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3 years ago

Given the following matrices, find 3A - 2B. (Hint: find 3A and 2B first and then subtract.)

1 answer:

3 0

$A= \left[\begin{array}{cc}1&3\\5&2\end{array}\right] 
\\
\\3A=\left[\begin{array}{cc}3&9\\15&6\end{array}\right] 
\\
\\B= \left[\begin{array}{cc}4&9\\0&3\end{array}\right] 
\\
\\2B= \left[\begin{array}{cc}8&18\\0&6\end{array}\right] 
\\
\\3A-2B=\left[\begin{array}{cc}3&9\\15&6\end{array}\right]- \left[\begin{array}{cc}8&18\\0&6\end{array}\right]=\left[\begin{array}{cc}3-8&9-18\\15-0&6-6\end{array}\right]=
\\
\\=\left[\begin{array}{cc}-5&-9\\15&0\end{array}\right]$

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The region bounded by the curve y = 2/(1 + e−x), the x- and y-axes, and the line x = 10 is rotated about the x-axis. Use Simpson

sp2606 [1]

Answer:

21 darted to the nearest 10

Step-by-step explanation:

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3 years ago

A sock drawer contains eight navy blue socks and five black socks with no other socks. If you reach in the drawer and take two s

Rzqust [24]

Answer:

a. the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b. the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c. the probability of either 2 navy socks is picked or one black & one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

Step-by-step explanation:

A sock drawer contains 8 navy blue socks and 5 black socks with no other socks.

If you reach in the drawer and take two socks without looking and without replacement, what is the probability that:

Solution:

total socks = N = 8 + 5 + 0 = 13

a) you will pick a navy sock and a black sock?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick the black sock, total number of socks left is 12.

Let B be the probability of picking a black sock again.

P (B) = 5/12.

Then, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b) the colors of the two socks will match?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick another navy sock, total number of socks left is 12.

Let B be the probability of another navy sock again.

P (B) = 7/12.

Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

Let D be the probability of picking a black socks first.

Then P (D) = 5/13

without replacing the black sock, will pick another black sock, total number of socks left is 12.

Let E be the probability of another black sock again.

P (E) = 4/12.

Then, the probability of picking 2 black sock = P (D & E)

= (5/13 ) * (4/12) = 5/39 = 0.128

Now, the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c) at least one navy sock will be selected?

this means, is either you pick one navy sock and one black or two navy socks.

so, if you will pick a navy sock and a black sock, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

also, if you will pick 2 navy sock, Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

now either 2 navy socks is picked or one black one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

4 0

3 years ago

I surveyed students in my Math II classes to see how many hours of television they watched the night before our big test on tria

Oduvanchick [21]

Given the table below representing the number of hours of television nine Math II class students watched the night before a big test on triangles along with the grades they each earned on that test.

$\begin{center}
\begin{tabular}
{|c|c|}
Hours Spent Watching TV & Grade on Test (out of 100) \\ [1ex]
4 & 71 \\ 
2 & 81 \\ 
4 & 62 \\ 
1 & 86 \\ 
3 & 77 \\ 
1 & 93 \\ 
2 & 84 \\ 
3 & 80 \\ 
2 & 85
\end{tabular}
\end{center}$

Let the number the number of hours of television each of the students watched the night before the test be x while the grades they each earned on that test be y.

We use the following table to find the equation of the line of best fit of the regression analysis of the data.

$\begin{center} \begin{tabular} {|c|c|c|c|} x & y & x^2 & xy \\ [1ex] 4 & 71 & 16 & 284 \\ 2 & 81 & 4 & 162 \\ 4 & 62 & 16 & 248 \\ 1 & 86 & 1 & 86 \\ 3 & 77 & 9 & 231 \\ 1 & 93 & 1 & 93 \\ 2 & 84 & 4 & 168 \\ 3 & 80 & 9 & 240 \\ 2 & 85 & 4 & 170 \\ [1ex]\Sigma x=22 & \Sigma y=719 & \Sigma x^2=64 & \Sigma xy=1,682 \end{tabular} \end{center}$

Recall that the equation of the line of best fit of a regression analysis is given by
$y=a+bx$
where:
$a= \frac{(\Sigma y)(\Sigma x^2)-(\Sigma x)(\Sigma xy)}{n(\Sigma x^2)-(\Sigma x)^2}$
and
$b= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2)-(\Sigma x)^2}$

$y=\frac{(719)(64)-(22)(1,682)}{9(64)-(22)^2}+\frac{9(1,682)-(22)(719)}{9(64)-(22)^2}x \\ \\ = \frac{46,016-37,004}{576-484} + \frac{15,138-15,818}{576-484} x \\ \\ = \frac{9,012}{92} + \frac{-680}{92} x \\ \\ =97.95-7.391x$

Thus, the equation of the line of best fit is given by y = 97.95 - 7.391x

<span>A student that watched 1.5 hours of TV will have a score given by
y = 97.95 - 7.391(1.5) = 97.95 - 11.0865 = 86.8635

Therefore, </span><span>a student’s score if he/she watched 1.5 hours of TV to the nearest whole number is 87.</span>

6 0

3 years ago

Consider the functions p and q.

xz_007 [3.2K]

Answer:

q= -1x

coordinates of p are .5, 2

Step-by-step explanation:

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2 years ago

I need help with this

kompoz [17]

Answer:

Step-by-step explanation:

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3 years ago