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3 years ago

For question 7, please pick 1,2,3 or 4 for the answer. Thank You!

Mathematics

1 answer:

andrew-mc [135]3 years ago

5 0

Answer:

The second answer, and possibly the first answer as also true.

She did run a test that would indicate its an unbalanced dice, but this wasn't tried out with a different person throwing the dice.

Step-by-step explanation:

This is because from the computer generator results we see 11 of the 25 values are estimating at 1/5 when we know dice are 1/6 and more than 1/2 show just under 1/5 which balances this to be 1/6

But there are 9/25 tests that showed values under 10 throws found a 6 in 9/25 events = 1/3 approx out of 1/10 of the throws, and 1/3 is still a higher value than 1/6 of the multiple throws so indicates 100 throws would not be enough to tell as we cannot possibly assume her results are comparable with a computer generator.As the computer generator completed 25 x 100 throws and have just compared only x10 in relation to 1/10 of the events of the generated computer. This showing 9 of the 25 (100) throw events in relation scores 1/3 of the results a 6. The answer is she would need to throw somewhere between 1000 and 3000 to compare to the computers results.

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Vesna [10]

The answer is A, all that was needed was to do the math.

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4 years ago

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If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

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$\tan\dfrac x2=\dfrac15$

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3 years ago

$7230 for 3.25% in 12 years

IceJOKER [234]

$2,819.70 is the answer

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3 years ago

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p

tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

A ∝ D^2

- Hence, LGP ∝ D^2

- Now compare the two diameters given:

LGP_1 ∝ (D_1)^2

LGP ∝ (D_2)^2

- Take a ratio of both:

LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute: LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

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3 years ago

Help stuck on question number 10

Zolol [24]

Answer:

y = 2/3x - 5 or in standard form 3y = 2x - 5

Step-by-step explanation:

Remember this fact: Parallel lines have the same slope

Step 1 Solve for y so that the equation is in the slope- intersect form

2x - 3y = 6

-3y = -2x + 6

-3y/-3 = -2x/-3 + 6/-3

y = 2/3 x -2

now we know the slope is 2/3 or

when the equation is in Standard form Ax + By = C you can use this fact: slope = - A/B so the slope = -2/-3 = 2/3

Remember the Parallel lines have the same slope

Find the y-intersect "b" use the slope = 2/3 and point (6, -1)

y = mx + b

-1 = 2/3(6) + b

-1 = 4 + b

-5 = b

Now write the equation of line that is parallel to the given line and passes through point (6, -1)

y = 2/3x - 5 or in standard form 3y = 2x - 5

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3 years ago