Of 13 possible books, you plan to take 6 with you on vacation. How many different collections of 6 books can you take?
Taking into account the definition of combination, you can take 1716 different collections of 6 books.
CombinationCombinations of m elements taken from n to n (m≥n) are called all the possible groupings that can be made with the m elements in which the order in which the elements are chosen is not taken into account and repetition is not possible.
The combination is calculated by:
[tex]C=\frac{m!}{n!(m-n)!}[/tex]
The term "n!" is called the "factorial of n" and is the multiplication of all numbers from "n" to 1.
Collections of 6 books that you can takeIn this case it is possible to apply a combination, not all the elements enter, the order in which the books are chosen does not matter, and they are not repeated.
Being m= 13 and n=6, the combination is calculated by:
[tex]C=\frac{13!}{6!(13-6)!}[/tex]
Solving:
[tex]C=\frac{13!}{6!7!}[/tex]
C= 1716
Finally, you can take 1716 different collections of 6 books.
Learn more about combination:
https://brainly.com/question/25821700
Which of the following are solutions to the equation below? (4x - 1)2 = 11
Answer:
X = - sqrt 12 / 4
X = sqrt 11 + 1 / 4
Write an appropriate inverse variation equation if y = 9 when x = 3.
110 students are surveyed about their pets. The results are shown in the table. Which statement is true?
The only true statement is b. 40% of the boys surveyed have at least one pet.
To determine which statement is true, let's analyze the data provided in the table:
- Total number of boys surveyed: 45
- Total number of girls surveyed: 65
Now, let's break down the information based on the provided table:
1. **At least one pet:**
- Boys: 18
- Girls: 39
- Total: 57
2. **No pets:**
- Boys: 27
- Girls: 26
- Total: 53
Now, let's check each statement:
a. 27% of the boys surveyed have no pets.
- Percentage of boys with no pets = (Number of boys with no pets / Total number of boys surveyed) * 100%
- = (27 / 45) * 100% ≈ 60%
- This statement is false.
b. 40% of the boys surveyed have at least one pet.
- Percentage of boys with at least one pet = (Number of boys with at least one pet / Total number of boys surveyed) * 100%
- = (18 / 45) * 100% = 40%
- This statement is true.
c. 49% of the girls surveyed have no pets.
- Percentage of girls with no pets = (Number of girls with no pets / Total number of girls surveyed) * 100%
- = (26 / 65) * 100% ≈ 40%
- This statement is false.
d. 57% of the students surveyed have at least one pet.
- Percentage of students with at least one pet = (Total number of students with at least one pet / Total number of students surveyed) * 100%
- = (57 / 110) * 100% ≈ 52%
- This statement is false.
So, the only true statement is b. 40% of the boys surveyed have at least one pet.
The probable question may be:
110 students are surveyed about their pets. The results are shown in the table. Which statement is true?
Boys | Girls | Total
At least one pet | 18 | 39 | 57
No pets | 27 | 26 | 53
Total | 45 | 65 | 110
a. 27% of the boys surveyed have no pets.
b. 40% of the boys surveyed have at least one pet.
c. 49% of the girls surveyed have no pets.
d. 57% of the students surveyed have at least one pet.
57% of men consider themselves professional baseball fans. you randomly select 10 men and ask each if he considers himself a professional baseball fan. find the probability that the number who consider themselves baseball fans is (a) exactly five, (b) at least six, and (c) less than four.
(a) [tex]\( P(X = 5) \approx 0.234 \)[/tex]
(b) [tex]\[ P(X \geq 6) \approx 0.892 \][/tex]
(c) [tex]\[ P(X < 4) \approx 0.020 \][/tex]
To solve this problem, we can use the binomial probability formula since each man's response (considering themselves a baseball fan or not) is independent and there are only two possible outcomes (success or failure).
