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3 years ago

HELP ME NOOWW PLEASE

Mathematics

1 answer:

erma4kov [3.2K]3 years ago

6 0

Answer:

Step-by-step explanation:

what i did is i subtract 24 to 11 equals to 13 then add 13 to 72 .

i hope this helps

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Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

kondor19780726 [428]

Answer:

The lower class boundary for the first class is 140.

Step-by-step explanation:

The variable of interest is the length of the fish from the North Atlantic. This variable is quantitative continuous.

These variables can assume an infinite number of values within its range of definition, so the data are classified in classes.

These classes are mutually exclusive, independent, exhaustive, the width of the classes should be the same.

The number of classes used is determined by the researcher, but it should not be too small or too large, and within the range of the variable. When you decide on the number of classes, you can determine their width by dividing the sample size by the number of classes. The next step after getting the class width is to determine the class intervals, starting with the least observation you add the calculated width to get each class-bound.

The interval opens with the lower class boundary and closes with the upper-class boundary.

In this example, the lower class boundary for the first class is 140.

6 0

3 years ago

Does the table show a proportional relationship? If so, what is the value of y when x is 11

prisoha [69]

Answer:

Yes, the table shows a proportional relationship

The value of y when x is 11 is 1331

Step-by-step explanation:

Let us check the relation between x and y in the table

∵ x = 4 and y = 64

∵ 64 = 4³

∴ y = x³

∵ x = 5 and y = 125

∵ 125 = 5³

∴ y = x³

∵ x = 6 and y = 216

∵ 216 = 6³

∴ y = x³

∵ x = 10 and y = 1000

∵ 1000 = 10³

∴ y = x³

∵ All the values on the table give the same relation

∴ x and y are proportion

∴ The table shows a proportional relationship

∵ y = x³

∵ x = 11

∴ y = (11)³

∴ y = 1331

∴ The value of y when x is 11 is 1331

8 0

3 years ago

Read 2 more answers

Help with q25 please. Thanks.

Westkost [7]

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

$f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\$

Now apply the derivative to that to get the second derivative

$f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\$

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

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

Let's compute dy/dx. We'll use f(x) as defined earlier.

$y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\$

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\$

If you need a refresher on the quotient rule, then

$\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\$

where P and Q are functions of x.

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

This then means

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\$