which quadrilateral has four bisectors that are congruent and bisect eachother a. square b. rectangle c. trapezoid d. parallelog
2 answers:
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Answer:
Step-by-step explanation:
Hello!
So you have a new type of shoe that lasts presumably longer than the ones that are on the market. So your study variable is:
X: "Lifetime of one shoe pair of the new model"
Applying CLT:
X[bar]≈N(μ;σ²/n)
Known values:
n= 30 shoe pairs
x[bar]: 17 months
S= 5.5 months
Since you have to prove whether the new shoes last more or less than the old ones your statistical hypothesis are:
H₀:μ=15
H₁:μ≠15
The significance level for the test is given: α: 0.05
Your critical region will be two-tailed:
So you'll reject the null Hypothesis if your calculated value is ≤-1.96 or if it is ≥1.96
Now you calculate your observed Z-value
Z=<u>x[bar]-μ</u> ⇒ Z=<u> 17-15 </u> = 1.99
σ/√n 5.5/√30
Since this value is greater than the right critical value, i.e. Zobs(1.99)>1.96 you reject the null Hypothesis. So the average durability of the new shoe model is different than 15 months.
I hope you have a SUPER day!
Answer:
a) The difference is of 3.222 lbs.
b) 1.64 standard deviations.
c) Z = 1.64
d) Not significant, as the z-score of 1.64 is between -2 and 2.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The mean of all of the weights is x=2.088 lb, and the standard deviation of the weights is s=1.968 lb.
This means that
a. What is the difference between the weight of 5.31 lb and the mean of the weights?
This is
The difference is of 3.222 lbs.
b. How many standard deviations is that [the difference found in part (a)]?
This is the z-score. So
1.64 standard deviations.
c. Convert the weight of 5.31 lb to a z score.
Z = 1.64, as found above.
d. If we consider weights that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the weight of 5.31 lb significant?
Not significant, as the z-score of 1.64 is between -2 and 2.