All categories

3 years ago

The first two terms of the perfect square trinomial are given, find "c".

Mathematics

2 answers:

Naddik [55]3 years ago

7 0

Answer:

A.) 36

Step-by-step explanation:

For a perfect square trinomial, a = 1 b = 2x c = x². For an example, (x+3)² = x² + 6x + 9.

For this trinomial, divide "b" by 2 to get 6. Square this number and you will have 36.

x² + 12x + 36

kupik [55]3 years ago

7 0

A.)36 would be the correct answer

You might be interested in

The graph of g(x)=ax^2 opens downward and is narrower than the graph of f(x)=x^2. Which of the

Masteriza [31]

You’re answer is

b

you’re welcome :)

6 0

3 years ago

Read 2 more answers

How do you answer rhis

Igoryamba

You answer this problem bye cost i already got my answer though thanks for trying...

5 0

3 years ago

Helppppppppppp:)))))))))

Whitepunk [10]

Hi there!

We are given the set of ordered pairs below:

$\large \boxed{(3, - 1),(2, - 2),(0,2),(2,1)}$

1. What is the domain?

Domain is a set of all x-values in one set of ordered pairs. So what are the x-values that I am talking about? In ordered pairs, we define x and y which both have relation to each others which we can write as (x,y). That's right, the domain is set of all x-values from ordered pairs.

Therefore, we gather only x-values from (x,y). Hence, the domain is {3,2,0,2}. Whoops! Something is not right. As we learn in Set Theory that we don't write the same or repetitive in a set. Hence, the actual domain is {0,2,3}

2. What is the range?

Because domain is set of all x-values. Then what do you think the range is? That's right! The range is set of all y-values. If you got this right before looking up the underlined words then a handclap for you! So how do we find range? Simple, we just do like finding the domain in the Q1, except we gather the y-values in (x,y) instead and make sure that we don't write same number!

Therefore, gather y-values from the ordered pairs. Hence, the range is {-2,-1,1,2}

3. Is the relation a function?

All functions are relations but not all relations are functions. Function is a set of ordered pairs where domain is not repetitive or in a set, there cannot be more than one same value. Consider the following relation: (1,1),(1,2) - Oh, looks like in a set of ordered pairs, there are two same domains which make it only a relation, and not a function. On the other hand, (1,1),(2,2) - Looking good! No same or repetitive domain, making it indeed a function.

Consider the domain from Q1 and see if there are two same values of x in a set. Looks like the relation is not a function since there are same x-values which are 2 in a set, making it only a relation. Hence, the relation is not a function.

These are all 3 answers along with an explanation. Let me know if you have any doubts regarding Relations and Functions.

From the Q1's answer, there are two bold texts, please choose the second bold text to answer (the one with underline) and not the first one (the one with same 2's).

Good luck on your assignment, have a nice day!

4 0

3 years ago

A sample size 25 is picked up at random from a population which is normally

Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .