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3 years ago

What is the volume of a prism that is 11×13×21?

Mathematics

1 answer:

posledela3 years ago

8 0

Answer:

3003 units^3.

Step-by-step explanation:

Just multiply the dimensions :

11 * 13 * 21

= 143 * 21

= 3003 units^3.

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It is now quarter to ten. What time will it be in four hours and fifty-one minutes?

Veseljchak [2.6K]

Answer:it would be 2:22

Step-by-step explanation:

If there’s 25 minutes till 10 so therefore it would be 9:35 and since it’s 9:35 four hours and 51 minutes from then would be 2:22 because the minutes would add up to 82 minutes and four hours so your answers to 2:22

7 0

2 years ago

How do I solve this problem

erma4kov [3.2K]

10√3 (-7 - √5)

10√3 × -7 + 10√3 × -√5

-70√3 + 10√3 × -√5

-70√3 - 10√3√5

-70√3 -10√3 × 5

-70√3 - 10√15

hope that helps, God bless!

6 0

3 years ago

Solve by the matrix method , x+y/5 -1=y-x

PIT_PIT [208]

10x - 4y = 5 (if it requires shorten)

Step-by-step explanation:

x + y/5 - 1 = y - x

<=> (5x + y - 1.5)/5 = 5(y - x)/5

<=> 5x + y - 5 = 5y - 5x

<=> 5x + 5x + y - 5y = 5

<=> 10x - 4y = 5

6 0

3 years ago

The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

3 years ago

A triangle has vertices at A(–2, 4), B(–2, 8), and C(6, 4). If A prime has coordinates of (–0. 25, 0. 5) after the triangle has

babunello [35]

The true statements about the dilation are as follows;

The scale factor is One-eighth

The coordinates of the C prime are (0.75, 0.5).

The coordinates of B prime are (–0.25, 1).

Given

A triangle has vertices at A(–2, 4), B(–2, 8), and C(6, 4).

If A prime has coordinates of (–0. 25, 0. 5) after the triangle has been dilated with a center of dilation about the origin.

<h3>What is transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. If an object is transformed then all its points are also transformed. Types of transformation are reflection, dilation, rotation, and translation.

Dilation is the enlargement or reduction in the size of an object. If a point X(x, y) is dilated about the origin by a factor k, its new location is X'(kx, ky).

Triangle ABC with vertices at A(–2, 4), B(–2, 8), and C(6, 4) is dilated to give A'(-0.25, 0.5)

kA = A'

(-2k, 4k) = (-0.25, 0.5)

-2k = -0.25, hence k = 1/8

Therefore the scale factor is 1/8 (one- eight)

The new location of the vertices is B'(-0.25, 1), C'(0.75, 0.5)

To know more about Transformation click the link given below.

brainly.com/question/11709244

3 0

2 years ago