All categories

3 years ago

Need help ASAP please all the number

Mathematics

1 answer:

Naddik [55]3 years ago

7 0

Answer:

doss

Step-by-step explanation:

You might be interested in

Jeanine paid an annual premium of $1,200 in total liability coverage for her car, including up to $400,000 in bodily injury cove

Vinil7 [7]

The answer is B- Yes, the cost of the annual premium for 12 years was less than the accident claims.

3 0

3 years ago

4x^2-5x-15=0 how to find the vertex and y intercept

mezya [45]

$ax^2+bx+c=0\\\\the\ vertex:\left(\frac{-b}{2a};\frac{-(b^2-4ac)}{4a}\right)\\\\y-intrcept=c\\=======================================\\\\4x^2-5x-15=0\\\\a=4;\ b=-5;\ c=-15\\\\\frac{-b}{2a}=\frac{-(-5)}{2\cdot4}=\frac{5}{8};\ \frac{-(b^2-4ac)}{4a}=\frac{-[(-5)^2-4\cdot4\cdot(-15)]}{4\cdot4}=-\frac{265}{16}\\\\\boxed{the\ vertex:\left(\frac{5}{8};-\frac{265}{16}\right)}\\\\\boxed{y-intercept:-15}$

5 0

3 years ago

Darren drank 2 liters of water.How much millilliters of water did he drank

Elodia [21]

1 liter = 1,000 ML
2 X 1,000 = 2,000 ML
2,000 milliliter is your answer

4 0

3 years ago

The graph of f(x) is shown. What is the value of f (x) when x= -2?

Elden [556K]

C. 3.5
You have to go to -2 on the “X axis” and find the corresponding “Y” value on the line.

7 0

3 years ago

The Copy Shop has made 20 copies of a document for you. Since the defective rate is 0.1, you think there may be some defective c

Pepsi [2]

Answer:

Binomial

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

Step-by-step explanation:

For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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

Of the 20 copies, 2 are defective, so $p = \frac{2}{20} = 0.1$ .

What is the probability that you will encounter neither of the defective copies among the 10 you examine?

This is P(X = 0) when $n = 10$ .

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

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487$

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

8 0

3 years ago