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2 years ago

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An expression is given: 2 open parentheses square root of k minus 1 close parenthesis plus square root of 8. If on adding negati

ve 6 square root of 2to the expression results in a rational number, what is the value of k?

1 answer:

neonofarm [45]2 years ago

3 0

Answer:

Possible value of k is √2

Step-by-step explanation:

The information given are;

The expression, 2·(√k - 1) + √8 to which may be added -6·√2 to obtain a rational number, we therefore have;

2·(√k - 1) + √8 - 6·√2 = R

Therefore, simplifying gives;

2·√k - 2 + 2·√2 - 6·√2 = 2·√k - 2 - 4·√2 = R

2·√k - 2 - 4·√2 + 2= R + 2 = R

2·√k - 2+ 2 - 4·√2 = R

2·√k - 2+ 2 - 4·√2 = R

2·√k + 0 - 4·√2 = 2·√k - 4·√2 = 2·(√k - 2·√2) = R

(√k - 2·√2) = R/2 = R

Therefore, √2 is a factor of √k such that √k - 2·√2 = R

Which gives k = x·√2, where x = a rational number

When x = 1, k = √2.

Therefore, a possible value of k is √2

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Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b

horrorfan [7]

Answer:

a. $X\sim N(\mu = 6.1, \sigma = 1.9)$ b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is $z_{1} = (6.4-6.1)/1.9 = 0.1579$ and the z-score related to 7 kg is $z_{2} = (7-6.1)/1.9 = 0.4737$ , we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0

2 years ago

a total of 17100 seats are still available for the next hockey game.if 62% of the tickets are sold how many seats are in the are

BartSMP [9]

If 62% of the seats are sold;

100-62= 38% seats still available.

17100=38% seats available

17100/38= seats in 1%
450 seats

In 100%,
450×100=450000 seats total.

Hope I helped :)

3 0

2 years ago

Find a matrix representation of the transformation L(x, y) = (3x + 4y, x − 2y).

mario62 [17]

Answer:

$\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}3&1\\4&-2\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]$

Step-by-step explanation:

The general matrix representation for this transformation would be:

$\left[\begin{array}{cc}x&y\end{array}\right] * A = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]$

As the matrix A should have the same amount of rows as columns in the firs matrix and the same amount of columns as the result matrix it should be a 2x2 matrix.

$\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]$

Solving the matrix product you have that the members of the result matrix are:

3x+4y = a*x + c*y

x - 2y = b*x + d*y

So the matrix A should be:

$\left[\begin{array}{cc}3&1\\4&-2\end{array}\right]$

8 0

2 years ago

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

2 years ago

Employees in the marketing department of a large regional restaurant chain are researching the amount of money that households i

kap26 [50]

Answer:

The <em>z</em>-score for the group "25 to 34" is 0.37 and the <em>z</em>-score for the group "45 to 54" is 0.25.

Step-by-step explanation:

The data provided is as follows:

25 to 34 45 to 54

1329 2268

1906 1965

2426 1149

1826 1591

1239 1682

1514 1851

1937 1367

1454 2158

Compute the mean and standard deviation for the group "25 to 34" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [1329+1906+...+1454]=\frac{13631}{8}=1703.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1086710.875}=394.01$

Compute the <em>z</em>-score for the group "25 to 34" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1703.875}{394.01}=0.3734\approx 0.37$

Compute the mean and standard deviation for the group "45 to 54" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [2268+1965+...+2158]=\frac{14031}{8}=1753.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1028888.875}=383.39$

Compute the <em>z</em>-score for the group "45 to 54" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1753.875}{383.39}=0.25333\approx 0.25$

Thus, the <em>z</em>-score for the group "25 to 34" is 0.37 and the <em>z</em>-score for the group "45 to 54" is 0.25.

8 0

2 years ago