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3 years ago

Each of the triangular faces has an area of 32 m2.

Mathematics

1 answer:

WARRIOR [948]3 years ago

7 0

Answer:

128 if you don't count the square base

Step-by-step explanation:

because each side has an area of 32, there are 4 triangular sides in a square pyramid, therefore 4*32 = 128

I couldn't find the area of the base because there were no side lengths given.

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Two number cubes are rolled in a board game. What is the probability that both number cubes land on the same number?

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50% chance they will land on the same number and there’s also a 50% chance they won’t.

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The answer is Option B. It would be cheaper for Miguel to drive 200 miles with Option B than Option A.

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3 years ago

Read 2 more answers

Could someone help me understand how to solve this?

monitta

Answer:

Amount in 4% account: $550

Step-by-step explanation:

Use simple interest formula $I=P\cdot r\cdot t,$ where

I = interest

P = principal

r = rate (as decimal)

t = time.

4% account:

$P_1=x\\ \\r_1=0.04\\ \\t_1=1,$

then

$I_1=x\cdot 0.04\cdot 1=0.04x$

5% account:

$P_2=1,500-x\\ \\r_2=0.05\\ \\t_2=1,$

then

$I_2=(1,500-x)\cdot 0.05\cdot 1=0.05(1,500-x)$

You have earned $69.50 in total, so

$I_1+I_2=69.50\\ \\0.04x+0.05(1,500-x)=69.50\\ \\4x+5(1,500-x)=6,950\\ \\4x+7,500-5x=6,950\\ \\4x-5x=6,950-7,500\\ \\-x=-550\\ \\x=550\\ \\1,500-x=1,500-550=950$

Amount in 4% account: $550

Amount in 5% account: $950

8 0

3 years ago

On Monday, Jan made withdrawals of $15, $55, and $65 from her savings account. On the same day, her twin sister Julie made withd

jeyben [28]

Answer:

340

Step-by-step explanation:

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3 years ago

In tossing four fair dice, what is the probability of tossing, at most, one 3

11Alexandr11 [23.1K]

Answer-

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

Solution-

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

$P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48$ ( ∵ xxxx )

$P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38$ ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

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3 years ago