:
[tex]ln(e)^{2x}=ln(1.4)[/tex]
According to the logarithms property:
[tex]ln(b)^a=aln(b)[/tex]
Then, applying the property, you get:
[tex](2x)ln(e)=ln(1.4)[/tex]
You need to remember the following:
[tex]ln(e)=1[/tex]
Therefore:
[tex]2x(1)=ln(1.4)\\\\2x=ln(1.4)[/tex]
And finally, you must divide both sides of the equation by 2:
[tex]\frac{2x}{2}=\frac{ln(1.4)}{2}\\\\x=0.1682[/tex]
Rounded to the nearest thousand:
[tex]x=0.168[/tex]
e. Which of the following is NOT a possible probability? a. 25/100 b. 1.25 c. 1 d. 0
Answer:
B. 1.25
Step-by-step explanation:
Probability is as below
[tex]0 \leqslant p(a) \leqslant 1[/tex]
When P(A) = 0, it is an unlikely event
When P(A) = 1, it is a certain event
Mathematicians say that "Statement P is a sufficient condition for statement Q" if P → Q is true. In other words, in order to know that Q is true, it is sufficient to know that P is true. Let x be an integer. Give a sufficient condition on x for x/2 to be an even integer.
Answer:
If x is a multiple of 4, then x/2 is even.
Step-by-step explanation:
An integer is even, if it is equal to 2n for some integer n. We want x/2 to be even, so ...
x/2 = 2n
x = 4n . . . . . multiply by 2
That is, x will be equal to 4n for some integer n. x is a multiple of 4.
To determine a sufficient condition for x/2 to be an even integer, we need to find values of x for which x/2 results in an even integer.
To find a sufficient condition for x/2 to be an even integer, we need to determine for which values of x the expression x/2 results in an even integer. Since an even integer is divisible by 2 without a remainder, a sufficient condition for x/2 to be an even integer is that x itself is divisible by 2 without a remainder. In other words, x should be an even integer.
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As part of an annual fundraiser to help raise money for diabetes research, Diane joined a bikeathon. The track she biked on was 1,920 yards long. Diane biked 38.5 laps. Her sponsors agreed to donate an amount of money for each mile she biked. How many miles did she bike? First fill in the blanks on the left side using the ratios shown. Then write your answer.
Given Ratios: 5280ft / 1 mi , 1 mi /5280 ft , 1,920 yards / 1 lap , 1 lap / 1,920 yards , 3 ft / 1 yard , 1 yard / 3 ft.
Blanks: 38.5 laps / 1 yard x (blank) x (blank) x (blank) = (blank) miles
I'm really confused on how to do this, and the explanations aren't exactly helping. If you could walk me through how to do this, it would be greatly appreciated.
Answer:
[tex]38.5\,\text{laps}\times\dfrac{1920\,\text{yd}}{1\,\text{lap}}\times\dfrac{3\,\text{ft}}{1\,\text{yd}}\times\dfrac{1\,\text{mi}}{5280\,\text{ft}}[/tex]42 milesStep-by-step explanation:
You know that fractions with the same value in numerator and denominator reduce to 1. This is true whether the value is a number, a variable expression, or some mix of those. That is ...
[tex]\dfrac{1760}{1760}=1\\\\\dfrac{3\,\text{mi}}{1\,\text{mi}}=\dfrac{3}{1}\cdot\dfrac{\text{mi}}{\text{mi}}=3[/tex]
This example should show you that you can treat units as if they were a variable.
So, the unit conversion process is the process of choosing combinations of numerator and denominator units so that all the units you don't want cancel, leaving only units you do want.
You're starting with a number than has "laps" in the numerator. To cancel that, you need to find a conversion factor with "lap" in the denominator. On the list you are given, the one that has that is ...
