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3 years ago

Pls help and look at the screenshot will give brainlist

Mathematics

2 answers:

mr Goodwill [35]3 years ago

7 0

Answer is D
Hope it help u

kumpel [21]3 years ago

6 0

Answer:

Step-by-step explanation:

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The gravitational force exerted on a baseball is 2.24 N down. A pitcher throws the ball horizontally with velocity 17.0 m/s by u

Fittoniya [83]

To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. 
b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. 
f = (ma), = .226 x 105.88, = 23.92N.

4 0

4 years ago

What is the quotient of (-6) divided by (-7)

stepladder [879]

Answer:

6/7 or .857

Step-by-step explanation:

Since they are both negative, it cancels each other and becomes positive

5 0

3 years ago

Use the expression 5a – 2b – 3 + 2b – 6a.

ella [17]

Answer:

5a + (-6a) + (-2b) + 2b + (-3) = -a -3 or -(a + 3) so first answer is correct because -(a + 3) is the same as -(3 + a)

Step-by-step explanation:

4 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:

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