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melamori03 [73]
3 years ago
5

PLSSSSSSSSSSSSSSSSSSSSSSSS HELP ME!!

Mathematics
1 answer:
zzz [600]3 years ago
3 0

Answer:

For the top table:

[x] | 3.1 | 2.5 | 1.2 | 0.9 | 0.14 | 0.06 | 0.02 |

[y] | 15.5 | 12.5 | 6 | 4.5 | 0.7 | 0.3 | 0.1 |

For the bottom table:

k = 5

[x] | 3.1 | 2.5 | 1.2 | 0.9 | 0.14 | 0.06 | 0.02 |

[y] | 15.5 | 12.5 | 6 | 4.5 | 0.7 | 0.3 | 0.1 |

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!
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{y}^{2}  + 49 \\  \\   = {y}^{2}  - ( - 49) \\  \\   =  {y}^{2} -  {(7i) }^{2} \\  \\   = (y + 7i)(y - 7i)
5 0
4 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

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).

4 0
3 years ago
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