[tex]\dfrac{1920\,\text{yd}}{1\,\text{lap}}[/tex]
Now, you have canceled laps, but you have yards. Also on your list of conversion factors is a ratio with yards in the denominator:
[tex]\dfrac{3\,\text{ft}}{1\,\text{yd}}[/tex]
This will cancel the yards in the numerator from the previous result, but will give you feet in the numerator. You want miles, so you look for a conversion factor between feet and miles, with miles in the numerator. The one you find is ...
[tex]\dfrac{1\,\text{mi}}{5280\,\text{ft}}[/tex]
These three conversion factors go into the blanks. When you form the product, you will get ...
[tex]\dfrac{38.5\cdot 1920\cdot 3}{1\cdot 1\cdot 5280}\cdot\dfrac{\text{laps$\cdot$yd$\cdot$ft$\cdot$mi}}{\text{lap$\cdot$yd$\cdot$ft}}=42\,\text{mi}[/tex]
Answer:
Step-by-step explanation:
Expand the logarithm. log 5x/4y
Answer:
[tex] log ( \frac { 5 x } { 4 y} ) \implies [/tex] [tex] log ( 5 ) + log ( x ) - log ( 4 ) + log ( y ) [/tex]
Step-by-step explanation:
We are given the following for which we are to expand the logarithm:
[tex] log ( \frac { 5 x } { 4 y} ) [/tex]
Expanding the log by applying the rules of expanding the logarithms by changing the division into subtraction:
[tex] log ( 5 x ) - log ( 4 y ) [/tex]
[tex] log ( 5 ) + log ( x ) - log ( 4 ) + log ( y ) [/tex]
[tex]\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log\left( \cfrac{5x}{4y} \right)\implies \log(5x)-\log(4y)\implies [\log(5)+\log(x)]-[\log(4)+\log(y)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \log(5)+\log(x)-\log(4)-\log(y)~\hfill[/tex]
Find the power set of each of these sets, where a and b are distinct elements. a) {a} b) {a, b} c) {1, 2, 3, 4} show steps
Answer:
a) {{}, {a}}.
b) {{}, {a}, {b}, {a, b}}.
c) {{}, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, {4}, {1, 4}, {2, 4}, {1, 2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}.
Step-by-step explanation:
The power set of a set is the set of all subset of the set in question. The number of power sets (including the empty set) of a set with [tex]n[/tex] (where [tex]n \in \mathbb{Z}[/tex]) unique elements is [tex]2^{n}[/tex].
In other words, there shall be
2 items in the power set of a), 4 items in the power set of b), and16 items in the power set of c).This explanation shows how to find the power set using binary numbers (only 0 and 1.) (Credit: Mathsisfun.)
a)List all the binary numbers that are equivalent to decimals ranging from 0 to [tex]2 - 1 = 1[/tex].
[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0 \\ 1 & 1\end{array}[/tex].
Reverse the original set. Each digit in the binary number corresponds to a member of the original set (i.e. a letter in a) and b) or a number in c).) 0 means that the element is absent in the subset and 1 means that the element is present.
[tex]\begin{array}{c|l}a & \text{Element of the Power Set}\\ 0 & \{\}\\ 1 & \{a\}\end{array}[/tex].
The power set of a) thus contains:
{} and{a}.b)Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]4 - 1 = 3[/tex].
[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 00 \\ 1 & 01 \\ 2 & 10 \\ 3 & 11\end{array}[/tex].
[tex]\begin{array}{cc|l}b & a & \text{Element of the Power Set}\\ 0 & 0 & \{\}\\ 0 & 1 & \{a\}\\ 1 & 0 & \{b\} \\ 1 & 1 & \{a, b\}\end{array}[/tex].
The power set of b) thus contains:
{},{a}, {b}, and{a, b}.c)Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]16 - 1 = 15[/tex].