Given:
- Probability of success (considering themselves a baseball fan) [tex]\( p = 0.57 \)[/tex]
- Probability of failure (not considering themselves a baseball fan) [tex]\( q = 1 - p = 1 - 0.57 = 0.43 \)[/tex]
- Number of trials [tex]\( n = 10 \)[/tex]
We'll calculate the probabilities for each case:
(a) To find the probability that exactly five men consider themselves baseball fans:
[tex]\[ P(X = 5) = \binom{10}{5} \times (0.57)^5 \times (0.43)^{10 - 5} \][/tex]
(b) To find the probability that at least six men consider themselves baseball fans, we can find the probability of six, seven, eight, nine, and ten men being baseball fans, and then sum them up.
(c) To find the probability that less than four men consider themselves baseball fans, we need to find the probabilities of zero, one, two, and three men being baseball fans, and then sum them up.
Let's calculate each probability:
(a) [tex]\[ P(X = 5) = \binom{10}{5} \times (0.57)^5 \times (0.43)^{5} \][/tex]
(b) To find the probability of at least six men being baseball fans:
[tex]\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \][/tex]
(c) To find the probability of less than four men being baseball fans:
[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
We'll use these formulas to find the probabilities for each case. Let me do the calculations.
(a) To find the probability that exactly five men consider themselves baseball fans:
[tex]\[ P(X = 5) = \binom{10}{5} \times (0.57)^5 \times (0.43)^{5} \][/tex]
Using the binomial coefficient formula [tex]\(\binom{n}{k} = \frac{n!}{k!(n - k)!}\)[/tex], where [tex]\(n = 10\)[/tex] and [tex]\(k = 5\)[/tex]:
[tex]\[ \binom{10}{5} = \frac{10!}{5!(10 - 5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \][/tex]
Now, we plug in the values:
[tex]\[ P(X = 5) = 252 \times (0.57)^5 \times (0.43)^{5} \][/tex]
[tex]\[ P(X = 5) \approx 0.234 \][/tex]
(b) To find the probability of at least six men being baseball fans:
[tex]\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \][/tex]
For each [tex]\(k = 6, 7, 8, 9, 10\)[/tex], we calculate [tex]\(P(X = k)\)[/tex] using the binomial formula and sum them up.
(c) To find the probability of less than four men being baseball fans:
[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
Similarly, for each [tex]\(k = 0, 1, 2, 3\)[/tex], we calculate [tex]\(P(X = k)\)[/tex] using the binomial formula and sum them up. Let me do the calculations.
(a) [tex]\( P(X = 5) \approx 0.234 \)[/tex]
(b) To find the probability of at least six men being baseball fans:
[tex]\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \][/tex]
Using the binomial formula for each [tex]\( k = 6, 7, 8, 9, 10 \)[/tex] and summing the probabilities:
[tex]\[ P(X \geq 6) \approx 0.892 \][/tex]
(c) To find the probability of less than four men being baseball fans:
[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
Using the binomial formula for each [tex]\( k = 0, 1, 2, 3 \)[/tex] and summing the probabilities:
[tex]\[ P(X < 4) \approx 0.020 \][/tex]
Kim uses decals to decorate 5 cars and 2 motorbikes. She uses 2/3 of the decals on the cars and 2/5 of the remaining on the motorbikes. She has 6 decals left. How many decals does Kim use on each car?
Average maintenance costs are $1.50 per machine-hour at an activity level of 8,000 machine-hours and $1.20 per machine-hour at an activity level of 13,000 machine-hours. assuming that this activity is within the relevant range, total expected maintenance cost for a budgeted activity level of 10,000 machine-hours would be closest to:
Explain how the phrase "oh heck another hour of algebra" can help a student recall the trigonometric ratios
The phrase "oh heck another hour of algebra" can help a student recall the trigonometric ratios by associating key words in the phrase with math concepts. The phrase contains words related to math, such as "algebra" and "hour," as well as a word similar to "angle." By linking these words to the trigonometric ratios, a student can better remember and understand them.
