If x=-2 solve 4x^3-6x^2+x-1
2 answers:
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Complete question is;
Use a t-test to test the claim about the population mean μ at the given level of significance α using the given sample statistics. Assume the population is normally distributed.
Claim: μ ≥ 8300,α= 0.10
Sample statistics: ¯x = 8000, s = 440, n = 24
a. What are the null and alternative hypotheses?
b. What is the value of the standardized test statistic? (Round to 2 decimal places as needed.)
c. What is the p-value? (Round to three decimal places as needed.)
d. Decide whether to reject or fail to reject the null hypothesis.
Answer:
A) Null hypothesis:H0: μ ≥ 8300
Alternative Hypothesis:Ha:μ < 8300
B) t = -3.34
C) p-value = 0.001
D) we will reject the null hypothesis
Step-by-step explanation:
A) We are told that the claim is: μ ≥ 8300. Thus, the null hypothesis would be the claim. So;
Null hypothesis:H0: μ ≥ 8300
Also, alternative hypothesis would be;
Alternative Hypothesis:Ha:μ < 8300
B)Formula for standardized test statistic with a t-test is;
t = (¯x - μ)/√(s/n)
Plugging in the relevant values, we have;
t = (8000 - 8300)/√(440/24)
t = -3.34
C) From online p-value from t-score calculator attached using t = -3.34, n = 24, significance level = 0.01, DF = 24 - 1 = 23 and a one - tailed test, we have;
p-value = 0.001421 ≈ 0.001
D) The p-value of 0.001 is less than the significance value of 0.01,thus we will reject the null hypothesis
Answer: 4 sixth-grade homerooms.
Step-by-step explanation:
According to the data given, each homeroom should have the same number of boys and the same number of girls. Since there are 32 boys and 36 girls, you must find the Greatest Common Divisor between theses numbers.
The steps are:
1. Descompose 36 and 32 into their prime factors:
2. Multiply the commons and non commons with the lowest exponent. In this case 2 is the only common one. So:
Therefore, the greatest possible of sixth-grade homeroom is 4.