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Ganezh [65]
4 years ago
14

I WILL GIVE BRAINLIEST

Mathematics
1 answer:
adelina 88 [10]4 years ago
7 0
The area of the fabric is height*width.

A=(5/8)((3/4)=15/32 yd^2  or if you wanted the answer is in^2

A=(15yd^2/32)(36^2in^2/yd^3)=607.5 in^2

...


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Ben’s father is paying for a $20 meal he has a 50% off coupon for the meal after the discount a 7% sales tax is applied what doe
Zina [86]
$20 = 100%
50%= $10
Making $10 the new 100%
$10/100=0.1
0.1=1%
0.1x7=0.7
Meal after discount = $10.70
I think this is right? I hope it helped :)
7 0
3 years ago
Simplify:
Alika [10]
{ e }^{ \ln { \left( B \right) } }{ \left( \frac { 1 }{ R } \right) }^{ -1 }\int { \frac { dA }{ dx } dx } \sqrt { { I }^{ 2 } } \left| -2N+N \right| { \pi }^{ \log _{ \pi }{ \left( L \right) } }\int { \frac { 1 }{ \frac { dx }{ dY } } } dx=\\
BRAI|-N|L\int \frac{dY}{dx}\,dx=\\
BRAINLY
4 0
3 years ago
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
The average price of a home in a certain town was $74,000 in 2012, but home prices
xxTIMURxx [149]

Answer:

$47,360

Step-by-step explanation:

8 percent X 8years is equal to 64percent

74,000 X 64/100=47,360 dollars

8 0
3 years ago
there are 19 pieces of candy in the candy jar Mrs Breen lost her mind and started eating all these mindlessly she realize she ha
kkurt [141]
The percent of candy she ate is 42%
4 0
3 years ago
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