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3 years ago

Please help me I need this ASAP

Mathematics

1 answer:

kobusy [5.1K]3 years ago

8 0

Answer:

cmnv m vbn,fdm,mnggnndfmd g vxhcb xngvzc bnxbkbhj cb jn lm,lkxjncxb

Step-by-step explanation:

dn xbvghhbgfd,m bv nm,bvvc nzncxcvcxccm bgfdvbcnbmvnbvchdbcsfsvsvga

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What is the greatest common factor of the terms in the excerpt 21 x -18 x + 28 x y

Mandarinka [93]

X(21-18+28y). axis the greatest common factor

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3 years ago

The average of a b c and d is p. If the average of a and c is q, what is the average of b and d in terms of p and q?

vodka [1.7K]

P - q i believe because
p = total average
q = average of ac
x = average of bd
p - q = x

4 0

4 years ago

Consider the probabilities of people taking pregnancy tests. Assume that the true probability of pregnancy for all people who ta

Valentin [98]

Using conditional probability, it is found that there is a 0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

<h3>What is Conditional Probability?</h3>

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this problem, the events are:

Event A: Positive test result.

Event B: Pregnant.

The probability of a positive test result is composed by:

99% of 10%(truly pregnant).

2% of 90%(not pregnant).

Hence:

$P(A) = 0.99(0.1) + 0.02(0.9) = 0.117$

The probability of both a positive test result and pregnancy is:

$P(A \cap B) = 0.99(0.1)$

Hence, the conditional probability is:

$P(B|A) = \frac{0.99(0.1)}{0.117} = 0.8462$

0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

You can learn more about conditional probability at brainly.com/question/14398287

7 0

2 years ago

The polynomial of degree 4, P ( x ) has a root of multiplicity 2 at x = 4 and roots of multiplicity 1 at x = 0 and x = − 4 . It

Annette [7]

Answer: P(x) = {(x-4)^2} (x) (x+4)

Step-by-step explanation:

Let's start with the multiplicity of 2;

At multiplicity of 2; x=4.

Therefore, x - 4 is a factor of the function P(x).

Since it has a multiplicity of 2, we will rewrite the factor as (x-4)^2

Now for the multiplicity of 1.

At this multiplicity of 1, x= 0 and - 4.

Therefore, the factors are x-0 and x+4

Since multiplicity of 1, the factors remain as they are without any additional root on top.

Therefore, the factors of the polynomial p(x) are (x-4)^2 and x and x+4.

And solution of P(x) in factor form will be: P(x) = {(x-4)^2} (x) (x+4)

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3 years ago

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