You are solving 5 – 83 = y3. Which equation will you get if you multiply both sides by 3?

Q3. Stefan has 4 1/2 cups of oats. He uses 3 1/4 cups to make granola bars for

AleksAgata [21]

Answer:

Days can Stefan feed

Coco with the oats he has left is 5 days

Step-by-step explanation:

Total Oats = 4 1/2 cups

He uses 3 1/4 cups to make granola bars for

a camping trip

Remaining oats after making granola bars = Total oats - oats used for granola bars

= 4 1/2 - 3 1/4

.= 9/2 - 13/4

= 18-13/4

= 5/4 cups

Remaining oats after making granola bars is 5/4 cups

Coco needs 1/4 of a cup of oats each day, how many days can Stefan feed

Coco with the oats he has left?

Days can Stefan feed

Coco with the oats he has left = Remaining oats / 1/4 cups

= 5/4 cups ÷ 1/4 cups

= 5/4 × 4/1

= 20/4

= 5 days

Days can Stefan feed

Coco with the oats he has left is 5 days

5 0

3 years ago

Use the table to write a proportion.

umka21 [38]

Answer:

it's probably ours it's just a lucky guess

Step-by-step explanation:

7 0

2 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

5 0

3 years ago

Provide an example of optimization problem

Mashutka [201]

Answer:

a. Convex solutions ,GO Methods

b. market efficiency

Explanation :

Step-by-step explanation:

A globally optimal solution is one where there are no other feasible solutions with better objective function values. A locally optimal solution is one where there are no other feasible solutions "in the vicinity" with better objective function values. You can picture this as a point at the top of a "peak" or at the bottom of a "valley" which may be formed by the objective function and/or the constraints -- but there may be a higher peak or a deeper valley far away from the current point.

In convex optimization problems, a locally optimal solution is also globally optimal. These include LP problems; QP problems where the objective is positive definite (if minimizing; negative definite if maximizing); and NLP problems where the objective is a convex function (if minimizing; concave if maximizing) and the constraints form a convex set. But many nonlinear problems are non-convex and are likely to have multiple locally optimal solutions, as in the chart below. (Click the chart to see a full-size image.) These problems are intrinsically very difficult to solve; and the time required to solve these problems to increases rapidly with the number of variables and constraints.

GO Methods

Multistart methods are a popular way to seek globally optimal solutions with the aid of a "classical" smooth nonlinear solver (that by itself finds only locally optimal solutions). The basic idea here is to automatically start the nonlinear Solver from randomly selected starting points, reaching different locally optimal solutions, then select the best of these as the proposed globally optimal solution. Multistart methods have a limited guarantee that (given certain assumptions about the problem) they will "converge in probability" to a globally optimal solution. This means that as the number of runs of the nonlinear Solver increases, the probability that the globally optimal solution has been found also increases towards 100%.

Where Multistart methods rely on random sampling of starting points, Continuous Branch and Bound methods are designed to systematically subdivide the feasible region into successively smaller subregions, and find locally optimal solutions in each subregion. The best of the locally optimally solutions is proposed as the globally optimal solution. Continuous Branch and Bound methods have a theoretical guarantee of convergence to the globally optimal solution, but this guarantee usually cannot be realized in a reasonable amount of computing time, for problems of more than a small number of variables. Hence many Continuous Branch and Bound methods also use some kind of random or statistical sampling to improve performance.

Genetic Algorithms, Tabu Search and Scatter Search are designed to find "good" solutions to nonsmooth optimization problems, but they can also be applied to smooth nonlinear problems to seek a globally optimal solution. They are often effective at finding better solutions than a "classic" smooth nonlinear solver alone, but they usually take much more computing time, and they offer no guarantees of convergence, or tests for having reached the globally optimal solution.

5 0

3 years ago

out of 30 students surveyed 17 have a dog based on these results predict how many of the 300 students in the school have a dog

Ira Lisetskai [31]

The whole school would have 170 students that have a dog

7 0

3 years ago

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