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Lyrx [107]
3 years ago
14

The equation K = mv2 represents the energy an object has based on its motion. The kinetic energy, K, is based on the mass of the

object, m, and the velocity of the object, v. Lashandra is given K and v for 10 different objects. In order to make solving more efficient, she solves the equation for m : m = . After attempting to determine the mass of a few objects, Lashandra realizes there must be something wrong with her formula. What is Lashandra’s error?
Mathematics
2 answers:
erastovalidia [21]3 years ago
7 0
Kinetic energy<span> is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has </span>kinetic energy<span>. It is expressed as:

KE = mv^2 / 2

Therefore, her error would be that he lack the factor 1/2 in her formula. Hope this answers the question. Have a nice day.</span>
pashok25 [27]3 years ago
6 0

Answer: A

Step-by-step explanation:

She should have multiplied by 2 instead of dividing by 2

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The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
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Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

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3 years ago
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Step-by-step explanation:

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Using calculator :

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The test statistic :

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