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Reil [10]
3 years ago
8

99 POINTS! PLZ HELP!!

Mathematics
2 answers:
yan [13]3 years ago
7 0

Answer:

If you are on usatestprep, the answer is C.m might cause n, or n might cause m.

bixtya [17]3 years ago
6 0

Answer:

The answer is C. M might cause N, N might cause M

Step-by-step explanation:

The slope of the line produced by Lisa’s graphing calculator has nothing to do with whether m causes n, n causes m, or neither (One exception to this would be if the slope were 0. This would imply no correlation, and thus, no causation.) Since the correlation coefficient (r) is 0.962 in both cases, a correlation probably exists between m and n. However, correlation does not imply causation, nor does it imply a lack of causation. Therefore, the conclusion of Lisa’s that is the most valid is m might cause n, or n might cause m.

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Write an equation, and express the product in standard form.
rosijanka [135]

Answer: 3 copies of 5 tenths = 3*0.5 = 1.5

Step-by-step explanation:

Ok, if we have:

"3 copies of a number B"

This is written as:

3*B

And we have that the number B is "5 tenths"

Now, a tenth is the first digit after the decimal point.

Then 5 tenths = 5*0.1 = 0.5

B = 0.5

3 copies of 5 tenths = 3*0.5 = 1.5

7 0
3 years ago
A croissant shop produces two products: bear claws (B) and almond filled croissants (C). Each bear claw requires 6 oz of flour,
Kruka [31]

Answer:

letter A: B = 400; C = 1000; Max Z = $380

Step-by-step explanation:

bear claws (B) need 6 oz of flour (f), 1 oz of yeast (y), 2 ts of paste (p)

croissants (C) need 3 oz of flour (f), 1 oz of yeast (y), 4 ts of paste (p)

putting that information in equations, we have:

B = 6f + y + 2p

C = 3f + y + 4p

The total number of resources (R) are:

R = 6600f + 1400y + 4800p

Let's call M our total profit, "k1" the number of B produced and "k2" the number of C produced.

So, we can state that:

M = k1*0.2 + k2*0.3

The number of resources R2 will demand is calculated like this:

R2 = k1*B + k2*C = (6k1+3k2)f + (k1+k2)y + (2k1+4k2)p

using R2 and R, we can make some inequations:

6k1+3k2 <= 6600 -> 2k1+k2 <= 2200

k1+k2 <= 1400

2k1+4k2 <= 4800 -> k1+2k2 <= 2400

if we try to maximize k2 (as it worths more), we will have k1 = 0 and k2 = 1200 (limited by p), but looking at the resources R, we will still have resources to use (f and y). Looking at B and C expressions, we see that removing one C gives enough 'p' to make 2 B, which is a good trade (as 2B worths 0.4 cents, and 1C worths 0.3 cents). we have 200 'y' remaining, so doing this 200 times give us k1 = 400 and k2 = 1000, and the only resource remaining will be some of 'f'.

calculating the profit M, we have:

M = 400*0.2 + 1000*0.3 = 380$

the right answer is letter A.

3 0
3 years ago
Mike owns a deli. He orders the deli meats listed in the table. Which deli meat is the most expensive per pound? Need Step by St
UkoKoshka [18]

Answer:

Im trying to say the same thing

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Someone plz help because I am using my last amount of points and I don't want it to be wasted from no answer.I need help with qu
ser-zykov [4K]
Wheres the questions???
7 0
2 years ago
1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.
xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}

(If you were to plot the actual curve, you would have both y=x^{2/3} and y=-x^{2/3}, but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx

We have

y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}

Then in the integral,

\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx

Substitute

u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx

This transforms the integral to

\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du

and computing it is trivial:

\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)

We can simplify this further to

\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}

7 0
3 years ago
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