(20 PTS) WILL GIVE BRAINLIST could you help me w this?

blagie [28]

1,2,3,4,6,9,12,18 and 36. 1,2,3,6,9,18,27, and 54 .Although the numbers in bold are all common factors of both 36 and 54 ,18 is the greatest common factor. The second method to find the greatest common factor is to list the prime factors, then multiply the common prime factors

4 0

3 years ago

How much money do I need to make to still have 1,000,000 left after 27.4% tax is removed?

Rama09 [41]

let's say the amount is "x", hmmm how much is 27.4% of "x"? well

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{27.4\% of x}}{\left( \cfrac{27.4}{100} \right)x}\implies 0.274x$

so then, we know that x - 0.274x = 1000

$\bf x - \stackrel{\textit{27.4\% of "x"}}{0.274x}=1000\implies 0.726x=1000\\\\\\ x = \cfrac{1000}{0.726}\implies x\approx 1377.41$

4 0

3 years ago

What is the solution to the equation 5x + 2(x-4)= 5x + x -10? 03 02 0-2 0-3

NISA [10]

Answer:

5x+2x(x-4)=5x+x-10

5x+2x-8=5x+x-10

5x-5x+2x+x=-10+8

x+3x=-2

4x=-2

x=-2

8 0

2 years ago

Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th

Anton [14]

Answer:

a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) 0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

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.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

$P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312$

Probability of having an emergency locator and not being discovered:

20% of 28%. So

$P(A cap B) = 0.2*0.28 = 0.056$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105$

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

$P(A) = 1 - 0.5312 = 0.4688$

Probability of not having an emergency locator and being discovered:

34% of 72%. So

$P(A \cap B) = 0.34*0.72 = 0.2448$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522$

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

66% of 72%. So

$P(A \cap B) = 0.66*0.72 = 0.4752$

Probability of being discovered, given that it has the emergency recorder:

$p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946$

This question asks for P(X = 7) when $n = 10$ . So

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

$P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064$

0.064 = 6.4% probability that 7 of them are discovered.

8 0

3 years ago

June is working on an addition problem and starts with 17,985. After she adds she still has 17,985. Which property of addition d

julsineya [31]

U start with 17,985...u add a number and the result is still 17,985. There is only 1 way this can happen and that is if the number u added was 0.

17,985 + 0 = 17,985....this is the additive identity property...which simply states that u can add 0 to a number and it still keeps its identity.

6 0

3 years ago

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