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levacccp [35]
3 years ago
11

According to data from the U.S. Department of Education, the average cost y of tuition and fees at public four-year institutions

in
year x is approximated by the equation
y = 0.024x* -0.87x + 9.6x² +97.2x + 2196
where x = 0 corresponds to 1990. If this model continues to be accurate, during what year will tuition and fees reach $4000?
a. 2000
C. 2016
b. 2010
d. 2004
Mathematics
2 answers:
taurus [48]3 years ago
8 0

Answer:

D

Step-by-step explanation:

edge2020

lianna [129]3 years ago
6 0

Answer:

D

Step-by-step explanation:

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99x78<br> 123x56<br> 98x172<br> 38x900
yanalaym [24]

Answer:

99x78=7722

123x56=6888

98x172=16856

900x38=34200

Step-by-step explanation:

8 0
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seven students took a bus to the State Park. The round-trip tickets cost $45 per student. They stayed for a few nights at a camp
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Hacksaw’s Boats of St. Lucia takes tourists on a daily whale/dolphinwatch cruise. Their brochure claims an 80% chance of siting
Fittoniya [83]

Answer:

The probability you see a dolphin or a whale on both days is 0.64

Step-by-step explanation:

Consider the provided information.

It is given that the brochure claims an 80% chance of siting a dolphin or a whale.

The provided claims can be written as: 80% or 0.80

Let say first day you see a dolphin or a whale and same happen on the next day.

Thus, the probability is:

0.80 × 0.80 = 0.64

Probability of see them on both days is 0.64

Hence, the probability you see a dolphin or a whale on both days is 0.64

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525$deposit. 4% interest. 1 year. calculate the simple interest earnef
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6 0
3 years ago
The manufacturer of a CD player has found that the revenue R​ (in dollars) is Upper R (p )equals negative 5 p squared plus 1 com
AleksAgata [21]

Answer:

The maximum revenue is $1,20,125 that occurs when the unit price is $155.

Step-by-step explanation:

The revenue function is given as:

R(p) = -5p^2 + 1550p

where p is unit price in dollars.

First, we differentiate R(p) with respect to p, to get,

\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550

Equating the first derivative to zero, we get,

\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155

Again differentiation R(p), with respect to p, we get,

\dfrac{d^2(R(p))}{dp^2} = -10

At p = 155

\dfrac{d^2(R(p))}{dp^2} < 0

Thus by double derivative test, maxima occurs at p = 155 for R(p).

Thus, maximum revenue occurs when p = $155.

Maximum revenue

R(155) = -5(155)^2 + 1550(155) = 120125

Thus, maximum revenue is $120125 that occurs when the unit price is $155.

6 0
3 years ago
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