The picture below shows a shaded rectangular region inside a large rectangle.

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

8 0

Answer:

the answer is d

Step-by-step explanation:

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A home pregnancy test is not always accurate. suppose the probability that the test indicates that a woman is pregnant when she

Gnom [1K]

Set Events:
T=tests positive~T=tests negativeP=subject is pregnant~P=subject is not pregnant
We are givenP(T n ~P)=0.02P(~T n P)=0.03P(P)=0.7
recall by definition of conditional probabilityP(A|B)=P(A n B)/P(B)

Need to find P(P|~T)
First step: make a contingency diagram of probabilities (intersection, n)
P ~P sum
T 0.67 0.02 0.69=P(T)
~T 0.03 0.28 0.31=P(~T)
sum 0.70 0.30 1.00
=P(P) =P(~P)

therefore
P(P|~T)=P(P n ~T)/P(~T)=0.03/0.31 [ both read off the contingency table ]
=0.0968

4 0

2 years ago

What can be multiplied to get 40 but also added to get 29

Tresset [83]

Answer:

is that even possible.

Step-by-step explanation:

4 0

2 years ago

Identify the coefficient(s) and constant(s) in this expression. 9x - 8y + 6

Art [367]

Answer:

the coefficient are 9 and -8

the constant is 6........

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2 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .