If f(x) contains point (-3, 7), which point is definitely on f^-1(x)?

The claim is that for the population of adult males, the mean platelet count is μ>217. The sample size is n=50 and the test

Deffense [45]

Step-by-step explanation:

First, find the degrees of freedom:

df = n − 1

df = 49

To find the p-value manually, use a t-score table. Find the row corresponding to 49 degrees of freedom. Then find the α column that corresponds to a t-score of 1.421. You'll find it's between α = 0.10 (t = 1.299) and α = 0.05 (t = 1.677). Interpolating, we get an approximate p-value of 0.084. For a more accurate answer, you'll need to use a calculator.

5 0

3 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

These two triangles are similar .Find side lengths a and b .The two figures are not drawn to scale

ziro4ka [17]

Answer:

a = 6, b = 22.5

Step-by-step explanation:

One triangle is 2.5 times bigger than the other.

9*2.5=22.5 which means b = 22.5

15/2.5=6 which means a = 6

7 0

3 years ago

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What is the unit multipliers for weeks and months

mash [69]

Answer:

Count the number of days in the month and divide that number by 7, which is the number of days in one week. For example, if March has 31 days, there would be a total of 4.43 weeks in the month

Step-by-step explanation:

5 0

2 years ago

Help me please!! first write the subtraction with a common denominator then subtract

tatiyna

Answer:

hope it helps you

Step-by-step explanation:

pls kindly mark it brainliest answer

3 0

3 years ago

Read 2 more answers