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Sunny_sXe [5.5K]

3 years ago

14

Barnabus caught 84 fish in 4 days. Each day he caught 10 more than on the day before. How many fish did he catch each of the 4 d

ays

1 answer:

balandron [24]3 years ago

5 0

X + x + 10 + x + 20 + x + 30 = 84.
4x + 60 = 84.
84 − 60 = 24
4x = 24.
24 / 4 = 6 = x.

Answer:
Day One: 6 fish
Day Two: 16 fish
Day Three: 26 fish
Day Four: 36 fish

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Answer:

0.0259,0.0406

Step-by-step explanation:

Given that in a poll taken in December 2012, Gallup asked 1006 national adults whether they were baseball fans: 48% said they were. Almost five years earlier, in February 2008, only 35% of a similar-size sample had reported being baseball fans.

a) p is the proportion of national adults who are baseball fans.

b) For 2012 poll, p=0.48

Std error = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.48*0.52}{1006} } \\=0.0158$

Margin of error at 90% = 1.645 * std error

= 0.0259

c) Margin of error is the admissible limit on either side of the mean where the sample mean can lie

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Consider the following equation. 4x + 2 = 8x - 5 Which equation and explanation could represent a step in finding the value of x

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Step-by-step explanation:

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2 years ago

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A bus covers 270km distance in 6 hours.

pentagon [3]

Answer:

(i)= 45 KPH, (ii) = 251 miles

Step-by-step explanation:

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1 year ago

(x3 + 2x² – 46x + 19) = (x + 8)<br> Synthetic division

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Answer:

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Step-by-step explanation:

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2 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

$\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0$

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

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Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

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3 years ago