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2 years ago

Please help

Mathematics

1 answer:

MissTica2 years ago

3 0

Answer:

this is not an answer but I will post one, could you tell us what the answer choices are?

Step-by-step explanation:

it would help with the answers.

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Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$

What is the probability that the person actually does have cancer?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213$

There is a 12.13% probability that the person actually does have cancer.

3 0

2 years ago

I need help with this 2(n+3) - 4 < 6 n

Harman [31]

Answer:

Step-by-step explanation:

2(n+3) - 4 < 6n

2n+6-4 < 6n

2n + 2 < 6n

2 < 4n

½ < n

7 0

2 years ago

The height of a triangle is two more than three times the base. Determine the dimensions that will give a total area of 28 yards

jonny [76]

H = 3b+2
A = (h*b)/2 28 = (3b+2)b/2 56 = 3b²+2b 0 = 3b² + 2b - 56

⊕ $\left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta \\ \\ \\ x^{2} \sqrt{x} \sqrt[n]{x} \frac{x}{y} x_{123} x^{123} \leq \geq \pi \alpha \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] x_{123} \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}}$
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8 0

2 years ago

A box is to be constructed with a rectangular base and a height of h cm. The length, l, of the box is 10 cm, and the width, w, i

Svetradugi [14.3K]

<span>The formula for the volume of a box is: Volume = L*W*H We know that the length is 10 cm. We know that the width is twice the height, so W = 2H. Then our volume equation is: Volume = 10*2H*H Volume = 20*H^2 Since H^2 is in the equation, this is a quadratic equation.</span>

8 0

2 years ago

A water storage tank has the shape of a cylinder with diameter 14 ft. It is mounted so that the circular cross-sections are vert

Gala2k [10]

Answer:

"85.7%" is the right answer.

Step-by-step explanation:

According to the question,

The total capacity of water tank will be:

⇒ $v_1=A\times 14$