Answer:
Explanation:
<u>1) Find the z-scores:</u>
a) z-score for 22.6 inches length
- z = [ 22.6 - 20 ] / 2.6 = 1.00
b) z-score for 17.4 inches length
- z = [ 17.4 - 20 ] / 2.6 = - 1.00
<u>2) Probability</u>
Then, you have to find the probability that the length of an infant is between - 1.00 and 1.00 standards deviations (σ) from the mean (μ).
That is a well known value of 68%, which is part of the 68-95-99.7 empirical rule.
The most exact result is obtained from tables and is 68.26%:
- 1 - P (z ≥ 1.00) - P (z ≤ - 1.00) = 1 - 0.1587 - 0.1587 = 0.6826 = 68.26%
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Answer:
no idea with the answer pls check with otherr
We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.