[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0000 \\ 1 & 0001 \\ 2 & 0010 \\ 3 & 0011\\4 & 0100 \\ 5 & 0101\\ 6 & 0110 \\ 7 & 0111\\ 8 & 1000\\ 9 & 1001\\ 10 & 1010\\ 11 & 1011\\ 12& 1100 \\13 & 1101 \\ 14 & 1110\\ 15 & 1111 \end{array}[/tex].
[tex]\begin{array}{cccc|l}4 & 3 & 2 &1& \text{Element of the Power Set}\\ 0 & 0 & 0 & 0 &\{\}\\ 0 & 0 & 0 & 1 & \{1\}\\ 0 & 0 & 1 & 0 & \{2\} \\ 0 & 0 &1 & 1 & \{1, 2\} \\ 0 & 1 & 0 & 0 & \{3\} \\ 0 & 1 & 0 & 1& \{1, 3\}\\ 0 & 1 & 1 & 0& \{2, 3\}\\ 0 & 1 & 1 & 1 & \{1, 2, 3\} \\ 1 & 0 & 0 & 0 & \{4\} \\ 1 & 0 & 0 & 1 & \{1, 4\}\\ 1& 0 & 1 &0&\{2, 4\}\\ 1 & 0 & 1 & 1 &\{1, 2, 4\}\\ 1 & 1 & 0 & 0 & \{3, 4\} \\ 1 & 1 & 0 & 1 & \{1, 3, 4\} \\ 1 & 1 & 1 & 0 & \{2, 3, 4\} \\ 1 & 1 & 1 & 1 & \{1, 2, 3, 4\}\end{array}[/tex].
The power set of c) thus contains:
{},{1}, {2}, {1, 2},{3},{1, 3},{2, 3},{1, 2, 3}{4},{1, 4},{2, 4},{1, 2, 4},{3, 4}, {1, 3, 4},{2, 3, 4}, and{1, 2, 3, 4}.Use the ratio test to determine whether ∑n=14∞n+2n! converges or diverges. (a) Find the ratio of successive terms. Write your answer as a fully simplified fraction. For n≥14, limn→∞∣∣∣an+1an∣∣∣=limn→∞.
Answer:
The sum [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex] diverges ∵ of the Ratio Test.
General Formulas and Concepts:
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]Special Limit Rule [Coefficient Power Method]: [tex]\displaystyle \lim_{x \to \pm \infty} \frac{ax^n}{bx^n} = \frac{a}{b}[/tex]Series Convergence Tests
Ratio Test: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{a_{n + 1}}{a_n} \bigg|[/tex]Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex]
Step 2: Find Convergence
[Series] Define: [tex]\displaystyle a_n = n + 2n![/tex][Series] Set up [Ratio Test]: [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n! \rightarrow \lim_{n \to \infty} \bigg| \frac{n + 1 + 2(n + 1)!}{n + 2n!} \bigg|[/tex][Ratio Test] Evaluate Limit [Coefficient Power Method]: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{n + 1 + 2(n + 1)!}{n + 2n!} \bigg| = \infty[/tex][Ratio Test] Define conclusiveness: [tex]\displaystyle \infty > 1[/tex]Since infinity is greater than 1, the Ratio Test defines the sum [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex] to be divergent.
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Topic: AP Calculus BC (Calculus I + II)
Unit: Taylor Series
The 8 rowers in a racing boat stroke so that all of the angles formed by their oars with the side of the boat stay equal. Explain why the oars on either side of the boat remain parallel
Explanation:
Assuming the side of the boat is a straight line, it would constitute a transversal crossing the lines of the oars. When corresponding angles at a transversal are congruent, the lines being crossed are parallel. Since the oars are those lines, the oars are parallel.
Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 32feet above the ground, the function h(t)=−16t2+48t+32 models the height, h, of the ball above the ground as a function of time, t. At what time will the ball reach a height of 64feet?
