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lukranit [14]
3 years ago
11

A recent survey by the American Cancer Society shows that the probability that someone is a smoker is P(S) = 0.19. It has also d

etermined that the probability that someone has lung cancer, given that he or she is a smoker, is P(LC|S) = 0.158. What is the probability (rounded to the nearest hundredth) that a random person is a smoker and has lung cancer P(S ∩ LC)?
0.03
0.83
0.35
0.02
Mathematics
1 answer:
Rom4ik [11]3 years ago
8 0
P(LC / S) = P(S intersect LC) / P(S)
P(S intersect LC) = P(S)*P(LC / S) = 0.19 * 0.158 = 0.03
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Need Help Please I'm Out Of Time ;( The point (-2, 7) is graphed on a coordinate plane. The point is located 2 units ______ the
zysi [14]

Answer:

Left, Up

Step-by-step explanation:

Since -2 is to the left of 0 on the x-axis, we are moving left 2 units.

Since 7 is above 0 on the y-axis, we are moving up 7 units.

7 0
3 years ago
Find a decomposition of a=⟨−5,−1,1⟩ into a vector c parallel to b=⟨−6,0,6⟩ and a vector d perpendicular to b such that c+d=a.
dezoksy [38]

The projection of vector A <em>parallel</em> to vector B is \langle -3, 0, 3\rangle and the projection of vector A <em>perpendicular</em> to vector B is \langle -2, -1, -2\rangle.

In this question, we need to determine all projections of a vector with respect to another vector. In this case, the projection of vector A <em>parallel</em> to vector B is defined by this formula:

\vec a_{\parallel , \vec b} = \frac{\vec a \,\bullet\,\vec b}{\|\vec b\|^{2}}\cdot \vec b (1)

Where \|\vec b\| is the norm of vector B.

And the projection of vector A <em>perpendicular</em> to vector B is:

\vec a_{\perp, \vec b} = \vec a - \vec a_{\parallel, \vec b} (2)

If we know that a = \langle -5, -1, 1 \rangle and \vec b = \langle -6, 0, 6 \rangle, then the projections are now calculated:

\vec a_{\parallel, \vec b} = \frac{(-5)\cdot (-6)+(-1)\cdot (0)+(1)\cdot (6)}{(-6)^{2}+0^{2}+6^{2}} \cdot \langle -6, 0, 6 \rangle

\vec a_{\parallel, \vec b} = \frac{1}{2}\cdot \langle -6, 0, 6 \rangle

\vec a_{\parallel, \vec b} = \langle -3, 0, 3\rangle

\vec a_{\perp, \vec b} = \langle -5, -1, 1 \rangle - \langle -3, 0, 3 \rangle

\vec a_{\perp, \vec b} = \langle -2, -1, -2\rangle

The projection of vector A <em>parallel</em> to vector B is \langle -3, 0, 3\rangle and the projection of vector A <em>perpendicular</em> to vector B is \langle -2, -1, -2\rangle.

We kindly invite to check this question on projection of vectors: brainly.com/question/24160729

7 0
3 years ago
The question is on the picture.
Maslowich
Expression: 8m+10m+10m+18m
Perimeter: 46
4 0
3 years ago
Find the value of x. Round your answer to the nearest whole number.
worty [1.4K]

Answer:

x = 49.45839812°

Step-by-step explanation:

Formula:

arcos( \frac{adjacent}{hypotenuse})

Replace with given values, so

arcos ( \frac{6.5}{10})

Put in calculator, remember it is is DEGREES not RADIANS

Answer:

x = 49.45839812

7 0
3 years ago
The relation Q is described as a list of ordered pairs, shown below. Q = { (-2, 4), (0, 2), (-1, 3), (4, -2) whats the domain an
tester [92]

Answer:

Domain: {-2,0,-1,4}

Range: {4,2,3,-2}

Step-by-step explanation:

1. Given the The relation Q={ (-2, 4), (0, 2), (-1, 3), (4, -2)}, you can determine the domain and the range as following:

DOMAIN:

The domain is the x-coordinate of each ordered pair. Therefore, you have:

Domain: {-2,0,-1,4}

RANGE:

The range is the y-coordinate of each ordered pair. Therefore, you have:

Range: {4,2,3,-2}

3 0
3 years ago
Read 2 more answers
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