Explanation:The phrase "oh heck another hour of algebra" can help a student recall the trigonometric ratios by focusing on the key words within the phrase. The phrase contains the words "algebra" and "hour," which are related to math, and the word "heck," which is similar to the word "angle." By associating these words with the phrase, a student can remember the trigonometric ratios, which involve angles and algebraic calculations. For example, the phrase can remind a student that the sine ratio involves the ratio of opposite and hypotenuse sides, similar to finding lengths in algebraic equations.
Learn more about Recalling trigonometric ratios here:https://brainly.com/question/25122832
#SPJ3
In how many ways can 5 starting positions on a basketball team be filled with 8 men who can play any of the positions?
8 men, 5 positions
so multiply 8*7*6 =336
divide by 3*2 =6
336/6 =56
56 different ways
how do you write an interger whose absolute value is greater than itself.
Jack and Andrea want to create a right triangle together using values of x and y and the polynomial identity to generate Pythagorean triples. If Andrea picks a value of x = 2, and the hypotenuse of the resulting right triangle is 5, what natural number value of y did Jack pick? y = 1 y = 2 y = 3 y = 4
A right triangle can be considered as a special type because the relationship of its sides can be described using the hypotenuse formula:
c^2 = a^2 + b^2
or
c^2 = x^2 + y^2
where,
c is the hypotenuse of the triangle and is the side opposite to the 90° angle
while a and b are the sides adjacent to the 90° angle
In the problem statement, we are given that one of the side has a measure of 2 = x, while the hypotenuse is 5 = c, therefore calculating for y:
y^2 = c^2 – x^2
y^2 = 5^2 – 2^2
y^2 = 21
y = 4.58
The natural number is the number before the decimal. Therefore the answer is:
y = 4
Answer:
it is now y=4, i swear!! I put this, and it was wrong!!!
Step-by-step explanation:
IHELP ME OUT WITH THIS MATH QUESTION
f DE is a mid segment of the triangle, then the measure of AC:
7.5
15.
30.
None of the choices are correct.
The population of a town is modeled by the equation P = 16,581e0.02t where P represents the population t years after 2000. According to the model, what will the population of the town be in 2020?
2020 is 20 years after 2000, so put 20 where t is in the equation and evaluate.
P = 16,581e^(0.02·20) = 16,581e^0.4
P ≈ 24,736
The model predicts the population in 2020 will be about 24,736.
what is the 31st term of the sequence below?
-108, -90, -72, -54, ...
Answer:
432Step-by-step explanation:
The form for finding the nth term of an arithmetic sequence is a(n) = b +(n-1)d, where a is the function of the nth term, b is the first number in the sequence, n is the nth term, and d is the common difference. Plugging in what we know we get:
a(31) = -108 + 30 x 18
multiply
a(31) = -108 + 540
add
a(31) = 432
PLEASE MARK BRAINLIEST
In a literal question what does f and c represent
Given p(a)=0.40, p(b)=0.50. if a and b are independent, what is the value of p(a intersection b)?
Final answer:
The value of P(A intersection B) is 0.20.
Explanation:
Given that events A and B are independent, the probability of their intersection can be found by multiplying their individual probabilities. So, P(A ∩ B) = P(A) * P(B). From the given information, P(A) = 0.40 and P(B) = 0.50. Therefore, P(A ∩ B) = 0.40 * 0.50 = 0.20.
A single fair die is tossed. Find the probability of rolling a number greater than 4.
The probability of rolling a number greater than 4 with a single fair die is 1/3 or approximately 0.33. This is calculated as the number of desired outcomes (a roll of 5 or 6) divided by the total possible outcomes (1, 2, 3, 4, 5, or 6). The actual results of a short series of rolls may vary, but over time the results should average out to this probability.