Answer:
There are two times for the ball to reach a height of 64 feet:
1 second after thrown ⇒ the ball moves upward
2 seconds after thrown ⇒ the ball moves downward
Step-by-step explanation:
* Lets explain the function to solve the problem
- h(t) models the height of the ball above the ground as a function
of the time t
- h(t) = -16t² + 48t + 32
- Where h(t) is the height of the ball from the ground after t seconds
- The ball is thrown upward with initial velocity 48 feet/second
- The ball is thrown from height 32 feet above the ground
- The acceleration of the gravity is -32 feet/sec²
- To find the time when the height of the ball is above the ground
by 64 feet substitute h by 64
∵ h(t) = -16t² + 48t + 32
∵ h = 64
∴ 64 = -16t² + 48t + 32 ⇒ subtract 64 from both sides
∴ 0 = -16t² + 48t - 32 ⇒ multiply the both sides by -1
∴ 16t² - 48t + 32 = 0 ⇒ divide both sides by 16 because all terms have
16 as a common factor
∴ t² - 3t + 2 = 0 ⇒ factorize it
∴ (t - 2)(t - 1) = 0
- Equate each bracket by zero to find t
∴ t - 2 = 0 ⇒ add 2 to both sides
∴ t = 2
- OR
∴ t - 1 = 0 ⇒ add 1 to both sides
∴ t = 1
- That means the ball will be at height 64 feet after 1 second when it
moves up and again at height 64 feet after 2 seconds when it
moves down
* There are two times for the ball to reach a height of 64 feet
1 second after thrown ⇒ the ball moves upward
2 seconds after thrown ⇒ the ball moves downward
The police chief wants to know if the city’s African Americans feel that the police are doing a good job. Identify the management problem (I.e dependent variable) and identify the independent variable.
What is 3 root 17 in a decimal
Answer:
=16.492
Step-by-step explanation:
3√17 is a surd that can be broken into 3 × √17
√17 = 4.123 ( to 4 S.F )
3 × 4.123 = 16.492
The decimal form is an estimation as it has many digits that can only be rounded off or truncated.
Solve the inequality and complete a line graph representing the solution. In a minimum of two sentences, describe the solution and the line graph.
8 3x + 5
Answer:
The solution is [tex]x\leq 1[/tex]
All real numbers less than or equal to 1
The graph in the attached figure
Step-by-step explanation:
we have
[tex]8\geq 3x+5[/tex]
Subtract 5 both sides
[tex]8-5\geq 3x[/tex]
[tex]3\geq 3x[/tex]
Divide by 3 both sides
[tex]1\geq x[/tex]
Rewrite
[tex]x\leq 1[/tex]
The solution is the interval ------> (-∞,1]
All real numbers less than or equal to 1
In a number line the solution is the shaded area at left of x=1 (close circle)
see the attached figure
The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected.What is the probability that the sample mean will be larger than 1224? Round your answer to three decimal places.
Answer: 0.008
Step-by-step explanation:
Given: Mean : [tex]\mu=1200[/tex]
Standard deviation : [tex]\sigma = 60[/tex]
Sample size : [tex]n=36[/tex]
The formula to calculate z-score is given by :_
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 1224, we have
[tex]z=\dfrac{1224-1200}{\dfrac{60}{\sqrt{36}}}=2.4[/tex]
The P-value = [tex]P(z>2.4)=1-P(z<2.4)=1-0.9918024=0.0081976\approx0.008[/tex]
Hence, the probability that the sample mean will be larger than 1224 =0.008
A card is selected at random from a standard deck of playing cards.
Compute the probability that the card is a 7. (Enter the probability as a fraction.)
[tex]|\Omega|=52\\|A|=4\\\\P(A)=\dfrac{4}{52}=\dfrac{1}{13}[/tex]
The probability of getting the card of 7 in a deck will be 1 / 13.
What is probability?Probability is defined as the ratio of the number of favourable outcomes to the total number of outcomes in other words the probability is the number that shows the happening of the event.
Probability = Number of favourable outcomes / Number of sample
Given that a card is selected at random from a standard deck of playing cards. The probability of getting a number 7 will be,
There are four 7 cards in the deck of 52 cards.
