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kozerog [31]
3 years ago
7

Of the 81 people who answered "yes" to a question, 6 were male. of the 70 people that answered "no" to the question, 6 were male

. if one person is selected at random from the group, what is the probability that the person answered "yes" or was male?
Mathematics
1 answer:
Stells [14]3 years ago
7 0

there is about a 4% chance that they answer yes and about an 8% chance they are male.

Yes probability- 6/151≈.039 which rounds to .4 or 4%

Male probability- 12/151≈.079 which rounds to .8 or 8%

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Xsquare + 10x + 16 equals 0
inn [45]
X^2+10x+16=0

First, you find factors of 16.
1 x 16
2 x 8
4 x 4

Next, you find which of the factor pairs adds up to 10 (from the 10x). In this case, 2 x 8 because 2 times 8 is 16 and 2 plus 8 is 10.

Then, the equation will be written out as: (x+2) (x+8). Take those two equations and set them equal to 0, and then solve.

x+2=0
-2 -2
x=-2

x+8=0
-8 -8
x=-8

So, your answers are x=-2 and x=-8. You can check by plugging in those two numbers as x.

7 0
2 years ago
Assume that T is a linear transformation. Find the standard matrix of T. T: R^3 right arrow R^2 , T(e 1) =(1,2), and T(e2 ) =( -
irina [24]

Answer:

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

Step-by-step explanation:

Given

T:R^3->R^2

T(e_1) = (1,2)

T(e_2) = (-4,6)

T(e_3) = (2,-6)

Required

Find the standard matrix

The standard matrix (A) is given by

Ax = T(x)

Where

T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

Ax = T(x) becomes

Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

The x on both sides cancel out; and, we're left with:

A = [T(e_1)\ T(e_2)\ T(e_3)]

Recall that:

T(e_1) = (1,2)

T(e_2) = (-4,6)

T(e_3) = (2,-6)

In matrix:

(a,b) is represented as: \left[\begin{array}{c}a\\b\end{array}\right]

So:

T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]

T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]

T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]

Substitute the above expressions in A = [T(e_1)\ T(e_2)\ T(e_3)]

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

Hence, the standard of the matrix A is:

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

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2 years ago
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tatyana61 [14]

Answer:

d he didn't make a mistake

Step-by-step explanation:

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hram777 [196]

Answer:

D. Y = x + 2

Step-by-step explanation:

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