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3 years ago

HELP ME PLEASE !!!!!!!!

Mathematics

2 answers:

lord [1]3 years ago

5 0

Answer:

B and D

Step-by-step explanation:

A is false as it would be 3.08 when rounded.

C is false as it is rounded to the tenth not hundredth.

Bogdan [553]3 years ago

5 0

Answer:

B and D

pls make this the brainliest

Step-by-step explanation:

A is false as it would be 3.08 when rounded.

C is false as it is rounded to the tenth not hundredth.

hope this helps

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What is 26/32 in simplest form

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Solve the following system of equations:

PSYCHO15rus [73]

If we multiply the first equation by 2 we get
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but the second equation is
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2 years ago

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B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

1 year ago

Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minim

iren [92.7K]

Answer:

We accept H₀

Step-by-step explanation:

Normal Distribution

size sample n = 69

sample mean 18.94

standard deviation 8.3

Is a one tailed-test to the left we are traying of find out is we have enough evidence to say that the mean is less than 20 min.

1.-Test hypothesis H₀ ⇒ μ₀ = 20

Alternative hypothesis Hₐ ⇒ μ₀ < 20

2.- Critical value

for α = 0.1 we find from z Table

z(c) = - 1.28

3.-We compute z(s)

z(s) = [ ( μ - μ₀ ) / (σ/√n) ⇒ z(s) = [( 18.94 - 20 )*√69)/8.3]

z(s) = ( -1.06)*8.31/8.3

z(s) = - 1.061

4.- We compare

z(c) and z(s) -1.28 > -1.061

Then z(c) > z(s)

z(s) in inside acceptance region so we accept H₀

5 0

2 years ago