About 115 are pansies when you round down from the original number (115.44)
If the length of a rectangle is 3m longer than its width, then:
L=W+3
Is the area really 154^2? Or is it 154m^2? If yes, then:
A=LW
154=(W+3)(W)
154=(W^2+3W)
0=W^2+3W-154
0=(W-11)(W+14)
This means either (W-11) or (W+14) is equal to zero so:
W=11 and W=-14
To find out let's substitute the numbers:
154=(11+3)(11)
154=154
Therefore, the width of the rectangle is 11m
Pls. see attachment.
We need to solve for the angles of the smaller triangle in
order to solve for the angle of the larger triangle which would help us solve
the missing measurement of a side.
Given:
51 degrees.
Cut the triangle into two equal sides and it forms a right
triangle. All interior angles of a triangle sums up to 180 degrees.
180 – 51 – 90 = 39 degrees
39 degrees * 2 = 78 degrees.
Angle Q is 78 degrees.
In the bigger triangle, 4.3 is the hypotenuse. We need to
solve for the measurement of the long leg which is the opposite of the 78
degree angle.
We will use the formula:
Sine theta = opposite / hypotenuse
Sin(78 deg) = opposite / 4.3
Sin(78 deg) * 4.3 = opposite
4.21 = opposite. This is also the height of the triangle.
Area of a triangle = ½ * base * height
A = ½ * 3units * 4.21units
A = 6.315 square units.
The answer is B. 10 9/12 + 2 1/12= 12 10/12. 12÷2=6 10÷2=5 making the answer 12 5/6
Answer:
The correct option is (b).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

The confidence interval for population mean can be computed using either the <em>z</em>-interval or <em>t</em>-interval.
The <em>t</em>-interval is used if the following conditions are satisfied:
- The population standard deviation is not known
- The sample size is large enough
- The population from which the sample is selected is normally distributed.
For computing a (1 - <em>α</em>)% confidence interval for population mean , it is necessary for the population to normally distributed if the sample selected is small, i.e.<em>n</em> < 30, because only then the sampling distribution of sample mean will be approximated by the normal distribution.
In this case the sample size is, <em>n</em> = 28 < 30.
Also it is provided that the systolic blood pressure is known to have a skewed distribution.
Since the sample is small and the population is not normally distributed, the sampling distribution of sample mean will not be approximated by the normal distribution.
Thus, no conclusion can be drawn from the 90% confidence interval for the mean systolic blood pressure.
The correct option is (b).