9514 1404 393
Answer:
"and"
Step-by-step explanation:
Properly written, the word "and" separates the integer portion from the fractional portion of the number. If there is no "and", then there is no integer portion, so the number is a fraction less than 1.
<u>Examples</u>:
one <em>and</em> three tenths = 1 3/10
thirty-one thousandths = 31/1000
Answer: All the ways I solved this the answer is the same,
x>5 or (5,∞) That being the Interval Notation {The second one}
Step-by-step explanation:
Answer:True, becuase every time energy is converted to another form, is some of it is bled off and it made unusable.
Step-by-step explanation:
The given equation of the ellipse is x^2
+ y^2 = 2 x + 2 y
At tangent line, the point is horizontal with the x-axis
therefore slope = dy / dx = 0
<span>So we have to take the 1st derivative of the equation
then equate dy / dx to zero.</span>
x^2 + y^2 = 2 x + 2 y
x^2 – 2 x = 2 y – y^2
(2x – 2) dx = (2 – 2y) dy
(2x – 2) / (2 – 2y) = 0
2x – 2 = 0
x = 1
To find for y, we go back to the original equation then substitute
the value of x.
x^2 + y^2 = 2 x + 2 y
1^2 + y^2 = 2 * 1 + 2 y
y^2 – 2y + 1 – 2 = 0
y^2 – 2y – 1 = 0
Finding the roots using the quadratic formula:
y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1
y = 1 ± 2.828
y = -1.828 , 3.828
<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828)
and (1, 3.828).</span>
Answer:
95% confidence interval for the difference between the average mass of eggs in small and large nest is between a lower limit of 0.81 and an upper limit of 2.39.
Step-by-step explanation:
Confidence interval is given by mean +/- margin of error (E)
Eggs from small nest
Sample size (n1) = 60
Mean = 37.2
Sample variance = 24.7
Eggs from large nest
Sample size (n2) = 159
Mean = 35.6
Sample variance = 39
Pooled variance = [(60-1)24.7 + (159-1)39] ÷ (60+159-2) = 7619.3 ÷ 217 = 35.11
Standard deviation = sqrt(pooled variance) = sqrt(35.11) = 5.93
Difference in mean = 37.2 - 35.6 = 1.6
Degree of freedom = n1+n2 - 2 = 60+159-2 = 217
Confidence level = 95%
Critical value (t) corresponding to 217 degrees of freedom and 95% confidence level is 1.97132
E = t×sd/√(n1+n2) = 1.97132×5.93/√219 = 0.79
Lower limit = mean - E = 1.6 - 0.79 = 0.81
Upper limit = mean + E = 1.6 + 0.79 = 2.39
95% confidence interval for the difference in average mass is (0.81, 2.39)