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4 years ago

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Mathematics

1 answer:

Harrizon [31]4 years ago

7 0

Let's take a look at D:

D) y = (x-1)^2 - 16 Compare this to
y = (x-h)^2 + k This is the std. equation of a parabola in vertex form.

You can see, by comparison, that h=1 and k= -16; these are the coordinates of the vertex, clearly shown in the diagram.

Since the coefficient of (x-h)^2 is +1, the graph opens upward (which the given graph confirms), and is neither compressed nor stretched vertically.

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An article in Medicine and Science in Sports and Exercise ["Maximal Leg-Strength Training Improves Cycling Economy in Previously

Sholpan [36]

Answer:

[300.202 , 329.798]

Step-by-step explanation:

The 95% confidence interval is given by the interval

$\large [\bar x-t^*\frac{s}{\sqrt n}, \bar x+t^*\frac{s}{\sqrt n}]$

where

$\large \bar x$ is the sample mean

s is the sample standard deviation

n is the sample size (n = 7)

$\large t^*$ is the 0.05 (5%) upper critical value for the Student's t-distribution with 6 degrees of freedom (sample size -1), which is an approximation to the Normal distribution for small samples (n<30).

Either by using a table or the computer, we find

$\large t^*= 2.447$

and our 95% confidence interval is

$\large [315-2.447*\frac{16}{\sqrt{7}}, 315+2.447*\frac{16}{\sqrt{7}}]=\boxed{[300.202,329.798]}$

7 0

3 years ago

What is this? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bf%7D%7B4%7D%20%20-%205%20%3D%20%20-%209" id="TexFormula1" title=

tigry1 [53]

Hey there!

<h2>ANSWER: $f=-16$ </h2><h2>EXPLANATION:</h2>

$\frac{f}{4}-5=-9$

Simplify both sides and you get:

$\frac{1}{4}f-5=9$

Now add 5 to both sides and you get:

$\frac{1}{4}f=-4$

Now you have to multiply both sides and you get:

$f=-16(ANSWER)$

Hope this helps!

$\text {-TestedHyperr}$

6 0

3 years ago

Find the value of..

Zina [86]

Step-by-step explanation:

[1/2]^-2= 4

[1/3]^-2=9

[1/4]^-2=16

So sum is 4+9+16=29

Hope this helps you.

6 0

3 years ago

The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

3 years ago

What are some methods for determining if a polynomial is prime?

ArbitrLikvidat [17]

Answer:

Polinomios irreducibles (primos) Un polinomio con coeficientes enteros que no pueden ser factorizados en polinomios de grado menor, también con coeficientes enteros, es llamado un polinomio irreducible o primo

Step-by-step explanation:

7 0

3 years ago