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-Dominant- [34]

3 years ago

7

Evaluate the expression \dfrac{7^2}{x^2-2} x 2 −2 7 2 start fraction, 7, squared, divided by, x, squared, minus, 2, end fracti

on for x=3x=3x, equals, 3

1 answer:

sertanlavr [38]3 years ago

3 0

Answer:

7

Step-by-step explanation:

We want to evaluate the fraction below for x = 3. We will put the value of x to be 3:

$\dfrac{7^2}{x^2-2}\\\\= \dfrac{7^2}{3^2-2}\\\\= \dfrac{49}{9-2}\\\\= \dfrac{49}{7} = 7$

The answer is 7.

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Gloria is an artist. She sets a goal to paint 2 paintings every month. The number of paintings Gloria creates depends on the num

BartSMP [9]

Answer:

we get the all equation of given condition $M = 2P+1$ & $P=2M+1$ Step-by-step explanation:

Given that,

Number of painting made by Gloria every month is 2.

Gloria creates depends on the number of paintings p, Gloria paints over m months if she meet her goal.

we have to check all the apply.

According to question,

M is the independent variable and P is dependent variable.

So, Relation formed by given statement is $M =2P$

Case(1): P is increased by 2 as M is increased by 1.

Then, $M+1= 2\times (P+2)$

$M+1= 2P+2$

$M = 2P+1$

This is Equation of the given case(1).

Again, P is the independent variable and M is dependent variable.

∴ $P=2M$

Case (2) M is increased by 2 as P increased by 1.

Then, $P+1=2\times(M+2)$

$P+1=2M+2$

$P=2M+1$

Hence,

we get the all equation of given condition $M = 2P+1$ & $P=2M+1$

4 0

3 years ago

A textile mill wishes to establish a control procedure on flaws in towels it manufactures. Using an inspection unit of 50 units,

Natasha_Volkova [10]

Answer:

•A c-chart is the appropriate control chart

• c' = 8.5

• Control limits, CL = 8.5

Lower control limits, LCL = 0

Upper control limits, UCL = 17.25

Step-by-step explanation:

A c chart is a quality control chart used for the number of flaws per unit.

Given:

Past inspection data:

Number of units= 100

Total flaws = 850

We now have:

c' = 850/100

= 8.5

Where CL = c' = 8.5

For control limits, we have:

CL = c'

UCL = c' + 3√c'

LCL = c' - 3√c'

The CL stands for the normal control limit, while the UCL and LCL are the upper and lower control limits respectively

Calculating the various control limits we have:

CL = c'

CL = 8.5

UCL = 8.5 + 3√8.5

= 17.25

LCL = 8.5 - 3√8.5

= -0.25

A negative LCL tend to be 0. Therefore,

LCL = 0

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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:

$\lambda \geq 6.63835$

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Find the Quadratic Formula: m2 - 5m - 14 = 0

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Answer: Do want to solve it using the quadratic formula because if so the answer is m=7,-2

Step-by-step explanation:

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