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QveST [7]
3 years ago
5

NEED HELP Find values for a, b, c, and d so that the following matrix product equals the 2X2 identity matrix. Explain or show ho

w you obtained these values.

Mathematics
1 answer:
ehidna [41]3 years ago
4 0

We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:

  • b = 1
  • a = -1
  • c = -2
  • d = -1

<h3>Presenting the equation:</h3>

Basically, we want to solve:

\left[\begin{array}{cc}-1&2\\a&1\end{array}\right]*\left[\begin{array}{cc}b&c\\1&d\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]

The matrix product will be:

\left[\begin{array}{cc}-b + 2&-c + 2d\\a*b + 1&a*c + d\end{array}\right]

Then we must have:

-b + 2 = 1

This means that:

b = 2 - 1 = 1

  • b = 1

We also need to have:

a*b + 1 = 0

we know the value of b, so we just have:

a*1 + b = 0

  • a = -1

Now the two remaining equations are:

-c + 2d = 0

a*c + d = 1

Replacing the value of a we get:

-c + 2d = 0

-c + d = 1

Isolating c in the first equation we get:

c = 2d

Replacing that in the other equation we get:

-(2d) + d = 1

-d = 1

  • d = -1

Then:

c  = 2d = 2*(-1) = -2

  • c = -2

So the values are:

  • b = 1
  • a = -1
  • c = -2
  • d = -1

If you want to learn more about systems of equations, you can read:

brainly.com/question/13729904

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The average amount parents and children spent per child on back-to-school clothes in Autumn 2010 was $527. Assume the standard d
Ad libitum [116K]

Answer: a. 0.14085

b. 3.826 x 10^{-3}

c.  0.5437

d. 0.0811  

Step-by-step explanation:

Given average amount parents and children spent per child on back-to-school clothes in Autumn 2010 , \mu = $527

Given standard deviation , \sigma = $160

Let X = amount spent on a randomly selected child

Also Z = \frac{X-\mu}{\sigma}

a. Probability(X>$700) = P(\frac{X-\mu}{\sigma} > \frac{700-527}{160}) = P(Z>1.08125) = 0.14085 {Using Z % table}

b. P(X<100) = P( Z < \frac{100-527}{160}) = P(Z< -2.66875) = P(Z > 2.66875) = 3.826 x 10^{-3}

c. P(450<X<700) = P(X<700) - P(X<=450)

   P(X<700) = 1 - P(X>=700) = 1 - 0.14085 = 0.8592

   P(X<=450) = P(Z<= \frac{450-527}{160}) = P(Z<= -0.48125) = P(Z<=0.48125) = 0.3155

   So final   P(450<X<700) =  0.8592 - 0.3155 = 0.5437

d. P(X<=300) = P(Z<= \frac{300-527}{160}) = P(Z<= -1.4188) = P(Z>=1.4188) = 0.0811                                                                        

All the above probabilities are calculated using Z % table along with interpolation between two values.

7 0
3 years ago
Two pitchers, Jerry and Louis, want to find out who has the faster pitch when compared to each of their teams. Jerry has a speed
saul85 [17]

Answer:

Louis has faster pitch when compared to each of their teams.

Step-by-step explanation:

We have two pitchers which we need to compare to each of their teams.

To calculate this, we will approximate the distributions to a normal distribution, and calculate the z-score, to know what proportion of players of their team fall below their score.

For Jerry, he has a speed of 86 and his team has a mean speed of 93 and standard deviation of 3.

We can calculate the z-score for Jerry speed as:

z=\dfrac{X-\mu}{\sigma}=\dfrac{86-93}{3}=\dfrac{-7}{3}= -2.333

The proportion of players that are below Jerry speed is approximated by the standard normal distribution:

p=\Phi(z

For Louis, his speed is 84 and his team has a mean speed of 89 and standard deviation of 3.5.

We can calculate the z-score for Jerry speed as:

z=\dfrac{X-\mu}{\sigma}=\dfrac{84-89}{3.5}=\dfrac{-5}{3.5}= -1.429

The proportion of players that are below Louis speed is approximated by the standard normal distribution:

p=\Phi(z

As the proportion of players of Louis team that are below Louis speed is much bigger than the proportion of players of Jerry's team that are below Jerry speed, we can say that Louis has faster pitch when compared to each of their teams.

6 0
3 years ago
Ammeters produced by a manufacturer are marketed under the specification that the standard deviation of gauge readings is no lar
AURORKA [14]

Answer:

0.101

Step-by-step explanation:

Given that :

Standard deviation, s = 0.2

Sample variance, s² = 0.0653

Sample size, n = 10

Population variance = σ² = 0.04

We use the Chi square distribution :

(n - 1)s² / σ²

For P(s² ≥ 0.065)

(n - 1)s² / σ² = (10 - 1)*0.0653 / 0.04

(n - 1)s² / σ² = 0.585 / 0.04 = 14.625

P(s² ≥ 14.625) = 0.101 ( chi squee calculator).

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ratelena [41]
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anzhelika [568]

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