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Vesna [10]
2 years ago
15

Find the particular solution of the differential equation dydx+ycos(x)=5cos(x) satisfying the initial condition y(0)=7.

Mathematics
1 answer:
natulia [17]2 years ago
8 0
\dfrac{\mathrm dy}{\mathrm dx}+y\cos x=5\cos x
e^{\sin x}\dfrac{\mathrm dy}{\mathrm dx}+ye^{\sin x}\cos x}=5e^{\sin x}\cos x
\dfrac{\mathrm d}{\mathrm dx}\left[e^{\sin x}y\right]=5e^{\sin x}\cos x
e^{\sin x}y=5\displaystyle\int e^{\sin x}\cos x\,\mathrm dx
e^{\sin x}y=5e^{\sin x}+C
y=5+Ce^{-\sin x}

With y(0)=7, we have

7=5+Ce^{-\sin 0}\implies 7=5+C\implies C=2

so that the particular solution is

y=5+2e^{-\sin x}
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Answer:

Step-by-step explanation:

We would determine the mean an standard deviation first

Mean = (107 + 92 + 97 + 95 + 105 + 101 + 91 + 99 + 95 + 104)/10 = 98.6

Standard deviation = √(summation(x - mean)/n

n = 10

Summation(x - mean) = (107 - 98.6)^2 + (92 - 98.6)^2 + (97 - 96.6)^2 + (95 - 98.6)^2 + (105 - 98.6)^2 + (101 - 98.6)^2 + (91 - 98.6)^2 + (99 - 98.6)^2 + (95 - 98.6)^2 + (104 - 98.6)^2 = 274

Standard deviation = √(274/10) = 5.23

The confidence interval is used to determine the range of values that could possibly contain a population parameter (population mean)

Confidence interval is written in the form,

(Point estimate - margin of error, Point estimate + margin of error)

The sample mean, x is the point estimate for the population mean.

We will use the t distribution because the sample size is small and the population standard deviation is not given.

Degree of freedom, df = 10 - 1 = 9

α = 1 - 0.99 = 0.01

z α/2 = 0.01/2 = 0.005

The area to the right of z 0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995. Using the t distribution table,

z = 3.25

Margin of error = z × s/√n

Where

s = sample standard Deviation = 5.23

Margin of error = 3.25 × 5.23/√10 = 5.38

Confidence interval = sample mean(point estimate) ± z × σ/√n

The lower boundary of the confidence interval is

98.6 - 5.38 = 93.22

The upper boundary of the confidence interval is

98.6 + 5.38 = 103.98

We estimate with 99% confidence that the mean weekly child care cost of their employees is between $93.22 and $103.98

Also,

99% of the confidence intervals constructed in this way would contain the true value for the population mean of mean weekly child care cost of their employees.

6 0
2 years ago
The grocery store is having a sale on frozen vegetables. 4 bags are being sold for $11.96. At this rate what is the cost of: 9 b
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Answer:

$26.91

Step-by-step explanation:

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4 0
1 year ago
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Masteriza [31]

Answer:

\large\boxed{43^o}

Step-by-step explanation:

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Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

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