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Olenka [21]
3 years ago
6

You meet with the financial aid office to discuss your costs for attending MGA next semester. Tuition is $113.67 per credit hour

, and fees are a flat cost of $660. You have a grant of $350 and a scholarship of $400. If you are taking 15 credit hours, what amount will you need to pay for your classes next semester?
Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0

Answer:

$1615.05

Step-by-step explanation:

First the cost. We know that tuition per credit hour is 113.67 and there is a flat fee of 660. This can be represented by 113.67h + 660, where h is equal to the number of credit hours.

Next, the amount paid. There are two constants of 350 and 400 already there. We can represent the remaining amount that needs to be paid by another variable, x.

Now, the cost needs to be equal to the amount you pay. We can replace h with 15 for 15 hours and solve for x.

113.67h + 660 = 350 + 400 + x

113.67(15) + 660 = 350 + 400 + x

1705.05 + 660 = 750 + x

2365.05 = 750 + x

1615.05 = x

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Type the correct answer in the box
TiliK225 [7]

The expression of X in terms of l is X = √5 l

<h3>How to calculate the diagonal of a rectangle</h3>

According to the given information:

  • Length = w
  • If length<u> l is twice as long </u>as the width, then l = 2w

Determine the diagonal using the Pythagoras theorem:

X² = w²+(2w)²
X² = w² + 4w²

X =√5w²
X = √5 w

Replace w with l

X = √5 l

Hence the expression of X in terms of l is X = √5 l

Learn more on diagonals here: brainly.com/question/26154016

5 0
3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
Solve for xxx.<br> −9x+2&gt;18 OR 13x+15≤−4
yanalaym [24]
-9x + 2 > 18.

-9x > 16

x < -16/9

13x ≤ -19

x ≤ -19/13

x < -16/9 or. x ≤ -19/13
3 0
3 years ago
Read 2 more answers
Jay was reaching into her purse and accidentally spilled her coin purse. 10 pennies fell on the floor. Jay noticed that only 2 o
kompoz [17]

Answer:

4.39%

Step-by-step explanation:

There are no marking to identify the coins, so the order is not important. Since the order not important we use combination instead of permutation.

A coin have 2 possible outcome, head or tails. Assuming the coin is fair each outcome will be 50% or 0.5 chance. If heads= A and tails =B, then the probability of 2 pennies will be:

P(A=2) = 2C10 * A^2 * B^(10-2)

P(A=2) = 10*9/2 * 0.5^2 * 0.5^8

P(A=2) = 0.0439= 4.39%

6 0
3 years ago
Which statements are true of functions?
ollegr [7]

 I don't know what your asking



5 0
3 years ago
Read 2 more answers
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