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2 years ago

8

La hormiguita Fidencia se encontro una hojita, la cargo y se llevo al hormiguero. Le dio a su amiguita 1/5 y ella se comió 1/3 d

e la hojita. ?Quién comió más?

1 answer:

vredina [299]2 years ago

7 0

Answer:

The ant Fidencia ate more

Step-by-step explanation:

The question in English

The ant Fidencia found a leaf, the charge and took the anthill. She gave her friend 1/5 and she ate 1/3 of the leaf. Who ate more?

Her friend

Multiply (1/5) by (3/3)

(1/5)*(3/3)=3/15

Ant Fidencia

Multiply (1/3) by (5/5)

(1/3)*(5/5)=5/15

Compare

5/15 > 3/15 -----> two fractions with the same denominator the fraction with the highest numerator is greater

therefore

The ant Fidencia ate more

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Margie has a $50 budget to purchase a $45 pair of boots. If there is an 8% sales tax rate, then how much under budget will Margi

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1.40 under budget I think, hope this is some help

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2 years ago

Read 2 more answers

What pages will be favored for the given search? Search terms: Michael OR Jordan A. Pages about Michael Jordan B. Pages about Mi

ad-work [718]

Answer:

A. Pages about Michael Jordan.

Step-by-step explanation:

Michael Jordan, MJ, is a great basketball player, and has achieved one the best records in his career. He is a National Basketball Association (NBA) player with great skills and energy.

From the search item, the two names Michael OR Jordan is for one person, Michael Jordan. The search would combine the two names because it is a well known one and give an output on Michael Jordan. Thus, the pages that would be favored are pages about Michael Jordan.

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2 years ago

Use lagrange multipliers to find the points on the given surface that are closest to the origin. y2 = 64 + xz

Inessa [10]

The distance between an arbitrary point on the surface and the origin is

$d(x,y,z)=\sqrt{x^2+y^2+z^2}$

Recall that for differentiable functions $g(x)$ and $h(x)$ , the composition $g(h(x))$ attains extrema at the same points that $h(x)$ does, so we can consider an augmented distance function

$D(x,y,z)=x^2+y^2+z^2$

The Lagrangian would then be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)$

We have partial derivatives

$\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}$

Set each partial derivative to 0 and solve the system to find the critical points.

From the second equation it follows that either $y=0$ or $\lambda=-1$ . In the first case we arrive at a contradiction (I'll leave establishing that to you). If $\lambda=-1$ , then we have

$\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0$

This means $y^2=64\implies y=\pm8$

so that the points on the surface closest to the origin are $(0,\pm8,0)$ .

8 0

2 years ago

Gavin likes biking. He never misses a chance to go for a ride when the weather is nice. This week his goal is to bike about 65 t

andreyandreev [35.5K]

Answer:

How far should he ride on each of the four days to reach his goal?

1st day: $8$ miles

2nd day: $12$ miles

3rd day: $18$ miles

4th day: $27$ miles

Step-by-step explanation: As the problem says, $x$ is the number of miles he rides on the first day. Let's start off with that.

1st day: $x$ miles

He want to ride 1.5 times as far as he rode the day before... no 1.5 more, but 1.5 <em>times</em> as far as he rode the day before; you would multiply 1.5 with the previous day's length.

2nd day: $1.5x$

Then you multiply $1.5$ to $1.5x$ to get the third day's.

3rd day: $2.25x$

4th day: $3.375x$

----------------------------

Phew! Gavin wants to ride a total of 65 miles over these four days, so if Gavin added all the miles of the four days, he should get 65...

1st+2nd+3rd+4th=65

$x+1.5x+2.25x+3.375x=65$

$8.125x=65$

$\frac{8.125x}{8.125}=\frac{65}{8.125}$

$x=8$

Yes! Now that we've got the hard part done... substitute 8 for ever single $x$ .

1st day: $8$ miles

2nd day: $1.5(8)=12$ miles

3rd day: $2.25(8)=18$ miles

4th day: $3.375(8)=27$ miles

-------------------------

Checking my answer:

Just add the miles!

$8+12+18+27=65$

$65=65$

✓

-----------------------

Hope that helps! :D

5 0

2 years ago

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0

2 years ago