Explanation:The question is asking to calculate the probability of rolling a die and getting a number greater than 4. In a fair six-sided die, the possible outcomes of a single toss are 1, 2, 3, 4, 5, or 6, making the total possible outcomes 6. The outcomes greater than 4 are 5 and 6, which gives us 2 desired outcomes.
Remember, probability is calculated as the number of desired outcomes over the number of total possible outcomes. So in this case, the probability is 2 (the number of outcomes greater than 4) divided by 6 (the total number of outcomes when rolling a die) which gives a probability of 1/3 or approximately 0.33.
This suggests that if you throw the single fair die repeatedly over time, about one third of the results would show a number greater than 4. It's essential to remember this is a theoretical probability, and actual short-term results may vary, but they should correspond to this probability in the long run per the law of large numbers.
Learn more about Probability here:https://brainly.com/question/22962752
#SPJ3
What is the answer to this problme
Which set or sets does the number 15 belong to?
what is the unit rate
How do i Divide z4 by z-3
A box contains 5 plain pencils and 7 pens. A second box contains 3 color pencils and 3 crayons. One item from each box is chosen at random. What is the probability that a plain pencil from the first box and a color pencil from the second box are selected?
Which of the following are true statements about any regular polygon? Check all that apply.
A. All of its angles measure 90.
B. It is a quadrilateral.
C. It is a closed figure.
D. It is a hexagon.
E. All of its angles have equal measures.
F. Its sides are congruent line segments.
Answer:
Statement C, E and F are true.
Step-by-step explanation:
We have to find the true statements about a regular polygon.
Regular Polygon
A regular polygon is a polygon that is equiangular that is all angles are equal in measure and equilateral that is all sides have the same length.A) False
This is not a necessity that all angles measure 90 degrees.
B) False
It may or may not be a a quadrilateral.
C) True
All regular polygons are closed figure.
D) False
All regular polygons are not hexagon but hexagon is a polynomial.
E) True
All the angles of a regular polygon are equal.
F) True
Since all the sides of a regular polygon are equal, thus, its sides are congruent line segments.
3. A carpenter is framing a window with wood trim where the length of the window is 6 and 2\3 feet. If the width of the window is 7 and3\4 feet, how many feet of the wood is needed to frame the window?
,,,Help Please.. Question in the file.
a cell phone company offers two different monthly plans.plan a charges $41 for unlimited cell phone minutes plus $0.10 per text message.Plan b charges $31 for unlimited cell phone minutes plus $0.15 per text message.How many text messages must a customer send in order for the cost of plan a to be equal to cost of plan b?
41+0.10x = 31+0.15x
10 +0.10x=0.15x
10=0.05x
x=10/.05
x= 200 text messages
check 200*0.10 = 20 +41 =61
200*0.15 = 30+31 = 61
they equal each other so number of texts is 200
Let r(t)=⟨t2,1−t,4t⟩. calculate the derivative of r(t)⋅a(t) at t=5, assuming that a(5)=⟨−4,4,−5⟩ and a′(5)=⟨−5,9,3⟩
Tatiana wants to give friendship bracelets to her 32 classmates. She has 5 bracelets now. She can buy more bracelets in packages of 4. If p is the number of packages Tatiana needs to buy to have at least 32 bracelets, the inequality representing the problem is: 4p+5≥32 What is the minimum number of packages Tatiana needs to buy?
Let
p--------> the number of packages Tatiana needs to buy
we know that
[tex] 4p+5 \geq 32\\ 4p \geq (32-5)\\ 4p \geq 27\\\\ p \geq \frac{27}{4} \\ \\ p \geq 6.75 [/tex]
therefore
the minimum number of packages does Tatiana needs to buy is [tex] 7 [/tex]
let's check
[tex] 7*4+5=33 [/tex] bracelets
[tex] 33 \geq 32 [/tex] ------> is ok
the answer is
the minimum number of packages is [tex] 7 [/tex]
What is the value 6,035
DE is parallel to XY. What is the value of X?