The dot on -1 is open, so the number -1 is not included. You need all real numbers greater than -1.
Answer: D
Answer:
The value of x = 10.
Step-by-step explanation:
From the given triangle,
- The value of angle M = (3x + 28)°
- The value of angle K = (5x - 18)°
- The value of angle L = 90°
We know that the sum of a triangle is 180°.
i.e.
K + L + M = 180°
(5x - 18)° + 90° + (3x + 28)° = 180°







Therefore, the value of x = 10
Answer:
<u>A. p(hat) = .139</u>
We divide our sample population by the amount who tested positive. 14851/107109 = .139.
<u>B. 1.62 million</u>
We just multiply the p times the population. 11.69 M * .139 = 1.62 M
<u>C. No</u>
It depends upon the sample method. From what I can tell, I assume all conditions are met and it was not biased.
If it wasn't random, that is a problem, but we aren't given this information.
We can test if it's small enough. It can't be larger than 10% of the population. 107109 * 10 < 11.69 million, so it's small enough.
We can also test if it's large enough. np and nq must be greater than 10. 107100 * .139 > 10, 107100 * .861 > 10.
Vs<vu
cause smaller sides have smaller angels in front ofthem