All categories

3 years ago

Which value must be added to the expression x2 + 12x to make it a perfect-square trinomial?

Mathematics

2 answers:

stellarik [79]3 years ago

8 0

Answer:

D) 81

Step-by-step explanation:

Alexxx [7]3 years ago

7 0

Answer:36

When you factor X^2+12x+36 you get a perfect square trinomial which is (x+6)^2

Hope this helps!

You might be interested in

Write an

vitfil [10]

Hopefully, this helps

5 0

3 years ago

Determine the measure of ∠A. ANSWWER: 1) 192° 2) 88° 3) 84° 4) 168°

hichkok12 [17]

Answer:

3) 84°

Step-by-step explanation:

The angle formed by a chord and a tangent measures half of the intercepted arc.

m<A = (1/2)m(arc) = (1/2) * (168 deg) = 84 deg

7 0

3 years ago

A farmer sows 100 seeds of a new type of corn and wants to quickly determine the yield, or total number of ears of corn, for the

snow_lady [41]

Answer:

the fewest number is 5,6,7

Step-by-step explanation:

hope it helps

7 0

2 years ago

What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 ml at 850 mm Hg?

REY [17]

Answer:

850 x 0.3 /1.7 = 150ltr

We round up to 200ltr. for 1sf

Step-by-step explanation:

8 0

3 years ago

The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .1

snow_lady [41]

We assumed in this answer that the question b is, Are the events V and M independent?

Answer:

(a). The probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ . (b). The events V and M are not independent.

Step-by-step explanation:

The key factor to solve these questions is to know that:

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

We already know from the question the following probabilities:

$\\ P(V) = 0.73$

$\\ P(M) = 0.18$

The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So

$\\ P(V \cap M) = 0.03$

The probability that a student has either a Visa card or a MasterCard

We can interpret this probability as $\\ P(V \cup M)$ or the sum of both events; that is, the probability that one event occurs OR the other.

Thus, having all this information, we can conclude that

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

$\\ P(V \cup M) = 0.73 + 0.18 - 0.03$

$\\ P(V \cup M) = 0.88$

Then, the probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ .

Are the events V and M independent?

A way to solve this question is by using the concept of conditional probabilities.

In Probability, two events are independent when we conclude that

$\\ P(A|B) = P(A)$ [1]

The general formula for a conditional probability or the probability that event A given (or assuming) the event B is as follows:

$\\ P(A|B) = \frac{P(A \cap B)}{P(B)}$

If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that

$\\ P(M|V) = \frac{P(M \cap V)}{P(V)}$

$\\ P(M|V) = \frac{0.03}{0.73}$

$\\ P(M|V) \approx 0.041$

If M were independent from V (according to [1]), we have

$\\ P(M|V) = P(M) = 0.18$

Which is different from we obtained previously;

That is,

$\\ P(M|V) \approx 0.041$

So, the events V and M are not independent.

We can conclude the same if we calculate the probability

$\\ P(V|M)$ , as follows:

$\\ P(V|M) = \frac{P(V \cap M)}{P(M)}$

$\\ P(V|M) = \frac{0.03}{0.18}$

$\\ P(V|M) = 0.1666.....\approx 0.17$

Which is different from

$\\ P(V|M) = P(V) = 0.73$

In the case that both events were independent.

Notice that

$\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)$

$\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03$

$\\ 0.03 = 0.03 = 0.03 = 0.03$

3 0

4 years ago