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olya-2409 [2.1K]
3 years ago
7

Classify the model as exponential growth or exponential decay. Identify the growth or decay factor AND the percent of increase o

r decrease per time period.
y=100(1.05)^t

Plz help!
Mathematics
1 answer:
MArishka [77]3 years ago
4 0
See also brainly.com/question/8975313

The growth factor is 1.05.

The increase is (1.05 -1)*100% = 5% per period.
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Find the general solution of the differential equation. (use c for the constant of integration.) dr dt = 8et 1 − e2t
natulia [17]
Try this solution:
if given dr/dt=8e^t -e^(2t), then
\frac{dr}{dt}=8e^t-e^{2t}; \   \int\ {dr} \, =\int\ (8e^t-e^{2t})dt;  \ r=8e^t- \frac{1}{2}e^{2t}+C.
5 0
3 years ago
Lee Jenkins worked the following hours as a manager for a local Pizza Hut: 82,61,7
emmainna [20.7K]

Answer:

223 hours

Step-by-step explanation:

Total hours worked by Lee = sum of all the hours worked

That’s

82 + 61 + 7 + 73 = 223 hours

Therefore, Lee worked 223 hours as manager for the local Pizza Hut

5 0
3 years ago
HELP PLEASE!!! WILL GIVE BRAINLIEST
Nadusha1986 [10]
The equation looks like this \frac{ x^{2} }{16} + \frac{ y^{2} }{25} =1.  In an ellipse, a is always the bigger value, so a^2 = 25.  This bigger value also tells us which axis is the major one.  Sine the bigger value a is under the y^2 of the equation, the major axis is the y-axis.  This is a vertical ellipse.  The center is always found within a set of parenthesis that exist with the x^2 and the y^2.  Since there are no parenthesis with either, there is no side to side movement, nor is there any up or down movement.  So the center doesn't move from the origin (0, 0).  The vertex is also along the major axis, and if a^2 is 25, then a = 5, so the vertices go up 5 from the center and down 5 from the center.  Vertices are (0, 5) and (0, -5).  The foci follow the formula c^{2}= a^{2}- b^{2}.  c is the distance that the foci are from the center.  c^{2}=25-16 and c = 3.  The foci also lie on the major axis, so the coordinates for the foci are (0, 3) and (0, -3).  There you go!
8 0
3 years ago
One year Ron had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched)
Arada [10]

Answer:

Ron's ERA has a z-score of -2.03.

Karla's ERA has a z-score of -1.86.

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

Step-by-step explanation:

Z-score:

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

Since the lower the ERA, the better the pitcher, whoever's ERA has the lower z-score had the better year relative to their peers.

Ron

ERA of 3.06, so X = 3.06

For the males, the mean ERA was 5.086 and the standard deviation was 0.998. This means that \mu = 5.086, \sigma = 0.998

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.06 - 5.086}{0.998}

Z = -2.03

Ron's ERA has a z-score of -2.03.

Karla

ERA of 3.28, so X = 3.28

For  the females, the mean ERA was 4.316 and the standard deviation was 0.558. This means that \mu = 4.316, \sigma = 0.558

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.28 - 4.316}{0.558}

Z = -1.86

Karla's ERA has a z-score of -1.86.

Which player had the better year relative to their peers, Ron or Karla?

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

5 0
3 years ago
A car traveling east at 40.0 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 25.0 m/s in
Anni [7]

Step-by-step explanation:

It is given that,

Initial speed of the car, u = 40 m/s

Final speed of the car, v = 25 m/s

Time taken, t = 3.5 s

(a) We need to find the magnitude and direction of the car’s acceleration as it slows down. It can be calculated using formula as :

a=\dfrac{v-u}{t}

a=\dfrac{25-40}{3.5}

a=-4.28\ m/s^2

The acceleration is in the opposite direction of motion i.e. west.

(b) Let s is the distance the car travel in the 3.5-s time period. It can be calculated using the third equation of motion as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{25^2-40^2}{2\times (-4.28)}

s = 113.9 meters

Hence, this is the required solution.

6 0
3 years ago
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