Answer:
Detailed steps:
y – 4 = –1 (x – 2)
y – 7 = –1 (x + 1)
y = –x + 6
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We opt to accept the null hypothesis. Given the following details provided: the Sample mean equals 28.8 miles per gallon, Sample size n is 120, and Alpha α is 0.01 with a sample standard deviation of 6.89 miles per gallon. Initially, we set up the null and alternative hypotheses. Utilizing a two-tailed t-test facilitates this hypothesis testing. By substituting the relevant values we calculate, and eventually conclude that we fail to reject the null hypothesis, endorsing that the average MPG for the Toyota Highlander Hybrid vehicles is indeed 28 miles per gallon.
Answer:
Table B demonstrates a linear function.
Step-by-step explanation:
The reason Table B is classified as a linear function is that for each increment of one in x, y increases by a consistent rate of 4.
Imagine having a pile of pennies and you want to determine if the count is odd or even without tallying them one by one. You can separate them into pairs by placing one coin in the left pile and one in the right.
If no coin is left unmatched, then the total number is even; if there is one coin remaining unpaired, the number is odd.
(a) The likelihood that all 5 eggs chosen are unspoiled is 0.0531. (b) The probability that 2 or fewer out of the 5 eggs are unspoiled is 0.3959. (c) The probability that more than 1 of the selected 5 eggs are unspoiled is 0.8747. Step-by-step explanation: The complete query is: A subpar carton of 18 eggs has 8 that are spoiled. An unsuspecting chef selects 5 eggs at random for his “Mega-Omelet Surprise.” Calculate the probability of receiving (a) exactly 5 unspoiled eggs, (b) 2 or fewer, and (c) more than 1 unspoiled egg. Define X = number of unspoiled eggs. In the faulty carton, 8 eggs are spoiled. The probability of selecting an unspoiled egg is independent of others. Provided that a chef randomly picks 5 eggs, the variable X follows a Binomial distribution with parameters n = 5 and p = 0.556. Success is defined as selecting an unspoiled egg. The probability mass function of X is as follows: (a) Calculate the probability of selecting all unspoiled eggs. Thus, this probability is found to be 0.0531. (b) For 2 or less unspoiled eggs, the probability is computed: P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2), resulting in a probability of 0.3959. (c) For more than 1 unspoiled egg: P (X > 1) = 1 - P (X ≤ 1), yields a final probability of 0.8747.
