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4 years ago

THE VOLUME OF A SPHERE IS 38808cm^3. find the height of the cone

Mathematics

1 answer:

Mars2501 [29]4 years ago

8 0

I am not 100% positive but I think it is 4117.66

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Skye's two new aquariums each hold exactly 200 gallons of water. One aquarium will hold small fish and the other will hold large

Natasha_Volkova [10]

Answer=140 fish

200 gallons of water in the first aquarium.

200 gallons of water in the second aquarium.

The first will be for the small fishes.

5 small fish x fish
_____________=__________
10 gallons 200 gallons

cross multiply
10x=1000
divide both sides by 10
x=100 small fish

The first aquarium has 100 small fish.

The second aquarium holds large fish.

8 large fish x fish
____________=___________
40 gallons 200 gallons

cross multiply
40x=1600
divide both sides by 40
x=40 large fish

The second aquarium has 40 large fish.

100small fish+40 large fish=140 total fish

7 0

3 years ago

Read 2 more answers

X(x+1)+(x+1)(x+2)+(x+2)(x+3) solve the equation for x

viva [34]

Answer:

3x²+9x+8

Step-by-step explanation:

x²+x+x(x+2)+1(x+2)+x(x+3)+2(x+3)

x²+x+x²+2x+x+2+x²+3x+2x+6

x²+x²+x²+x+2x+x+3x+2x+2+6

3x²+9x+8

4 0

3 years ago

Read 2 more answers

The table below shows a correct representation that 15 pounds (lb) of sugar cost $3:

SIZIF [17.4K]

Answer: True

Step-by-step explanation: There are 15 pounds and 3 dollars. 15 over 3 is $1 for 5 pounds.

4 0

3 years ago

Our buyer client Heather just signed a purchase agreement for a $520,000 home. The LTVR is 60%. How much is Heather putting down

aivan3 [116]

312,000$.

I just found 60 percent of 520,000 so I hope that’s what you wanted, and I am truly sorry if it isn’t.

5 0

3 years ago

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with A random s

wel

Answer:

95%: (3278.354 ; 3270.083)

99% : (3221.646 ; 3278.354)

Step-by-step explanation:

Given :

Sample size, n = 12

Mean, xbar = 3250

Sample standard deviation = √1000

The 95% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.05, df=12-1 = 11 ;

Tcritical at 95% = 2.20

Hence,

Margin of Error = (2.20 * √1000/√12) = 20.083

Confidence interval : 3250 ± 20.083

Lower boundary = 3250 - 20.083 = 3229.917

Upper boundary = 3250 + 20.083 = 3270.083

2.)

The 99% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.01, df=12-1 = 11 ;

Tcritical at 99% = 3.106

Hence,

Margin of Error = (3.106 * √1000/√12) = 28.354

Confidence interval : 3250 ± 28.354

Lower boundary = 3250 - 28.354 = 3221.646

Upper boundary = 3250 + 28.354 = 3278.354

3 0

3 years ago