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2 years ago

Please answer this question

Mathematics

2 answers:

Rzqust [24]2 years ago

8 0

1) Associative

2) Reversal

3) Addition

4) Commutative

5) Multiplicative Inverse

6) Additive Inverse

7) Reversal

8) Substitution

Ksenya-84 [330]2 years ago

3 0

1) Associative

2) Reversal

3) Addition

4) Commutative

5) Multiplicative Inverse

6) Additive Inverse

7) Reversal

8) Substitution

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Help with 4 pls thx

m_a_m_a [10]

The answer is a. Because at approximately 17 weeks the two lines meet.

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3 years ago

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Un estudiante debe recorrer dos tercios de kilómetro para llegar al colegio, si avanza medio kilómetro en bus y lo demás a pie ¿

Murljashka [212]

Answer:

Al estudiante le falta caminar $\frac{1}{3}$ de kilómetro.

Step-by-step explanation:

Un estudiante debe recorrer dos tercios de kilómetro para llegar al colegio. Dos tercios equivale a la fracción $\frac{2}{3}$ .

El estudiante avanza medio kilómetro en bus. Un medio equivale a la fracción $\frac{1}{2}$ , por lo tanto un medio es la mitad de una cantidad. Para calcular la distancia recorrida en bus se realiza la siguiente multiplicación:

$\frac{1}{2} *\frac{2}{3}$

Resolviendo:

$\frac{1}{2} *\frac{2}{3} =\frac{1*2}{2*3} =\frac{2}{6} =\frac{1}{3}$

Entonces, la distancia recorrida en bus es $\frac{1}{3}$ de kilómetro.

Para calcular la distancia recorrida a pie se calcula la diferencia entre la distancia total a recorrer por el estudiante y la distancia recorrida en bus. Esto es:

$\frac{2}{3} - \frac{1}{3}$

Resolviendo:

$\frac{2}{3} - \frac{1}{3}=\frac{2-1}{3} =\frac{1}{3}$

Al estudiante le falta caminar $\frac{1}{3}$ de kilómetro.

6 0

3 years ago

What is the greatest common factor of 12 and 90?

dsp73

The answer is 6. Hopefully this helped!!

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3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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3 years ago

Write the quadratic function f(x) = x2 - 2x - 8 in factored form.

Ronch [10]

Answer:

Hello,

answer C

Step-by-step explanation:

$f(x)=x^2+2x-8\\\\=x^2-4x+2x-8\\\\=x(x-4)+2(x-4)\\\\=(x-4)(x+2)\\\\Answer\ C$

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2 years ago