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2 years ago

3

Radioactive waste is to be disposed of in fully enclosed lead cuboid boxes of inner volume V. The base of the box has the dimens

ions in the ratio n/1.Find the minimum inner surface area of the box.

2 answers:

Shalnov [3]2 years ago

8 0

The inner surface of the box is 1

Bond [772]2 years ago

5 0

The correct answer is 1

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If n - 2y = 3y- n/m,find the value of n when y = 5 and m = -3

Gnom [1K]

Answer:

37 1/2

Step-by-step explanation:

n - 2y = 3y- n/m,

find the value of n when y = 5 and m = -3

--------

n-2*5=3*5-n/-3

n-10=15+n/3

n-n/3=15+10

2n/3=25

2n=3*25

n=75/2

n=37 1/2 or 37.5

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2 years ago

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The angle whose sine is 0.39581 is

Lorico [155]

The angle whose sine is 0.39581 is 23.31650126° (round it how you want).
To calculate this, you need to do the inverse sine of 0.39581.

Inverse sine looks like $sin^{-1}$ , however, it is not the sine of the angle to the power of -1.

4 0

2 years ago

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Jeri finds a pile of money with at least $\$200$. If she puts $\$50$ of the pile in her left pocket, gives away $\frac23$ of the

Marizza181 [45]

Answer:

The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350

Step-by-step explanation:

Here we have, pile of money ≥ $200

Amount in put the left pocket = $50

Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100

Amount put in right pocket = ≥ $150 - $100 ≥ $50

Total amount remaining with Jeri = $50 +≥ $50 ≥ $100

Also original pile - $200 < $100

Therefore where maximum amount given away to have more money = $200 we have

2/3× (original pile - 50) = $200

Maximum amount for original pile = $350

Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.

6 0

2 years ago

Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll

JulijaS [17]

Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

$Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So, the possibility of getting a total of 7 = $\frac{6}{36}=\frac{1}{6}$

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = $\frac{2}{36}$ = $\frac{1}{18}$

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = $\frac{28}{36}$ = $\frac{7}{9}$

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

$\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11$

$\frac{11}{3}+\frac{11}{3}-\frac{77}{9}$

$\frac{22}{3}-\frac{77}{9}$

$\frac{66-77}{9}$

$-\frac{11}{9}$

$-1.22$

Therefore the expected payoff for this game is -$1.22.

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2 years ago

A nationwide survey of 100 boys and 50 girls in the first grade found that the daily average number of boys who are absent from

dusya [7]

Te difference of 2 standard deviation of a population n1 & n2 is given by the formula:
sigma (difference)=√(sigma1/n1 + sigma2/n2), Plug:

sigma(d)= √(49/100 + 36/50)

Sigma(d=difference) =1.1

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2 years ago

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