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8090 [49]
3 years ago
14

Solve the equation: [3 2 5 5] [x1 x2]+ [1 2]= [2 -3]

Mathematics
1 answer:
dsp733 years ago
7 0

9514 1404 393

Answer:

  (x1, x2) = (3, -4)

Step-by-step explanation:

As with any 2-step linear equation, subtract the constant, then multiply by the inverse of the coefficient of the variable.

  \left[\begin{array}{cc}3&2\\5&5\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]+\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}2\\-3\end{array}\right]\\\\\left[\begin{array}{cc}3&2\\5&5\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}1\\-5\end{array}\right]\\\\\left[\begin{array}{c}x\\y\end{array}\right]=\dfrac{1}{5}\left[\begin{array}{cc}5&-2\\-5&3\end{array}\right]\left[\begin{array}{c}1\\-5\end{array}\right]

Performing the multiplication of the matrix by the vector gives the solution.

  x = ((5)(1) +(-2)(-5))/5 = 15/5 = 3

  y = ((-5)(1) +(3)(-5))/5 = -20/5 = -4

Using your variables, x1, x2, the solution is ...

  (x1, x2) = (3, -4)

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Of a group of randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wa
Wittaler [7]

Answer:

We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

Step-by-step explanation:

We are given that in a group of randomly selected adults, 160 identified themselves as executives.

n = 160

Also we are given that 42 of executives preferred trucks.

So the proportion of executives who prefer trucks is given by

p = 42/160

p = 0.2625

We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.

We can use normal distribution for this problem if the following conditions are satisfied.

n×p ≥ 10

160×0.2625 ≥ 10

42 ≥ 10 (satisfied)

n×(1 - p) ≥ 10

160×(1 - 0.2625) ≥ 10

118 ≥ 10 (satisfied)

The required confidence interval is given by

$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $

Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.

Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96

$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $

$ 0.2625 \pm 1.96\times \sqrt{\frac{0.2625(1-0.2625)}{160} } $

$ 0.2625 \pm 1.96\times 0.03478 $

$ 0.2625 \pm 0.06816 $

0.2625 - 0.06816, \: 0.2625 + 0.06816

(0.1943, \: 0.3306)

(19.43\%, \: 33.06\%)

Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

5 0
3 years ago
Jane will draw one card from the regular deck of 52 cards. In how many ways can she choose a jack or a black card?
elena-s [515]

Answer:

28

Step-by-step explanation:

In a regular 52 card deck, 26 are black (clubs or spades).  Additionally, there are two red jacks (diamonds and hearts).  This equals 28/52.  As a probability, it would be 28/52=14/26=7/13.  This could also be expressed as a percentage or decimal.  0.538 or 53.8%.  But the question asks <em>how many ways</em> can she choose a jack or black card, so I’m assuming they don’t want the probability, just the number, which is 28.

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1 yard=36 inches
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