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3 years ago

Will give brainiest to best answer

Mathematics

1 answer:

cluponka [151]3 years ago

3 0

Answer:

1.V=288, A=288

2. V=460, A=494

3. V=672

4. A=898

Step-by-step explanation:

V= l×b×h

288

A=2(lb+lh+bh)

=288

V¹= larger cuboid =320

V²=smaller cuboid=140

V=460

A¹=328

A²=166

A=494

V=(Area of cross section)×length

=672

Area of cube=l³

³√294=6.65

add 3

9.65³=898

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Determine if this is a linearly dependent or independent set: (V, V, V ) where syon V = (1,2,2,-1), V = (4,9,9,-4), v = (5,8,9,-

marishachu [46]

Answer:

This is a linearly independent set.

Step-by-step explanation:

We have these following vectors:

$V_{1} = (1,2,2,-1)$

$V_{2} = (4,9,9,-4)$

$V_{3} = (5,8,9,-5)$

In a set of 3 vectors, if one of these vectors can be written as a linear combination of the 2 other vectors, they are linearly dependent. Otherwise, they are linearly independent.

We can verify this by solving the following system:

$xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)$

If the only solution is $(x,y,z) = (0,0,0)$ , they are L.I. Otherwise, they are L.D.

Solution:

$xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)$

$x(1,2,2,-1) + y(4,9,9,-4) + z (5,8,9,-5) = (0,0,0,0)$

We have the following system of equations:

$x + 4y + 5z = 0$

$2x + 9y + 8z = 0$

$2x + 9y + 9z = 0$

$-x -4y - 5z = 0$

I am going to solve this by the row-reduction of the augmented matrix.

This system has the following augmented matrix:

$\left[\begin{array}{cccc}1&4&5&0\\2&9&8&0\\2&9&9&0\\-1&-4&-5&0\end{array}\right]$

To reduce the first row, i am going to make these following operations:

$L_{2} = L_{2} - 2L_{1}$

$L_{3} = L_{3} - 3L_{1}$

$L_{4} = L_{4} + L_{1}$

So the augmented matrix now is:

$\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&1&-1&0\\0&0&0&0\end{array}\right]$

Now I reduce the second row, doing:

$L_{3} = L_{3} - L_{2}$

So the matrix is:

$\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&0&1&0\\0&0&0&0\end{array}\right]$

Now we can solve the system:

From the third line, we have that

$z = 0$

From the second line:

$y - 2z = 0$

$y - 2(0) = 0$

$y = 0$

From the first line

$x + 4y + 5z = 0$

$x + 4(0) + 5(0) = 0$

$x = 0$

The only solution for this system is $(x,y,z) = (0,0,0)$ . This means that we have a linearly independent set.

3 0

3 years ago

Canine Crunchies Inc. (CCI) sells bags of dog food to warehouse clubs. CCI uses an automatic filling process to fill the bags. W

KatRina [158]

Answer:

a) 0.9999 = 99.99% probability that a filled bag will weigh less than 49.5 kilograms

b) 0.0018 = 0.18% probability that a randomly sampled filled bag will weight between 48.5 and 51 kilograms.

c) 46.24 kilograms

d) The standard deviation would have to be of 3.41 kilograms.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 45 kilograms and a standard deviation of 1.2 kilograms.

This means that $\mu = 45, \sigma = 1.2$

a. What is the probability that a filled bag will weigh less than 49.5 kilograms?

This is the pvalue of Z when X = 49.5. So

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

$Z = \frac{49.5 - 45}{1.2}$

$Z = 3.75$

$Z = 3.75$ has a pvalue of 0.9999

0.9999 = 99.99% probability that a filled bag will weigh less than 49.5 kilograms

b. What is the probability that a randomly sampled filled bag will weight between 48.5 and 51 kilograms?

This is the pvalue of Z when X = 51 subtracted by the pvalue of Z when X = 48.5.

X = 51

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

$Z = \frac{51 - 45}{1.2}$

$Z = 5$

$Z = 5$ has a pvalue of 1

X = 48.5

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

$Z = \frac{48.5 - 45}{1.2}$

$Z = 2.92$

$Z = 2.92$ has a pvalue of 0.9982

1 - 0.9982 = 0.0018

0.0018 = 0.18% probability that a randomly sampled filled bag will weight between 48.5 and 51 kilograms.

c. What is the minimum weight a bag of dog food could be and remain in the top 15% of all bags filled?

This is the 100 - 15 = 85th percentile, which is X when Z has a pvalue of 0.85. So X when Z = 1.037

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

$1.037 = \frac{X - 45}{1.2}$

$X - 45 = 1.037*1.2$

$X = 46.24$

46.24 kilograms.

d. CCI is unable to adjust the mean of the filling process. However, it is able to adjust the standard deviation of the filling process. What would the standard deviation need to be so that no more than 2% of all filled bags weigh more than 52 kilograms?

X = 52 would have to be the 100 - 2 = 98th percentile, which is X when Z has a pvalue of 0.98, so X when Z = 2.054. We would need to find the value of $\sigma$ for this.

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

$2.054 = \frac{52 - 45}{\sigma}$