Answer: 9
Step-by-step explanation: so first you remove the ( ) because they’re useless in this then once you remove then Since two opposites add up to zero, remove them from the expression then you only get left with 9.
A. √(0.8^2) + (0.6^2) = √1 = 1 => OK
<span>b.(-2/3,√ 5/3) = √(-2/3)^2 + 5/9) = √(4/9 +5/9) = √1 = 1 => OK
c.(√ 3/2, 1/3) = √(3/4 + 1/9) < 1 => it is inside the unit circle
d.(1,1)
= √(1 + 1) = √2 > 1 => NO. This point is beyond the limits of the unit circle.</span>
Answer:
The sweat chloride reference value is less than 30 mmol/L. A value of more than 60 mmol/L of chloride in the sweat is consistent with a diagnosis of cystic fibrosis. The values of 30-60 mEq/L may represent heterozygous carriers, these carriers cannot be accurately identified with a sweat chloride test.
Total children = 6
1) Probability of 4 girls =4/6=0.66
Probability of at least 4 girls >_ 0.66
That is greater or equal to 0.66
2) probability of 5 girls = 5/ total children
=5/6=0.83
Probability of at most 5 girls <_ 0.83
That is less than equal to 0.83
