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3 years ago

Answer this correctly I’ll give brainalist + 10 points

Mathematics

2 answers:

Semmy [17]3 years ago

6 0

Answer:that means it d

Step-by-step explanation:

Effectus [21]3 years ago

5 0

I’m positive it’s 110 since it’s a little more than a right angle

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What dose the hypotenuse=?

Kitty [74]

Hypotenuse by Pythagoras theorem = 407.83.

3 0

3 years ago

Whats the square root of 1/5 minus the square root of 5

I am Lyosha [343]

You have to simplify the 1/5 first.
Then simplify the square root of 1 to equal 1
Then rationalize the denominator
Then simplify

8 0

3 years ago

Plssssssss help with this

Olenka [21]

Answer:

x=4

Step-by-step explanation:

If all the figures are similar, we'll have to shrink them at the same rate as the rest.

The ratio here is 2/5. So,

2/5 = 6.4/16

2/5 = x/10

x=4

3 0

3 years ago

Read 2 more answers

Solve x^2-3=0 using a quadratic equation

anzhelika [568]

Answer:

I think the answers are 3 & 0 .

Step-by-step explanation:

Good luck .

3 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