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ziro4ka [17]
3 years ago
8

Soren has two jobs. During the day he works as an office clerk, and in the evening he works as a cashier. His office job pays hi

m $11 per hour. His cashier job pays him $8.75 per hour. In one week, Soren worked 46 hours. He earned a total of $470.
How many hours did Soren work in each job?





A.

Office clerk: 16 hours; cashier: 30 hours



B.

Office clerk: 20 hours; cashier: 26 hours



C.

Office clerk: 30 hours; cashier: 16 hours



D.

Office clerk: 31 hours; cashier: 15 hours
Mathematics
1 answer:
svet-max [94.6K]3 years ago
6 0
A. 16x11 = 176 + 30x8.75 = 262.5. 176+262.5 = 438.5. Not 470
B. 20x11 = 220 + 26x8.75 = 227.5. 220+227.5 = 448. Not 470
C. 30x11 = 330 + 16x8.75 = 140. 330+140 = 470. C. is the correct answer
D. 31x11 = 341 + 15x8.75 = 131.25. 341+131.25 = 472.25 Not 470
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Mary needs a board 4 feet 8 inches longShe cut a board 56 inches longDid Mary cut the right length?
dmitriy555 [2]

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Step-by-step explanation:

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Please help me with both questions!!!!​​
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Answer:

Step-by-step explanation:

16.

points are (0,-3),(2,1),(4,-3)

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now eq of line through (2,1)and (4,-3) is

y-1=\frac{-3-1}{4-2}(x-2)

y-1=-2(x-2)

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17.

consider the points (0,4),(2,3),(4,4)

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y-4=\frac{3-4}{2-0}(x-0)

y-4=-1/2(x)

2y-8=-x

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put x=0,y=0

0≥8

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0≤-4

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5 0
3 years ago
The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .1
snow_lady [41]

We assumed in this answer that the question b is, Are the events V and M independent?

Answer:

(a). The probability that a student has either a Visa card or a MasterCard is<em> </em>\\ P(V \cup M) = 0.88. (b). The events V and M are not independent.

Step-by-step explanation:

The key factor to solve these questions is to know that:

\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)

We already know from the question the following probabilities:

\\ P(V) = 0.73

\\ P(M) = 0.18

The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So

\\ P(V \cap M) = 0.03

The probability that a student has either a Visa card or a MasterCard

We can interpret this probability as \\ P(V \cup M) or the sum of both events; that is, the probability that one event occurs OR the other.

Thus, having all this information, we can conclude that

\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)

\\ P(V \cup M) = 0.73 + 0.18 - 0.03

\\ P(V \cup M) = 0.88

Then, <em>the probability that a student has either a Visa card </em><em>or</em><em> a MasterCard is </em>\\ P(V \cup M) = 0.88.<em> </em>

Are the events V and M independent?

A way to solve this question is by using the concept of <em>conditional probabilities</em>.

In Probability, two events are <em>independent</em> when we conclude that

\\ P(A|B) = P(A) [1]

The general formula for a <em>conditional probability</em> or the probability that event A given (or assuming) the event B is as follows:

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

If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that

\\ P(M|V) = \frac{P(M \cap V)}{P(V)}

\\ P(M|V) = \frac{0.03}{0.73}

\\ P(M|V) \approx 0.041

If M were independent from V (according to [1]), we have

\\ P(M|V) = P(M) = 0.18

Which is different from we obtained previously;

That is,

\\ P(M|V) \approx 0.041

So, the events V and M are not independent.

We can conclude the same if we calculate the probability

\\ P(V|M), as follows:

\\ P(V|M) = \frac{P(V \cap M)}{P(M)}

\\ P(V|M) = \frac{0.03}{0.18}

\\ P(V|M) = 0.1666.....\approx 0.17

Which is different from

\\ P(V|M) = P(V) = 0.73

In the case that both events <em>were independent</em>.

Notice that  

\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)

\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03

\\ 0.03 = 0.03 = 0.03 = 0.03

3 0
4 years ago
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