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3 years ago

WILL MARK BRAINLIEST IF CORRECT ANSWER! HOW MANY SOLUTIONS???

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

6 0

The answer is 1 solution which is x=1

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When Arianna left her house,her cell phone was 45% charged started losing charge at a constant rate. 4 hours after leaving her h

4vir4ik [10]

Answer:

B = 3.75t

Step-by-step explanation:

Given;

initial charge of Arianna's phone = 45%

the charge remaining in Arianna battery = 15%

If after 4hours of leaving her house, the battery had 15% remaining battery life, then the battery is discharging at the rate of ( 15% / 4hours) 3.75% per hour.

Thus, the charge remaining in Arianna's battery after she left her house is given by;

B = 3.75t

where;

t is time in hours

B is the percentage of charge remaining

6 0

3 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-\frac{1}{1+\lambda}$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

4 0

3 years ago

Xavier said that the recursive formula for this sequence could

Sergeeva-Olga [200]

Answer

The answer is in the step by step explanation

Step-by-step explanation:

YOU ARE LAZY TO FIND AN ANSWER ON YOUR OWN SO GET OUT OF HERE CHEATING LIL KID

7 0

3 years ago

Write the equation of the sphere in standard form. x^2 + y^2 + z^2 + 2x − 4y − 6z = 22

madam [21]

The equation $x^{2} +y^{2} +z^{2} +2x-4y-6z=22$ in standard form looks like $(x+1)^{2} +(y-2)^{2} +(z-3)^{2} =(6)^{2}$ .

Given equation of sphere be $x^{2} +y^{2} +z^{2} +2x-4y-6z=22$ .

We are required to express the given equation in the standard form of the equation of sphere.

Equation is basically relationship between two or more variables that are expressed in equal to form. Equation of two variables look like ax+by=c. It may be linear equation,quadratic equation, cubic equation or many more depending on the powers of variables. The standard form of the equation of sphere looks like $x^{2} +y^{2} +z^{2} =r^{2}$ .