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3 years ago

Do the ratios 1:9 and 4:6 form a proportion?

Mathematics

1 answer:

Umnica [9.8K]3 years ago

7 0

We have: $1:9=\frac{1}{9}=\frac{1.4}{9.4} =\frac{4}{36}=4:36\neq 4:6$

So the answer is: No

Ok done. Thank to me :>

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Hey anybody plz help me ;C

jekas [21]

Answer:

her equation should be a² + 3² = 4²

Step-by-step explanation:

a² + 3² = 4²

a² + 9 = 16

a² = 7

a = $\sqrt{7}$

a ≈ 2.6 inches

3 0

3 years ago

Past records indicate that the probability of online retail orders

Tcecarenko [31]

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

The mean of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$p = 0.08, n = 20$

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

$E(X) = np = 20*0.08 = 1.6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21$

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

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

$P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887$

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

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

$P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282$

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \geq 2)$

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

In which

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

From itens b and c

$P(X \leq 1) = 0.1887 + 0.3282 = 0.5169$

$P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831$

48.31% probability that two or more online retail orders will turn out to be fraudulent.

4 0

4 years ago

Pls help ill give brainliest

Ymorist [56]

Answer:

$\boxed {\boxed {\sf (5g^2+2)(2g-5)}}$

Step-by-step explanation:

We are given the expression:

$10g^3-25g^2+4g-10$

There are no common factors between the four numbers, however the first two have a common factor and the last two do too. Therefore, we can factor by grouping.

Group the first two terms and the last two.

$(10g^3-25g^2)+(4g-10)$

Find the greatest common factor (GCF) of the first group. It is 5g². Factor it out of the first group. You can do this by dividing both terms by the GCF.