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MrRissso [65]
3 years ago
13

Fifty specimens of a new computer chip were tested for speed in a certain application, along with 50 specimens of chips with the

old design. The average speed, in MHz, for the new chips was 495.6, and the standard deviation was 19.4. The average speed for the old chips was 481.2, and the standard deviation was 14.3.Can you conclude that the mean speed of the new chips is greater than that of the old chips?
Mathematics
1 answer:
Murljashka [212]3 years ago
4 0

Answer:

Under 99% confidence level we can say that mean speed of the new chips is greater than that of the old chips

Step-by-step explanation:

H_{0}: Mean speed of the new chip is the same as the old chip

H_{a}: Mean speed of the new chip is greater than the old chip

to calculate the z-statistic we can use the formula:

z=\frac{M_{n}-M_{o}}{\sqrt{\frac{s_{n}^2}{N_{n}} +\frac{s_{o}^2}{N_{o}} } }

if we put the numbers then

z=\frac{495.6-481.2}{\sqrt{\frac{19.4^2}{65}} +\frac{14.3^2}{65} } } =4.8171

The probability of this z-statistic is < 0.001 Therefore Under 99% confidence level we can say that mean speed of the new chips is greater than that of the old chips

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Dahasolnce [82]

Answer:

The correct answer is C. 13

Step-by-step explanation:

It is 13 because all triangles equal 180

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5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B
riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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2 years ago
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The correct answer is:

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The lengths of the sides of KELY are:

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Each side of KELY is twice the length of the corresponding side on BRAD.  This makes the ratio of the sides 2:1 and the figures are similar.

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Ivahew [28]
If that is -2+1+4+13 I think it should be 16
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