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2 years ago

Please help ! I’m in such a rush! I have to be done before 3pm and it’s already 2:30pm

Mathematics

1 answer:

lutik1710 [3]2 years ago

6 0

Answer:

170°

Step-by-step explanation:

By inspection of the diagram, the angle is 170°

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Which graph representative this function

nalin [4]

Answer:

in graph A the vertex is (0,0)

in graph B the vertex is (0,3)

in graph C the vertex is (0,-3)

And in graph D the vertex is (0,-1)

The axis of symmetry for all of the is 0

And i don't remember maximum and minimum sorry

8 0

3 years ago

Which is 196 written as a power of a number? A) 98^2 B) 13^2 C) 49^4 D) 14^2

kap26 [50]

Its not A. because , 98^2 = 9,604
Its not B. because , 13^2 = 169
Its for sure not C. because, 49^4 = 5,764,801
It is most likey, D. because 14^2 = 196
Option: D 14^2 would most likely be your answer

Reason , because 14(14) = 196 = 14^2

☺

3 0

3 years ago

Sara made enough raspberry jam to fill 3

NeTakaya

Answer:

5/8 Jar

Step-by-step explanation:

I took the test...

3 0

2 years ago

Someone please help! Here's the question: Twenty boys signed up for the school play. The number of girls who signed up was 195%

kipiarov [429]

20*1.95 =39 girls
26girls/39girls =0.667
0.667*100 = 66.67%

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:

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

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

2 years ago