Number of favourable outcomes = 4
Number of sample = 52
The probability is,
P = 4 / 52
P = 1 / 13
Therefore, the probability of getting the card of 7 in a deck will be 1 / 13.
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How do I solve for the minimum and maximum of the function y=-1/2x^2 -5x+2
Try this solution:
There are several ways to find the max or min of the given function:
1. to use derivative of the function. For more details see the attachment (3 basic steps); the coordinates of max-point are marked with green (-5; 14.5)
2. to use formulas. The given function is the standart function with common equation y=ax²+bx+c, it means the correspond formulas are (where a<0, the vertex of this function is its maximum):
[tex]X_0=-\frac{b}{2a} ; \ X_0=-\frac{-5}{2*(-\frac{1}{2})} =-5.[/tex]
[tex]Y_0=-\frac{D}{4a}; \ Y_0=-\frac{25+4*2*0.5}{4*(-\frac{1}{2})} =14.5[/tex]
Finally: point (-5;14.5) - maximum of the given function.
3. to draw a graph.
Analyze the diagram below and answer the question that follows.
Answer:
The right statement is sin(J) = cos(L) ⇒ answer D
Step-by-step explanation:
* Lets describe the figure
- LKJ is a right triangle, where K is a right angle
∵ m∠K = 90°
∵ LJ is opposite to angle K
∴ LJ is the hypotenuse
∵ LJ = 219
∵ KJ = 178
- By using Pythagoras Theorem
∵ (LJ)² = (LK)² + (KJ)²
∴ (219)² = (LK)² + (178)² ⇒ subtract (178)² from both sides
∴ (LK)² = (219)² - (178)²
∴ (LK)² = 16277
∴ LK = √16277 = 127.58
* Lets revise how to find the trigonometry function
# sin Ф = opposite/hypotenuse
# cos Ф = adjacent/hypotenuse
# tan Ф = opposite/adjacent
∵ LK is the opposite side to angle J
∵ LJ is the hypotenuse
∵ sin(J) = LK/LJ
∵ LK = 127.58 , LJ = 219
∴ sin(J) = 127.58/219 = 0.583
∵ LK is the adjacent side to angle L
∵ LJ is the hypotenuse
∵ cos(L) = LK/LJ
∵ LK = 127.58 , LJ = 219
∴ cos(L) = 127.58/219 = 0.583
∴ sin(J) = cos(L)
* The right statement is sin(J) = cos(L)
A hemispherical tank of radius 2 feet is positioned so that its base is circular. How much work is required to fill the tank with water through a hole in the base when the water source is at the base? (The weight-density of water is 62.4 pounds per cubic foot.)
To fill the hemispherical tank with water through a hole in the base, the work required is 418.88 foot-pounds.
Explanation:To calculate the work required to fill the hemispherical tank with water through a hole in the base, we can use the concept of work done against gravity.
The volume of the tank can be calculated using the formula for the volume of a hemisphere, which is (2/3)πr^3. In this case, the radius is given as 2 feet.
The weight of the water can be found by multiplying the volume by the weight-density of water, which is 62.4 pounds per cubic foot.
The work done is then the weight of the water multiplied by the height it is lifted, which is equal to the radius of the hemisphere.
So, the work required to fill the tank with water is (2/3)π(2^3)(62.4)(2) = 418.88 foot-pounds.
The work required to fill the hemispherical tank with water through a hole in the base is [tex]$\frac{1248\pi}{5}$[/tex] foot-pounds.
To solve this problem, we need to calculate the work done against gravity to fill the tank with water. The work done to lift a small layer of water at a height [tex]$h$[/tex] is given by the force required to lift it times the height [tex]$h$[/tex]. The force is the weight of the water, which is the volume of the water times its weight-density.
