I think it might be 72+0?
Answer:
<h2>a) approximately 133 graduates</h2><h2>b) approximately 120°</h2>
Step-by-step explanation:
a) the number of graduates planning to continue studying :
= (37 1/2% + 12 1/2% + 16 2/3%) × 200

= (37.5 + 12.5 + 16.666666666667)×2
= 133.333333333334
…………………………………
b) the measurement of the angle representing those who plan to work :
= (360× 33 1/2)÷100
= (360× 33.333333333333)÷100
=119.999999999999
Answer:
Step-by-step explanation: it is C
Answer:
A: (-8, -1)
B: (2, -1)
C: (2, 5)
D: (-6, 5)
Step-by-step explanation:
If the trapezoid ABCD was able to translate 3 units to the left, we'll only need to calculate for all the input values!
Trapezoid in present:
A - -5
B - 5
C - 5
D - -3
Now move the units to the left!
Trapezoid in the future:
A - -8
B - 2
C - 2
D - -6
Then add the y values into the units
A: (-8, -1)
B: (2, -1)
C: (2, 5)
D: (-6, 5)
*************************************************
Answer:
The sample size required is 910.
Step-by-step explanation:
The confidence interval for population proportion is:

The margin of error is:

Given:

The critical value of <em>z</em> for 90% confidence level is:
*Use a standard normal table.
Compute the sample size required as follows:

Thus, the sample size required is 910.