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2 years ago

3 out of 20 boxes are damaged what percent is damaged?

2 answers:

8 0

'out of' generally indicates division

So 3 out of 20 boxes, is basically the same as 3 ÷ 20.

3 ÷ 20 = 0.15

To work out the percentage, you have to multiply this value by 100 (per CENT, cent means 100 in Latin).

0.15 × 100 = 15%.

The answer is 15% of the boxes are damaged.

Goshia [24]2 years ago

7 0

The answer is 15%.
Just divide 3 and 20.

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Which expression is equivalent to 4/9x7?

LiRa [457]

Is there an answer key?

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2 years ago

If 5 gallons cost $ 6.60 how much do 29 gallons of gasoline cost

ikadub [295]

Answer:

Should be 38.28

Step-by-step explanation:

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2 years ago

How tall is a building that casts a 78-ft shadow when the angle of elevation of the sun is 30 degrees?

Marta_Voda [28]

Answer: $45.03\ ft$

Step-by-step explanation:

Given

The length of the shadow is 78 ft

the angle of elevation is $\theta =30^{\circ}$

Suppose the height of the building is $h$

From the figure, we can write

$\tan 30^{\circ}=\dfrac{h}{78}\\\\h=78\tan 30^{\circ}\\\\h=45.03\ ft$

The height of the building is 45.03 ft

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2 years ago

The following triangles are similar by SAS~

svet-max [94.6K]

Step-by-step explanation:

sides:91/12=42/26

angles:LMN=LBA and MNL=BAL

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2 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago