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3 years ago

I really need help with this problem.

Mathematics

2 answers:

nekit [7.7K]3 years ago

5 0

It is d i think dont quote me

sukhopar [10]3 years ago

3 0

Ok I might be able to help you in the future.

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The base of the triangle is 5<img src="https://tex.z-dn.net/?f=%5Csqrt%7B3%7D" id="TexFormula1" title="\sqrt{3}" alt="\sqrt{3}"

Vsevolod [243]

Answer:

yes you are absolutely right

it will be the answer

4 0

3 years ago

Which of the functions below could have created this graph?

bearhunter [10]

Answer:

Step-by-step explanation:

Two facts need to guide your answer.

One

The highest power is odd: you know this because an even power would start on the left come down do it's squiggles if had any and wind up on the right going up.

This graph comes down on the left does it's squiggles and then goes further down on the right. That's the behavior of something whose highest power is odd.

Two

The leading coefficient, the number in front of the highest power must be minus. If it was positive as in y = x^3 the graph would be the mirror image of what it is.

Argument

B and D cannot be true. The highest power is even.

C is false because the leading coefficient is + 1.

So that leave A which is the answer.

The graph is included with this answer

7 0

3 years ago

The questions are in the picture

o-na [289]

Answer:

I can't see the picture

Step-by-step explanation:

I can't see it

4 0

3 years ago

Read 2 more answers

I donk know what to do

ohaa [14]

What’s the question?

7 0

3 years ago

A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou

Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let $z$ be the

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question, $x = 5$ and $\sigma = 2$ . The equation becomes

$\displaystyle z = \frac{5 - \mu}{2}$ .

To solve for $\mu$ , the mean of this distribution, the only thing that needs to be found is the value of $z$ . Since

The problem stated that $P(X \le 5) = 69.85\% = 0.6985$ . Hence, $P(Z \le z) = 0.6985$ .

The problem is that the normal distribution tables list only the value of $P(0 \le Z \le z)$ for $z \ge 0$ . To estimate $z$ from $P(Z \le z) = 0.6985$ , it would be necessary to find the appropriate

Since $P(Z \le z) = 0.6985$ and is greater than $P(Z \le 0) = 0.50$ , $z > 0$ . As a result, $P(Z \le z)$ can be written as the sum of $P(Z < 0)$ and $P(0 \le Z \le z)$ . Besides, $P(Z < 0) = P(Z \le 0) = 0.50$ . As a result: