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4 years ago

Reduce 738 and 7/32 by 48 and 9/32 .

Mathematics

1 answer:

Helga [31]4 years ago

8 0

Answer: 689 + 30/32

Step-by-step explanation:

A reduction of A by B, means that we need to solve:

A - B.

So if we want to reduce 738 and 7/32 by 48 and 9/32 we have to solve:

(738 + 7/32) - (48 + 9/32)

we can separate it into whole and fraction:

(738 - 48) + (7/32 - 9/32)

690 - 2/32

But we usually don't want a negative fraction, so we can use that:

1 = 32/32

690 - 2/32 = 689 + 1 - 2/32 = 689 +32/32 - 2/32 = 689 + 30/32

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

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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.

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3 years ago

A student compiled a list of daily low temperatures (in °F) that she observed at her college last semester. The

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3 years ago

Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for aluminum alloy sheets of a parti

neonofarm [45]

Answer:

a1) $\mu_{\bar{X}} = 70$

a2) $\sigma_{\bar{X}} = 0.4$

b1) $\mu_{\bar{X}} = 70$

b2) $\sigma_{\bar{X}} = 0.2$

c) X is more likely to be within 1 GPa of 70 GPa in the random sample of part b because of the largeness in sample size and less scattering of data

Step-by-step explanation:

Mean value, $\mu = 70$

Standard deviation, $\sigma = 1.6$

a1) sample size, n = 16

Mean of the sampling distribution of the sample mean = mean value, i.e.

$\mu_{\bar{X}} = \mu\\\mu_{\bar{X}} = 70$

a2) The standard deviation of the sampling distribution of the sample mean

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n} } \\\sigma_{\bar{X}} = \frac{1.6}{\sqrt{16} }\\\sigma_{\bar{X}} = 0.4$

b1) For sample size, n = 64

Mean of the sampling distribution of the sample mean = mean value, i.e.

$\mu_{\bar{X}} = \mu\\\mu_{\bar{X}} = 70$

a2) The standard deviation of the sampling distribution of the sample mean

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n} } \\\sigma_{\bar{X}} = \frac{1.6}{\sqrt{64} }\\\sigma_{\bar{X}} = 0.2$

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Step-by-step explanation:

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3 years ago

Read 2 more answers

The school that Scott goes to is selling tickets to a fall musical. On the first day of ticket sales the school sold 12 adult ti

harina [27]

Answer:

The price of one adult ticket = $13

The price of one student ticket = $4

Step-by-step explanation:

Let the price of 1 adult ticket = x

Let the price of 1 student ticket = y

On the first day of ticket sales the school sold 12 adult tickets and 10 student tickets for a total of $196.

12 × x + 10 × y = $196

12x + 10y = 196....... Equation 1

The school took in $59 on the second day by selling 3 adult tickets and 5 student tickets

3 × x + 5 × y = $59

3x + 5y = 59.......... Equation 2

Using Elimination method

We eliminate y, by Multiplying equation 1 by 5 and Equation 2 by 10

12x + 10y = 196....... Equation 1 × 5

3x + 5y = 59.......... Equation 2 × 10

60x + 50y = 980....... Equation 3

30x + 50y = 590......... Equation 4

Subtract Equation 4 from Equation 3

30x = 390

x = 390/30

x = $13

Therefore, the price of one adult ticket = $13

Remember: x = $13

3x + 5y = 59.......... Equation 2

Substitute

3(13) + 5y = 59

39 + 5y = 59

5y = 59 - 39

5y = 20

y = 20/5

y = $4

Therefore, the price of one student ticket = $4

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3 years ago