Determine the scale factor of the function f(x)=1/3root x

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) 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.

Taking into account the definition of combination, you can take 1716 different collections of 6 books.

Combinations 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.

In 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:

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Answer:

Step-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

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.

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.

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The probability of achieving exactly two successful calls in eight attempts in a telephone survey using a random digit dialing machine is approximately 0.324 or 32.4%.

To calculate the probability of achieving exactly two successful calls in eight attempts in a telephone survey using a random digit dialing machine, we can use the binomial distribution formula. The formula is:

P(x) = (nCx) * (p^x) * ((1-p)^(n-x))

Where:

Plugging in the values from the question, we have:

P(2) = (8C2) * (0.15^2) * ((1-0.15)^(8-2)) = 28 * 0.0225 * 0.411771 = 0.3239965

Therefore, the probability of achieving exactly two successful calls in eight attempts is approximately 0.324 or 32.4%.

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8 men, 5 positions

 so multiply 8*7*6 =336

divide by 3*2 =6

336/6 =56

 56 different ways

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 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:

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]

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

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.

Answer:

X = - sqrt 12 / 4

X = sqrt 11 + 1 / 4

(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]

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.

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.

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