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3 years ago

Find the area of this semi-circle with diameter 18 Give your answer rounded to 2 DP.

Mathematics

1 answer:

ANTONII [103]3 years ago

5 0

Step-by-step explanation:

diameter(d): 18

radius(r): 18 ÷ 2 = 9

area of a semi-circle:

$\frac{\pi {r}^{2}}{2}$

$= \frac{\pi \times 9 \times 9}{2} = 127.23$

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2 Cups

Step-by-step explanation:

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3 years ago

Every integer is negative true or false?

Elena L [17]

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3 years ago

Read 2 more answers

In order to solve the system of linear equations, the coefficients of one of the variables must be the same number but opposite

Mashutka [201]

Answer:

Step-by-step explanation:

because both equations do not have matching numbers for the x or y variable, and both equations are positive you are going to have to multiply each equation by a number so that there will be at least one variable with the same number but with opposite signs.

it does not matter which variable you choose.

lets use y because 2 and 3 are smaller then 2 and 5.

so lets multiply the first equation by 2 in order to get y equal to 6.

2(2x)+2(3y)=(2)6

(do not forget to multiply what the equation is equal to also)

4x+6y=12

now for the second equation we need y to equal negative 6

-3(5x)+-3(2y)=-3(4)

-15x-6y=-12

now lets put the 2 new equations next to each other and see what we can cancel out

4x+6y=12

-15x-6y=-12

-11x=0

x=0

now plug 0 in for x and solve for y (it does not matter which of the 4 equations you choose to solve.

2(0)+3y=6

3y=6

y=2

so your answer is x=0, y=2

8 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:

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