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2 years ago

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A bowl shaped like a sphere has a radius of 15 cm. Which is closest to the amount of air needed to fill the ball?

2 answers:

zhannawk [14.2K]2 years ago

6 0

The formula for a sphere is 4/3*pi*r^3 so we replace 15 with 5 since 15 is the radius and after multiplying everything you get 14,137.17 which can be rounded the 14,130 which is B

SIZIF [17.4K]2 years ago

4 0

I believe it is D not quite sure

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Irina-Kira [14]

I'd say B. but that's only because I divided the number of times he flipped the coin by the number of sides it has, 2, and came up with 246. The closest number on here to that is 231 or B. I hope I helped.

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A roof rises 24 feet vertically and 14 feet horizontally. What is the slope of the roof?

mixas84 [53]

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Step-by-step explanation:

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1 year ago

Help plzzz thank you !

VashaNatasha [74]

Answer:

a. Ameribank-$15,157.50

b. Capital Two-$4,646.25

Step-by-step explanation:

a. Tad's savings is $15,000, we calculate his total amount at the end of the year for each bank:

#Ameribank

$A=P+I=P+PRT\\\\=15000+15000\times 0.0105\times 1\\\\=\$15,157.50$

#Huffington( we use the effective rate to calculate the compound amount):

$i_m=(1+i/m)^m-1\\\\=(1+0.0095/12)^[12}-1=0.009541\\\\A=P(1+i_m)^n\\\\=15000(1.009541)^1\\\\=\$15,143.12$

#Sixth-Third, Take 1 yrs=52 weeks:

$i_m=(1+i/m)^m-1\\\\=(1+0.01/52)^{52}-1=0.01005\\\\A=15000(1.01005)^1\\\\=\$15,150.74$

#Hence, Ameribank is the best option as his money grows to $15,157.50 which is greater than all the remaining two options.

b. We use the compound interest formula $A=P(1+r/n)^{nt}$ to determine which bank gives the best option:

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$A=P(1+r/n)^{nt}\\\\=4000(1+0.0375/12)^{12\times4}\\\\=\$4,646.25$

#J.C Morgan, t=2, r=3.55% n=12

$A=P(1+r/n)^{nt}\\\\=4000(1+0.0355/12)^{12\times 2}\\\\=\$4,293.87$

#Silverman Slacks, n=12,t=3, r=3.65%

$A=P(1+r/n)^{nt}\\\\=4000(1+0.0365/12)^{12\times3}\\\\=\$4,462.14$

We compare the investment amounts after t years:

$Capital>Silver>Morgan=4646.25>4462.14>4293.87$

Hence, Capital two is the best option with an investment amount of $4,646.25

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