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allochka39001
1 month ago
8

QUESTION 2 - Parts A&B) A baseball coach is considering two different companies for ordering team jerseys. Guy's Graphix cha

rges $30 to create the artwork for the team logo plus $7.50 per jersey. Patty's Printing charges $50 to create the artwork for the team logo plus $5.50 per jersey.

Mathematics
1 answer:
lawyer [12.5K]1 month ago
4 0

Response:

The cost incurred by Patty's Printing is $20 less than that of Guy's Graphix for 20 jerseys.

Detailed breakdown:

1st. Calculate the total expense for Guy's Graphix: (7.50*20)+30=$180

2nd. Calculate the total expense for Patty's Printing: (5.50*20)+50=$160

3rd. Subtract to verify the outcome: 180-160=$20

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2x+7x=747
9x=747
x=83

83(2) = 166
83(7) = 581
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2 months ago
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HELP ME ASAP IF U CAN!!! a baker needs sugar syrup that is 40% sugar how many gallons of water shoud he add to 5 gallons of 70%
tester [12383]
The result is 15/4. The equation is 40(5+x)=70*5+0x. When solved: 200+40x=350, which simplifies to 40x=150, therefore x=15/4.
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A newspaper vendor sells three papers to 135 customers, the papers are the daily times, the observer and the new Nigeria. 70 cus
AnnZ [12381]

Answer:

Three customers purchased all three newspapers.

Step-by-step explanation:

Let D, O, and N stand for the three newspapers.

D = The Daily Times

O = The Observer

N = The New Nigeria.

According to the provided information,

n(D\cup O\cup N)=135, n(D)=70,n(O)=60,n(N)=50

n(D\cap O)=17, n(O\cap N)=15, n(D\cap N)=16

We can deduce that:

n(A\cup B\cup C) = n(A) + n(B) + n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A) + n(A\cap B\cap C)

Applying this formula, we discover

n(D\cup O\cup N) = n(D)+n(O) + n(N)-n(D\cap O)-n(O\cap N)-n(N\cap D) + n(D\cap O\cap N)

135 =70+60+50-17-15-16 +n(D\cap O\cap N)

135 =132+n(D\cap O\cap N)

135 -132=n(D\cap O\cap N)

3=n(D\cap O\cap N)

Hence, three customers bought all three papers.

5 0
2 months ago
Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e
babunello [11817]

Answer:

0.75 = 75% chance that only one tag is lost, provided at least one tag is lost

Step-by-step explanation:

Independent events:

If A and B are independent events, then:

P(A \cap B) = P(A)*P(B)

Conditional probability:

P(B|A) = \frac{P(A \cap B)}{P(A)}

Here

P(B|A) refers to the probability of event B occurring, given that event A has occurred.

P(A \cap B) is the probability of both A and B occurring together.

P(A) is the probability of event A occurring.

In this scenario:

Event A: At least one tag is missing

Event B: Only one tag is missing.

Each tag has a 40% likelihood of being lost, which is equal to 0.4.

Probability of at least one tag missing:

The events can be considered as either no tags are missing or at least one is. Their probabilities sum to 1. Thus

p + P(A) = 1

p is the probability that none are lost. Each tag has a 60% = 0.6 chance of not being lost, and since they are independent,

p = 0.6*0.6 = 0.36

Then

P(A) = 1 - p = 1 - 0.36 = 0.64

Intersection:

The intersection of at least one lost (A) and exactly one lost (B) is precisely one lost.

Then

Probability of at least one lost:

The first being lost (0.4 chance) and the second not lost (0.6 chance)

Or

The first not being lost (0.6 chance) and the second lost (0.4 chance)

So

P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48

Calculate the probability that exactly one tag is lost, given that at least one tag is lost (round to two decimal places).

P(B|A) = \frac{0.48}{0.64} = 0.75

0.75 = 75% likelihood that precisely one tag is lost, assuming at least one tag is lost

8 0
2 months ago
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