Answer:
distance = sqrt( 65 ) ~= 8.06
Step-by-step explanation:
To find the distance between two points, we can use the distance formula.
distance = sqrt( [x2 - x1]^2 + [y2 - y1]^2)
distance = sqrt( [-5 - 2]^2 + [3 - -1]^2)
distance = sqrt( [-7]^2 + [4]^2)
distance = sqrt( 49 + 16 )
distance = sqrt( 65 ) ~= 8.06
Cheers.
Answer:
The mean is the better method.
Step-by-step explanation:
The best way to meassure the average height is throught mean. The mean of a sample is the average of that sample's height, and it will be a good estimate for the population's average height.
The mode just finds the most frequent height. Even tough the most frequent height will influence the average height, knowing only what height is the most frequent one doesnt give you enough informtation about how the height is centrally distributed.
As for the median, it is fine to use the median of a sample to estimate the median of the population, but if you use the median to estimate the average height you may have a few issues. For example, if you include babies in your population, the babies will push the average height down a lot and they are far below te median height. This, as a result, will give you a median height of a sample way above the average height of the population, becuase median just weights every person's height the same, while average will weight extreme values more, in the sense that a small proportion of extreme values can push the average far from the median.
Answer:
5 is NOT a solution to the equation.
t = 6
Step-by-step explanation:
Step 1: Write out equation
2t - 1 = 11
Step 2: Add 1 to both sides
2t = 12
Step 3: Divide both sides by 2
t = 6
Answer:
There is not enough statistical evidence to state that the mean score on one-tailed hypothesis test questions is higher than the mean score on two-tailed hypothesis test questions.
Step-by-step explanation:
To solve this problem, we run a hypothesis test about the difference of population means.
The appropriate hypothesis system for this situation is:
![H_0:\mu_X-\mu_Y=0\\H_a:\mu_X-\mu_Y > 0\\\\$Difference of means in the null hypothesis is:\\\mu_X-\mu_Y=M_0=0\\\\$The test statistic is $Z=\frac{[( \bar X-\bar Y)-M_0]}{\sqrt{\frac{\sigma^2_X}{n_X}+\frac{\sigma^2_Y}{n_Y}}}\\](https://tex.z-dn.net/?f=H_0%3A%5Cmu_X-%5Cmu_Y%3D0%5C%5CH_a%3A%5Cmu_X-%5Cmu_Y%20%3E%200%5C%5C%5C%5C%24Difference%20of%20means%20in%20the%20null%20hypothesis%20is%3A%5C%5C%5Cmu_X-%5Cmu_Y%3DM_0%3D0%5C%5C%5C%5C%24The%20test%20statistic%20is%20%24Z%3D%5Cfrac%7B%5B%28%20%5Cbar%20X-%5Cbar%20Y%29-M_0%5D%7D%7B%5Csqrt%7B%5Cfrac%7B%5Csigma%5E2_X%7D%7Bn_X%7D%2B%5Cfrac%7B%5Csigma%5E2_Y%7D%7Bn_Y%7D%7D%7D%5C%5C)
![$$The calculated statistic is Z_c=\frac{[(7.79-7.64)-0]}{\sqrt{\frac{1.06^2}{80}+\frac{1.31^2}{80}}}=0.79616\\\\p-value = P(Z \geq Z_c)=0.42594\\\\](https://tex.z-dn.net/?f=%24%24The%20calculated%20statistic%20is%20Z_c%3D%5Cfrac%7B%5B%287.79-7.64%29-0%5D%7D%7B%5Csqrt%7B%5Cfrac%7B1.06%5E2%7D%7B80%7D%2B%5Cfrac%7B1.31%5E2%7D%7B80%7D%7D%7D%3D0.79616%5C%5C%5C%5Cp-value%20%3D%20P%28Z%20%5Cgeq%20Z_c%29%3D0.42594%5C%5C%5C%5C)
Since, the calculated statistic 
is less than critical 
, the null hypothesis do not should be rejected. There is not enough statistical evidence to state that the mean score on one-tailed hypothesis test questions is higher than the mean score on two-tailed hypothesis test questions.
