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3 years ago

PLEASE HELP I GIVE THANKS

Mathematics

1 answer:

Elden [556K]3 years ago

3 0

The answer of first one is A, because this is positive.

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What is the sum of 2x^2-x and -x-2x^2-2

Lorico [155]

<h2>Answer:</h2><h3>-2x-2</h3>

 $solution \\ {2x}^{2} - x + ( - x - {2x}^{2} - 2) \\ = {2x}^{2} - x - x - {2x}^{2} - 2 \\ = {2x}^{2} - {2x}^{2} - x - x - 2 \\ = - 2x - 2$ 

Hope it helps

Good luck on your assignment

5 0

3 years ago

Read 2 more answers

10% OFF! What was the original price of a set of dishes whose sale price is $216?

Romashka [77]

Answer:

240

Step-by-step explanation:

If we get 10% off, we still have to pay 100-10 = 90%

Let x be the original price

x *90% = 216

.90x = 216

Divide each side by .90

.90x/.90 = 216/.90

x =240

The original price is 240

4 0

3 years ago

Read 2 more answers

What is the sum of 1/9, 2/3, and 5/18? A. 121 B.4/15 C. 19/18 D. 8/30

sergey [27]

Answer is C...Use calculator

8 0

3 years ago

A financial talk show host claims to have a 55.3 % success rate in his investment recommendations. You collect some data over th

baherus [9]

Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

$p = 0.553, n = 10$

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$