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Pavel [41]
3 years ago
8

Answer quickly, please

Mathematics
1 answer:
V125BC [204]3 years ago
8 0
A . Serena's distance from home when she begins cycling

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What is the estimated difference?
icang [17]
5,714,688 rounded to the nearest thousand is
5,715,000

327,149 rounded to the nearest thousand is 
327,100

5,715,000
- 327,100
-----------------

<span>5388000</span>
7 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
PLEASE HELP!!!! 40PTS+ BRAINLIEST
antoniya [11.8K]

Step-by-step explanation:

Please refer to the attachment.

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Solve the simltaneous equtions: <br><br> 3x - 4y = 18<br><br><br> 9x + 2y = 12
hichkok12 [17]

Answer:

x=2 y= -3

Step-by-step explanation:

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The total weight w, in pounds, of a tractor-trailer capable of carrying 8 cars depends on the number of cars x on the trailer. T
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Answer: your the imposter you killed red then vented in admin.

Step-by-step explanation:

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