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3 years ago

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The sale bin in a clothing store contains an assortment of t-shirts in different sizes. There are 7 small, 8 medium, and 4 large

shirts. Alan is looking for a large shirt. He starts grabbing shirts one at a time and checking the size. After he checks each shirt, he leaves it outside the bin. What is the probability that at least one of the first four shirts he checks is a large?

1 answer:

umka21 [38]3 years ago

4 0

Answer:

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

Step-by-step explanation:

We have 7 small shirts, 8 medium shirts and 4 large shirts

Total number of shirts = 7 + 8 + 4 = 19 shirts

The probability that at least one of the first four shirts he checks is a large is given by

P(at least 1 large) = 1 - P(no large)

So first we need to find the probability that the none of the first four shirts he checks are large.

For the first check, there are 15 small and medium shirts and total 19 shirts so,

15/19

For the second check, there are 14 small and medium shirts and total 18 shirts left so,

14/18

For the third check, there are 13 small and medium shirts and total 17 shirts left so,

13/17

For the forth check, there are 12 small and medium shirts and total 16 shirts left so,

12/16

the probability of not finding the large shirt is,

P(no large) = 15/19*14/18*13/17*12/16

P(no large) = 0.352

Therefore, the probability of finding at least one large shirt is,

P(at least 1 large) = 1 - P(no large)

P(at least 1 large) = 1 - 0.352

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

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<h3>What is a decimal multiplier?</h3>

Increases of a% or decreases of a% re represented by decimal values, as follows:

The equivalent multiplier for an increase of a% is given by: $\frac{100 + a}{100}$

The equivalent multiplier for an decrease of a% is given by: $\frac{100 - a}{100}$

In this problem:

The price at the beginning of January was of a.

Increase of 20% in January, hence $1.2a$ .

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The price of gasoline at the end of April was the same as it had been at the beginning of January, hence:

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3 years ago

Which two equations are true?(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)=6.6×10−10 (6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.

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let me edit your question as:

Which two equations are true?

<u>Eq1:</u>

(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)

<u>Eq2:</u>

6.6×10−10(6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.7×103)

<u>Eq3:</u>

2.7×103(7.5×106)−(2.5×106)=5×100

Answer:

No one is true

Step-by-step explanation:

let's check each equation, if the values on both sides (left and right side) are equal then the equation is true otherwise false.

Using PEMDAS rule we are simplifying the equations as;

<u>Eq1:</u>

$(2*10-4)+(1.5*10-4)=3.5*10-4(3*10-5)+(2.2*10-5)\\(16)+(11)=35-4(25)+(17)\\27=35-100+17\\27=-48\\$

<u>Eq2:</u>

<u></u> $6.6*10-10(6.3*10-1)-(2.1*10-1)=3*10-1(5.4*103)-(2.7*103)\\66-10(62)-(20)=30-1(556.2)-278.1\\66-620-20=30-556.2-278.1\\-574=-804.1$ <u></u>

<u>Eq3:</u>

$2.7*103(7.5*106)-(2.5*106)=5*100\\221089.5-265=500\\220824.5=500\\$

<u>we observed that none of the equation has two same values on both sides thus none of the three equations is true.</u>

<u>Also, no value of Eq1, Eq2 or Eq3 are same thus none of the equation is true</u>

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