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Phoenix [80]
3 years ago
15

Please help I’ll give brainliest if correct

Mathematics
1 answer:
pochemuha3 years ago
6 0

the second one looks more relative to the graph so go with that one.

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The drama club had a car wash as a fundraiser where they washed a total of 41 vehicles. Members washed
Dimas [21]

Answer:

<em>Solution: (26,15) </em>

<em> </em>

<em>Cars washed=26 </em>

<em> </em>

<em>Trucks washed=15</em>

Step-by-step explanation:

3 0
2 years ago
(hg^(-3))/(w^(-4) k^0 )
Svetlanka [38]
One may note you never quite asked anything, now, assuming simplification,

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4 0
2 years ago
1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING
iogann1982 [59]

If

y=\displaystyle\sum_{n=0}^\infty a_nx^n

then

y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n

The ODE in terms of these series is

\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0

\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0

\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}

We can solve the recurrence exactly by substitution:

a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0

\implies a_n=\dfrac{(-2)^n}{n!}a_0

So the ODE has solution

y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}

which you may recognize as the power series of the exponential function. Then

\boxed{y(x)=a_0e^{-2x}}

7 0
3 years ago
How do we display and<br> summarize data<br> distributions?
photoshop1234 [79]

Answer:

A good display will help to summarize a distribution by reporting the center, spread, and shape for that variable. ... In these plots the horizontal axis represents the values of the variable and the height of the bar represents how many observations are equal to the particular value.

Step-by-step explanation:

6 0
3 years ago
According to the data, the mean quantitative score on a standardized test for female college-bound high school seniors was 500.
Sveta_85 [38]

Answer:

6.68% of the female college-bound high school seniors had scores above 575.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 500

Standard Deviation, σ = 50

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(scores above 575)

P(x > 575)

P( x > 575) = P( z > \displaystyle\frac{575 - 500}{50}) = P(z > 1.5)

= 1 - P(z \leq 1.5)

Calculation the value from standard normal z table, we have,  

P(x > 575) = 1 - 0.9332= 0.0668 = 6.68\%

6.68% of the female college-bound high school seniors had scores above 575.

7 0
3 years ago
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