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3 years ago

Factor out the minus sign - 1+30

Mathematics

2 answers:

True [87]3 years ago

6 0

Answer:

Your answer will be 29

Step-by-step explanation:

Add -1 and 30

madreJ [45]3 years ago

4 0

First, you need to find the greatest common factor, and because we are factoring out the minus sign, it should be a negative number.

For -1 & 30, the GCM is -1, and using the GCM, we can factor each of them out:
-1 30
/ \ / \
-1 x 1 -1 x 30

Then, we can use distributive property to factor out the minus with the GCM outside of the parentheses:

-1 + 30 =

-1(1 + 30)

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Jane must get at least three of the four problems on the exam correct to get an A. She has been able to do 80% of the problems o

NISA [10]

Answer:

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

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

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

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.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so $\pi = 0.8$ .

(a) What is the probability she gets an A?

There are four problems, so $n = 4$

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find $P(X \geq 3)$

$P(X \geq 3) = P(X = 3) + P(X = 4)$

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

$P(X = 3) = C_{4,3}.(0.80)^{3}.(0.2)^{1} = 0.4096$

$P(X = 4) = C_{4,4}.(0.80)^{4}.(0.2)^{0} = 0.4096$

$P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192$

There is n 81.92% probability that she gets an A.

(b) If she gets the first problem correct, what is the probability she gets an A?

Now, there are only 3 problems left, so $n = 3$

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find $P(X \geq 2)$

$P(X \geq 3) = P(X = 2) + P(X = 3)$

$P(X = 2) = C_{3,2}.(0.80)^{2}.(0.2)^{1} = 0.384$

$P(X = 4) = C_{3,3}.(0.80)^{3}.(0.2)^{0} = 0.512$

$P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896$

If she gets the first problem correct, there is an 89.6% probability that she gets an A.

3 0

3 years ago

Yyoooo free moneys but im deleting the question if u dont say the right thing in the answer.

Zielflug [23.3K]

UR MOM

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3 0

2 years ago

Read 2 more answers

ngozi earns \$24{,}000$24,000dollar sign, 24, comma, 000 in salary in the first year she works as an interpreter. each year, she

goblinko [34]

The function that Ngozi's salary s(t) in dollars, t years after she starts to work as an interpreter is given by $S = 24000 * (1.035)^t$ .

What is a function?

A function in mathematics from a set X(ax) to a set Y allocates exactly one element of Y to each element of X(ax).
A relation between a collection of inputs and labor is known as a function.

You can use the fact that the raise is done on the previous year's salary.

Let the initial salary be P = $24000

After 1 year, her salary increased by R = 3.5% which, thus,

Total salary = P + increment

Similarly going on, we get

$P_t = P(1+ \frac{R}{100})^t = S$ (we can say)