Let's break down the tank into thin horizontal slices. Each slice has a thickness [tex]$dh$[/tex] and is at a height [tex]$h$[/tex] from the base. The radius of each slice is given by [tex]$r = \sqrt{2^2 - h^2}$[/tex], where [tex]$2$[/tex] feet is the radius of the tank.
The volume of each slice is the area of the slice times its thickness [tex]$dh$[/tex]. The area of the slice is [tex]$\pi r^2$[/tex], so the volume [tex]$dV$[/tex] is [tex]\pi (2^2 - h^2) dh[/tex].
The weight of the water in each slice is the volume times the weight-density of water, which is [tex]$62.4$[/tex] pounds per cubic foot. Therefore, the weight of the water in each slice is [tex]$62.4\pi (2^2 - h^2) dh$[/tex].
The work done to lift this slice to the height [tex]$h$[/tex] is the weight of the water times the height [tex]$h$[/tex], which is [tex]$62.4\pi h (2^2 - h^2) dh$[/tex].
To find the total work done to fill the tank, we integrate this expression from [tex]$h = 0$[/tex] to[tex]$h = 2$[/tex] feet:
[tex]\[ W = \int_{0}^{2} 62.4\pi h (2^2 - h^2) dh \] \[ W = 62.4\pi \int_{0}^{2} h (4 - h^2) dh \] \[ W = 62.4\pi \int_{0}^{2} (4h - h^3) dh \] \[ W = 62.4\pi \left[ 2h^2 - \frac{h^4}{4} \right]_{0}^{2} \] \[ W = 62.4\pi \left[ 2(2)^2 - \frac{(2)^4}{4} \right] - 62.4\pi \left[ 2(0)^2 - \frac{(0)^4}{4} \right] \] \[ W = 62.4\pi \left[ 8 - 4 \right] \] \[ W = 62.4\pi \tims 4 \] \[ W = 249.6\pi \] \[ W = \frac{1248\pi}{5} \][/tex]
Therefore, the work required to fill the hemispherical tank with water through a hole in the base is [tex]$\frac{1248\pi}{5}$[/tex] foot-pounds.
Their perimeter of a triangle is 82' one side of the triangle is 2 times the 2nd side the 3rd side is 2' longer than the 2nd side find the length of each side
Answer:
side 1: 40; side 2: 20; side 3: 22
Step-by-step explanation:
It appears that sides 1 and 3 are based on side 2. So side 2 is our "control" and we'll call it x.
If side 1 is 2 times side 2, then
side 1 = 2x.
If side 3 is 2 feet longer than side 2, then
side 3 = x + 2
We are given the perimeter value. The perimeter of any shape is a measure around the outside of the shape. Since a triangle has 3 sides, we add them together and set their sum equal to the perimeter we were given. That looks like this:
x + 2x + x + 2 = 82
Combine like terms to get 4x + 2 = 82
Solve for x to get 4x = 80 and x = 20. That means that side 2 measures 20 feet; side 1 measures twice that, so side 1 = 40 feet, and side 3 measures 2 more than side 2, so side 3 measures 22. Let's add them all together and make sure the numbers work:
20 + 40 + 22 = 82
It works! So we're all done!
For a standard normal distribution (µ=0, σ=1), the area under the curve less than 1.25 is 0.894. What is the approximate percentage of the area under the curve less than -1.25?
Answer:
10.6%
Step-by-step explanation:
Normal curves are symmetrical. That means that on a standard normal distribution, the area less than -1.25 is the same as the area greater than +1.25. The total area under the curve is 1, so:
P = 1 - 0.894
P = 0.106
Approximately 10.6% of the area under the curve lies below -1.25.
For a standard normal distribution, the area under the curve to the left of a z-score of -1.25 is approximately 10.6% due to the fact that a normal distribution is symmetric around its mean.
Explanation:In a standard normal distribution, the properties of symmetry mean that the area on either side of the mean (µ=0) is identical. When looking at positive and negative z-scores that are the same magnitude but opposite in direction, the areas under the curve to their respective sides will be equivalent.
