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Hunter-Best [27]
2 years ago
4

Gina takes classes at both Westside Community College and Pinewood Community College. At Westside, class fees are $ 98 per credi

t hour, and at Pinewood, class fees are $ 115 per credit hour. Gina is taking a combined total of 13 credit hours at the two schools.At Westside she is taking w credit hours. What is an expression for the combined total dollar amount she paid for her class fees?
Mathematics
1 answer:
Katen [24]2 years ago
8 0

Answer:

98w+115(13-w)\\

Step-by-step explanation:

Given:

Gina takes classes at both West side Community College and Pinewood Community College.

At West side, class fees are $ 98 per credit hour,

At Pinewood, class fees are $ 115 per credit hour.

Gina is taking a combined total of 13 credit hours at the two schools.

At West side she is taking w credit hours.

Question asked:

What is an expression for the combined total dollar amount she paid for her class fees?

Solution:

Given that at West side, she is taking number of credit hours = w

As combined credit hour is 13,

Then, at Pinewood, class, she is taking number of credit hours = 13-w

<u>Now, we will find total amount she paid for </u>w<u> number of classes at West side Community College, by unitary method:</u>

She paid for 1 credit hour,  = $98

She paid for w credit hour = 98w

Thus, total amount she paid for w number of classes at West side Community College is 98w

Similarly, we will find total amount she paid for 13-w number of classes at Pinewood Community College, by unitary method:

She paid for 1 credit hour = $115

She paid for 13-w credit hour = 115(13-w)

Thus, total amount she paid for 13-w number of classes at Pinewood Community College is 115(13-w)

<em><u>The expression for the combined total dollar amount she paid for her classes:-</u></em>

98w+115(13-w)\\

,

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2 years ago
If f(x) = 3 – 2x and 1/x+5, what is the value of (f/g)(8)? –169 –1 13 104
Basile [38]
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The drama club at Del Rosa Middle School is having a production.
Firdavs [7]

Answer:

320 Student Tickets

180 Adult Tickets

Step-by-step explanation:

You can solve this problem by using system of equations. First, we need to figure out our equations.

Equation 1: x as students and y as adults

x+y=500

We get this equation because the total tickets sold was 500. The x represents the students sold to students, and the y represents the tickets sold to adults.

Equation 2:

3x+5y=1850

We get this equation based on the prices. Each student ticket costs $3, and each adult ticket costs $5. The total amount earned was $1850.

Now that we have out equations, we can use system of equations to find our students and adults.

x+y=500

3x+5y=1860

Typically elimination is the easiest strategy because you are able to cross out variables.

3(x+y=500)

3x+5y=1860

Becomes:

3x+3y=1500

3x+5y=1860

We see that both equations now have 3x. We can cancel out 3x.

-2y=-360

y=180

Now that we know y=180, we can plug it back into one of our equations to find x.

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6 0
2 years ago
A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm
AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, P\left(\frac{E}{A}\right)=0.02

Now, from the conditional probability,

P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}

\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)

\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

P\left(\frac{E}{A}\right)=0.40

and \Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)

This is the probability of disregarding the partially defective parts by inspection machine.

P\left(\frac{E}{A}\right)=0.98

and \Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

=P(E\cap C)

= 0.0686 [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)

\Rightarrow P(E)=0.018+0.012+0.0686

\Rightarrow P(E)=0.0986

So, the total probability that the part produced get shipped

=1-P(E)=1-0.0986=0.9014

The probability that the part is good (either defect free or slightly defective)

=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)

=(0.9-0.018)+(0.03-0.012)

=0.9

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}

=\frac{0.9}{0.9014}

=0.9984

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0
2 years ago
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