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3 years ago

HowSolve The D. E dy/dx=2x^5 y^3 - 4y/x

1 answer:

3 0

Dy/dx (2x⁵ y³ - 4y/x)

dy/dx (2y³ x⁵ - 4y/x)

dy/dx ( 2y³ x⁵ ) - dy/dx (4y/x)

= 2y³ 5x⁴ - (-4y × 1/x²)

=
10x⁶ y³ + 4y
-------------------
x²

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Can you help me solve this?

yuradex [85]

Answer:

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Step-by-step explanation:

hes the 16th president, duh

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3 years ago

JL = 28,<br> KL= 5x + 6, and<br> JK = 3x + 6,

Alexeev081 [22]

Answer:

KL = 16

Step-by-step explanation:

For this problem, JL is the sum of KL and JK. So we can say this:

JK + KL = JL

( 3x + 6 ) + ( 5x + 6 ) = 28

8x + 12 = 28

8x = 16

x = 2

So, now we can find KL:

5x + 6 = ?

5(2) + 6 = ?

10 + 6 = ?

16 = ?

So the length of KL is 16.

Cheers.

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3 years ago

The 9’s in 9,905,482

Sergeu [11.5K]

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Step-by-step explanation:

If this is wrong you can report it to remove I’m sure,

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3 years ago

What is the equivalent ratio to 5/10

dimaraw [331]

1/2
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3 years ago

A center for medical services reported that there were 295,000 appeals for hospitalization and other services. For this group, 4

pshichka [43]

Answer:

a) 0.0025 = 0.25% probability that none of the appeals will be successful.

b) 0.0207 = 2.07% probability that exactly one of the appeals will be successful.

c) 0.9768 = 97.68% probability that at least two of the appeals will be successful

Step-by-step explanation:

For each appeal, there are only two possible outcomes. Either it is succesful, or it is not. The probability of an appeal being succesful is independent of other appeals, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

45% of first-round appeals were successful.

This means that $p = 0.45$

Suppose 10 first-round appeals have just been received by a Medicare appeals office.

This means that $n = 10$

(a) Compute the probability that none of the appeals will be successful.

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.45)^{0}.(0.55)^{10} = 0.0025$

0.0025 = 0.25% probability that none of the appeals will be successful.

(b) Compute the probability that exactly one of the appeals will be successful.

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.45)^{1}.(0.55)^{9} = 0.0207$

0.0207 = 2.07% probability that exactly one of the appeals will be successful.

(c) What is the probability that at least two of the appeals will be successful

This is $P(X \geq 2)$

Either less than two appeals are succesful, or at least two are. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0207 = 0.0232$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0232 = 0.9768$

0.9768 = 97.68% probability that at least two of the appeals will be successful

6 0

3 years ago

HowSolve The D. E dy/dx=2x^5 y^3 - 4y/x​

HowSolve The D. E dy/dx=2x^5 y^3 - 4y/x