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3 years ago

(07.05 LC) How many solutions does the equation 4y + 7 = 5 + 2 + 4y have? One Two None Infinitely many

Mathematics

2 answers:

Elza [17]3 years ago

8 0

Answer: Infinitely many

Step-by-step explanation:

First simplify the right side by adding 5 and 2

now you have the equation 4y+7=7+4y

That is the exact same thing so any number works.

Reil [10]3 years ago

7 0

just to add to the great reply above.

$\bf \begin{matrix} 4y \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}+7=5+2~~\begin{matrix} +4y \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}\implies 7=7$

whenever you end up with something like 0 = 0, or 7 = 7, is a flag that both equations are exactly the same, in this case, the one on the right-hand-side is really the one one the left-hand-side <u>in disguise</u>.

So the graph of one, is the same as the graph of the other, or put in another words, let's say the graph the first one, the second one when graphed, will just be pancaked on top of the first one, and any point whatsoever on the second one, matches with the first one, and since both lines continue infinitely, then infinitely many solutions.

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3 years ago

Help me please it is literal equation

riadik2000 [5.3K]

Answer:

Step-by-step explanation:

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3 years ago

Suppose that 37% of college students own cats. If you were to ask random college students if they own a cat what would the proba

Likurg_2 [28]

Using the binomial distribution, the probabilities are given as follows:

a) 0.37 = 37%.

b) 0.5065 = 50.65%.

c) 0.3260 = 32.60%.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

For this problem, the fixed parameter is:

p = 0.37.

Item a:

The probability is P(X = 1) when n = 1, hence:

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

$P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37$

Item b:

The probability is P(X = 3) when n = 3, hence:

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

$P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065$

Item c:

The probability is P(X = 2) when n = 4, hence:

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

$P(X = 2) = C_{4,2}.(0.37)^{2}.(0.63)^{2} = 0.3260$

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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2 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

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2 years ago

What is bigger 7 1/8 or 7.025?

Stolb23 [73]

7.025 is greater than 7 1/8

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3 years ago

Read 2 more answers