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2 years ago

Plz help sense nobody wants to help

Mathematics

2 answers:

Klio2033 [76]2 years ago

5 0

Answer:

3/4

Step-by-step explanation:

1/2 divided by 2/3

first you flip the second fraction then multiply.

bulgar [2K]2 years ago

4 0

Answer:

3/4

Step-by-step explanation:

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15 + brainliest help me please

Fiesta28 [93]

2f +4f +2 -3
combine the like terms to get 6f-1

so yes they are equivalent

2f+4f+2-3 at f=3 would also be 17

8 0

3 years ago

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A⋅a/\2⋅a/\3=?? I need help ASAP

wolverine [178]

Answer:

a^6

Step-by-step explanation:

a^1 * a^2 * a^3 = 1+2+3(add all power when same variable)

a^6

7 0

3 years ago

A study found that a 95% confidence interval for the mean μ of a particular population was computed from a random sample of 1200

kotykmax [81]

Answer:

95% confidence interval for the mean μ is (6,14)

The Population mean μ lies between ( 6, 14 )

Step-by-step explanation:

Explanation:-

Given random sample 'n' = 1200

95% confidence interval for the mean μ is determined by

$(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} +Z_{\alpha } \frac{S.D}{\sqrt{n} } )$

Level of significance = 95% 0r 0.05

Z₀.₀₅ = 1.96

$(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} +Z_{\alpha } \frac{S.D}{\sqrt{n} } )$ = 10 ± 4

Mean of the small sample = 10

95% of confidence intervals are

( 10 ±4 )

( 10 -4 , 10+4)

( 6 , 14 )

95% confidence interval for the mean μ lies between ( 6, 14 )

8 0

3 years ago

A rubber ball is dropped from the top of a hole. Exactly 20 seconds later, the sound of the rubber ball hitting bottom is heard.

gayaneshka [121]

9514 1404 393

Answer:

4192.9 ft

Step-by-step explanation:

We assume your falling-distance formula is supposed to be ...

s = 16t²

So, the time required to fall distance s is ...

t = √(s/16) = (1/4)√s

The time required for sound to travel distance s is ...

s = 1100t

t = s/1100

Then the sum of the time for the ball to fall and the time for the sound to travel back is ...

(1/4)√s + s/1100 = 20

275√s = 22000 -s . . . . . multiply by 1100 and subtract s

75625s = s^2 -44000s +484,000,000 . . . square both sides

s^2 -119,625s +484,000,000 = 0 . . . . put in standard form

s ≈ (1/2)(119,625 ±√12,374,140,625) = {4192.943, 115432.057}

Only the smaller of these two solutions makes any sense in this problem.

The hole is about 4192.9 feet deep.

_____

Additional comment

The distance equation for the falling object presumes a vacuum. The sound transmission presumes the presence of air, so the question setup is self-contradictory.

7 0

3 years ago

The stadium is being courted to host weekly UFC events, with each event projected to generate $90,000 in revenue. If the cost fo

luda_lava [24]

$50,000 Profit * 52 (Weeks in a Year) = $2,600,000

4 0

3 years ago

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