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Nadya [2.5K]
3 years ago
15

What is the distance between (-5,8) and( 6,8)

Mathematics
1 answer:
nekit [7.7K]3 years ago
6 0

Answer: 11

Step-by-step explanation:Since the y coordinates are the same, the distance is the values between the x coordinates 6 - -5

6+5

11

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Shaneesha wants to display all of her 48 trophies, so her uncle said he would make her some shelves. She can fit 8 trophies on e
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Anne claims that a store-brand fertilizer works better than homemade compost as a soil enhancement when growing tomatoes. To tes
goldenfox [79]

Answer: Paired t interval for μdiff

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The confidence interval suitable for analysing the data described in the scenario above is the paired t test which is employed when there are two different sets of measured reading or observation for each subject. In the scenario above, the subjects are of of the different planters employed whereby the pair of measurement are the weight of tomatoes obtained for each of the two different fertilizer types applied to each of the two plants in a planter. With these data, we can measure the difference in the effectiveness of each of the store brand and homegrown compost.

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Is 3,9,27,81 an arithmetic sequence
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3 years ago
What is the product of -8/1 * -1/8​
spayn [35]

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1

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Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t
Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let \sigma_{1}^{2} be the variance for the population of weight gains for rats given a low dose, and \sigma_{2}^{2} the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2} vs H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}. We have that the sample standard deviation for n_{2} = 22 female control rats was s_{2} = 28 g and for n_{1} = 18 female low-dose rats was s_{1} = 51 g. So, we have observed the value

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As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

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