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Effectus [21]
4 years ago
10

The choices are:

Mathematics
1 answer:
Harrizon [31]4 years ago
7 0
Let's take a look at D:   

<span>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.</span>
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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 <em>is the sample mean  </em>

<em>s is the sample standard deviation  </em>

<em>n is the sample size (n = 7)  </em>

\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 <em>an approximation to the Normal distribution for small samples (n<30).</em>

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?<br><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..<br><br>​
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 <em>operating at a 6 sigma level, </em>the probability is:

\\ 1 - p = 1 - 0.999999998027 = 0.000000001973

Similarly, for finding <em>no defects</em> 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 <em>6 sigma level is markedly different when it comes to the number of defects present:</em>

\\ {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 <em>2 defects per billion</em> opportunities, whereas a five sigma process has <em>600 defects per billion</em> 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
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