The given z-score is 1.25 and we know the area under the curve to the left of this z-score is 0.894 or 89.4%. Because of symmetry, the z-score of -1.25 will have an equal area under the curve to the right, which also represents 89.4%. Therefore, the area under the curve to the left of a z-score of -1.25 is 1 - 0.894 = 0.106, or approximately 10.6%.
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A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 1000 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?
Answer:
The largest area is 125000 m²
The dimensions of the farmland are 250 m and 500 m
Step-by-step explanation:
* Lets pick the information from the problem
- The farmland is shaped a rectangle
- The farmland will be bounded on one side by a river
- The other three sides are bounded by a single-strand electric fence
- The length of wire is 1000 m
- Lets consider the width of the rectangle is x and the length is y
- The side which will be bounded by the river is y
∴ The perimeter of the farmland which will be bounded by the electric
fence = x + x + y = 2x + y
- We will use the wire to fence the farmland
∵ The length of the wire is 1000 m
∵ The perimeter of the farmland is equal to the length of the wire
∴ 2x + y = 1000
- Lets find y in term of x
∵ 2x + y = 1000 ⇒ subtract 2x from both sides
∴ y = 1000 - 2x
- Now lets find the area can enclose by the wire
∵ The area of the rectangle = length × width
∵ The width of the farmland is x and its length is y
∴ The area of the farmland (A) = x × y = xy ⇒ (2)
- Use equation (1) to substitute the value of y in equation (2)
∴ A = x (1000 - 2x) ⇒ simplify
∴ A = 1000 x - 2 x²
- To find the maximum area we will differentiate A with respect to x
and equate the answer by zero to find the value of x which will make
the enclosed area largest
* Lets revise the rule of differentiation
- If y = ax^n, then dy/dx = a(n) x^(n-1)
- If y = ax, then dy/dx = a
- If y = a, then dy/dx = 0 , where a is a constant
∵ A = 1000 x - 2 x² ⇒ (3)
- Differentiate A with respect to x using the rules above
∴ dA/dx = 1000 - 2(2) x^(2-1)
∴ dA/dx = 1000 - 4x
- Put dA/dx = 0 to find the value of x
∵ 1000 - 4x = 0 ⇒ add 4x to both sides
∴ 1000 = 4x ⇒ divide both sides by 4
∴ 250 = x
∴ The value of x is 250
- Lets substitute this value in equation 3 to find the largest area
∵ A = 1000 x - 2 x²
∴ A = 1000 (250) - 2(250)² = 125000 m²
* The largest area is 125000 m²
∵ The width of the farmland is x
∵ x = 250
∴ The width of the farmland = 250 m
- Substitute the value of x in the equation (1) to find y
∵ y = 1000 - 2x
∵ x = 250
∴ y = 1000 - 2(250) = 1000 - 500 = 500
∵ The length of the farm lend is y
∴ The length of the farm land = 500 m
* The dimensions of the farmland are 250 m and 500 m
To get the maximum area from a rectangular farmland bounded on one side by a river and on the other three sides by a single-strand electric fence with 1000 meters of wire, the dimensions should be 250m x 500m yielding a maximum area of 125,000 square meters.
Explanation:This problem can be approached as a classic calculus maximization problem. The scenario mentioned in your question is about maximizing the area of a rectangle with a constant perimeter, this occurs when the rectangle is a square.
The three sides of your plot will consume the 1000 m of wire, if the length of each adjacent sides are x and y (where x is the length of the fence along the river, and y is the length of the other two fences), then
y + 2x = 1000
Substituting y from this equation into the area equation, A = xy(which results in A = x(1000-2x)), you can differentiate this equation with regards to x and set the derivative equal to zero to find the x-value that will give the maximum area. This happens when x = 250m, y = 500m then the maximum area is 125,000 m².
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