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3 years ago

A positive and negative charge will attract each

1 answer:

8 0

In contrast to the attractive force between two objects with opposite charges, two objects that are of like charge will repel each other. That is, a positively charged object will exert a repulsive force upon a second positively charged object. ... Objects with like charge repel each other.

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(3x − 4)(2x)2 + 2x − 1).

zzz [600]

(3x−4)(2x)(2)+2x−1

Distribute:

=12x2+−16x+2x+−1

Combine Like Terms:

=12x2+−16x+2x+−1=(12x2)+(−16x+2x)+(−1)=12x2+−14x+−1

Answer:=12x2−14x−1

hope this helps! was there meant to be a parenthesis by the 2 + 2x - 1??

4 0

3 years ago

Gio is solving the quadratic equation by completing the square. 5x2 + 15x – 4 = 0. What should Gio do first

Vaselesa [24]

We observe that we have a polynomial of the form:
ax2 + bx + c
Therefore, to complete the square, we must use the following formula:
a (x + (b / 2a)) ^ 2 + c - (b ^ 2 / 4a)
Thus,
Step 1:
We define:
a = 5
b = 15
c = -4
Step 2:
we use the formula:
5 (x + (15 / (2 * 5))) ^ 2 + (-4) - ((15) ^ 2 / (4 * 5))
5 (x + (15 / (2 * 5))) ^ 2 - 15.25
5 (x + 1.5) ^ 2 - 15.25

5 (x + 3/2) ^ 2 - 61/4

Answer:
Gio should do first:
a (x + (b / 2a)) ^ 2 + c - (b ^ 2 / 4a)
a = 5
b = 15
c = -4

4 0

3 years ago

Read 2 more answers

The block factory has boxes to pack groups of one hundred books if maks has an order for 2,340 blocks how can he pack them using

SpyIntel [72]

He will use 24 boxes

4 0

3 years ago

Read 2 more answers

It is known that there only is 1% chance of getting a disease. a test is being devised to detect the disease. the probability th

Cerrena [4.2K]

Suppose $D$ is the event that a given patient has the disease, and $P$ is the event of a positive test result.

We're given that

$\mathbb P(D)=0.01$
$\mathbb P(P\mid D)=0.98$
$\mathbb P(P^C\mid D^C)=0.95$

where $A^C$ denotes the complement of an event $A$ .

a. We want to find $\mathbb P(P^C)$ . By the law of total probability, we have

$\mathbb P(P^C)=\mathbb P(P^C\cap D)+\mathbb P(P^C\cap D^C)$

That is, in order for $P^C$ to occur, it must be the case that either $D$ also occurs, or $D^C$ does. Then from the definition of conditional probability we expand this as

$\mathbb P(P^C)=\mathbb P(D)\mathbb P(P^C\mid D)+\mathbb P(D^C)\mathbb P(P^C\mid D^C)$

so we get

$\mathbb P(P^C)=0.01\cdot0.02+0.99\cdot0.95=0.9407$

b. We want to find $\mathbb P(D\mid P)$ . Now, we can use Bayes' rule, but if you're like me and you find the formula a bit harder to remember, we can easily derive it.

By the definition of conditional probability,

$\mathbb P(D\mid P)=\dfrac{\mathbb P(D\cap P)}{\mathbb P(P)}$

We have the probabilities of $P$ / $P^C$ occurring given that $D$ / $D^C$ occurs, but not vice versa. However, we can expand the probability in the numerator to get a probability in terms of $P$ being conditioned on $D$ :

$\mathbb P(D\cap P)=\mathbb P(D)\mathbb P(P\mid D)$

Meanwhile, the law of total probability lets us rewrite the denominator as

$\mathbb P(P)=\mathbb P(P\cap D)+\mathbb P(P\cap D^C)$

or in terms of conditional probabilities,

$\mathbb P(P)=\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)$

so that

$\mathbb P(D\mid P)=\dfrac{\mathbb P(D)\mathbb P(P\mid D)}{\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)}$

which is exactly what Bayes' rule states. So we get

$\mathbb P(D\mid P)=\dfrac{0.01\cdot0.98}{0.01\cdot0.98+0.99\cdot0.05}\approx0.1653$

6 0

3 years ago

10 pts Match the word to its definition or symbol.

earnstyle [38]

Answer:

1. Element

2.empty set

3.subset

4.infinite set

5.null set

6.finite set

7.not an element

Step-by-step explanation:

3 0

4 years ago