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n200080 [17]
3 years ago
15

Please help me with this question

Mathematics
1 answer:
adoni [48]3 years ago
5 0

Answer:

y = e^{x^2 - 3x}\\\\\\Let \ u = x^2 - 3x\\\\y = e^u\\\\\frac{dy}{du} = e^u\\\\\frac{du}{dx}  = 2x - 3\\\\y' = \frac{dy}{dx}  = \frac{dy}{du} *\frac{du}{dx} = e^u * (2x -3)\\\\Substitute \ u = x^2 -3\\\\y' = (2x -3)e^{x^2-3x}\\\\option A

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The kindergarten class has 10 girls and 10 boys. If two students are chosen at random to be in the school play, what is the prob
Akimi4 [234]
50% because there's equal value of both variables
6 0
3 years ago
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Two mechanics worked on a car. The first mechanic worked for 5 hours, and the second mechanic worked for 15 hours. Together they
Wewaii [24]
X = rate of mechanic who worked 5 hours

y = rate of mechanic who worked 15 hours

x + y = 200
5x + 15y = 1950

you find the first variable in one of the equations, then subsitute the result into the other equation. your answer would be:

the mechanic who worked for 5 hours charged $105 per hour, the one who worked 15 hours charged $95 per hour.

hope this helps :)


3 0
3 years ago
Over the first five years of owning her car, Gina drove about 12,200 miles the first year, 16,211 miles the second year, 12,050
weeeeeb [17]
(12,200 + 16,211 + 12,050 + 11,350 + 13,325) / 5 = 65136/5 = 
13027.2 <== this is the mean (average)

11,350 , 12,050 , 12,200 , 13,325, 16,211
median (middle number) = 12,200

there is no mode...a mode is a number that appears most often...all the numbers appear once in this data.
4 0
3 years ago
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Contact [7]

Answer:

c

Step-by-step explanation:

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5 0
3 years ago
In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. C
horrorfan [7]

Answer:

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

Step-by-step explanation:

<em>Step(i)</em>:-

<em>Given sample size 'n' =300</em>

Given data  random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.

<em>Given sample proportion </em>

<em>                        </em>p^{-}  = \frac{x}{n} = \frac{182}{300} =0.606

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

<em>90% confidence interval for the proportion is determined by</em>

(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} }  , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })

(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} }  ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })

(0.6066 -  0.0463  ,0.6066 +  0.0463)

(0.5603,0.6529)

<u>final answer</u>:-

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

5 0
3 years ago
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