150+40.50x=120+46.75x

Anna35 [415]

6z -4 - z + 10 = -9
Combine like terms
5z + 6 = -9
Subtract both sides by 6
5z = -15
Divide both sides by 5
z = -3
Have an awesome day! :)

7 0

3 years ago

An academic department with five faculty members—Anderson, Box, Cox, Cramer, and Fisher—must select two of its members to serve

Maru [420]

Answer:

(a) 0.10

(b) 0.70

(c) 0.30

Step-by-step explanation:

Items b and c are missing from the question:

(b) What is the probability that at least one of the two members whose name begins with C is selected?

(c) If the five faculty members have taught for 3, 6, 7, 10, and 14 years, respectively, at the university, what is the probability that the two chosen representatives have a total of at least 18 years teaching experience there?

The number of ways to select two representatives out of 5 people is given by the combination of picking two out of 5:

$n=\frac{5!}{(5-2)!2!}\\n=10\ ways$

(a) Since there is only one way for both Anderson and Box to be selected, the probability is:

$P = \frac{1}{10}=0.10$

(b) There are four ways for Cox to be selected, four ways for Cramer to be selected, and one where both are selected. The probability is:

$P =\frac{4+4-1}{10}=0.7$

(c) The only possible ways for the total of number of years to exceed 18 is by selecting: (14;6 14;7 14;10). Therefore the probability is:

$P = \frac{3}{10}=0.30$

8 0

3 years ago

Show that {(1,1,0),(1,0,1),(0,1,1)} is linearly independent subset of r^3

Keith_Richards [23]

Answer: Yes, the given set of vectors is a linearly independent subset of R³.

Step-by-step explanation: We are given to show that the following set of three vectors is a linearly independent subset of R³ :

B = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} .

Since the given set contains three vectors which is equal to the dimension of R³, so it is a subset of R³.

To check the linear independence, we will find the determinant formed by theses three vectors as rows.

If the value of the determinant is non zero, then the set of vectors is linearly independent. Otherwise, it is dependent.

The value of the determinant can be found as follows :

$D\\\\\\=\begin{vmatrix}1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1\end{vmatrix}\\\\\\=1(0\times1-1\times1)+1(1\times0-1\times1)+0(1\times1-0\times0)\\\\=1\times(-1)+1\times(1)+0\\\\=-1-1\\\\=-2\neq0.$

Since the determinant is not equal to 0, so the given set of vectors is a linearly independent subset of R³.

Thus, the given set is a linearly independent subset of R³.

7 0

3 years ago

Solve the system of equations. y=4x+8 x^2+7x-20

harina [27]

Can this help you our

3 0

4 years ago

Simplify the given expression. Assume that no variable equals 0. (14x20y10 / 28x12y14)4

liq [111]

Answer:

$\frac{1}{2^{4}*14^{3} *x^{28}*y^{46}}$

Step-by-step explanation:

To simplify the expression $\frac{14x^{20}y^{10}}{(2*14*x^{12} y^{14} )^{4}}$ first we need to raise every member of the denominator to the given exponent, number 4

Remember that $(x^{b} )^{c} =z^{b*c}$ where 'a' and 'b are natural numbers (any numbers you want

$\frac{14x^{20}*y^{10}}{2^{4}*14^{4}*x^{48}*y^{56}}$

Where we can make some operations, remember that $\frac{x^{a} }{x^{c} }=x^{a-c}$

$\frac{1}{2^{4}*14^{3} *x^{28}*y^{46}}$

We have assumed that nor x, nor y equals 0 in any moment, since if that would happen, we ca not simplify anything bacause error math 0/0.

7 0

4 years ago

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