Pretty sure ( -2,-7) because from -1 you go over one then each y is going down 3 so -4 - 3 is -7
Answer:
50 i think
Step-by-step explanation:
<h3>
Answer: 23.13 square meters</h3>
===============================================================
Explanation:
The top and bottom semicircles can be joined to form a full circle. The same can be said about the left and right semicircles. Each circle has diameter 3 and radius 1.5
The area of one circle is
A = pi*r^2
A = 3.14*(1.5)^2
A = 7.065
So two circles doubles to 2*7.065 = 14.13 square meters in area.
Then the last step is to add on the area of the 3 by 3 square (area 3*3 = 9) to get 14.13+9 = 23.13
This area is approximate because pi = 3.14 is approximate. Use more decimal digits in pi to get a more accurate area. However, your teacher wants you to use this specific value so it's best to stick with 23.13
Answer:
A person must get an IQ score of at least 138.885 to qualify.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
(a). [7pts] What IQ score must a person get to qualify
Top 8%, so at least the 100-8 = 92th percentile.
Scores of X and higher, in which X is found when Z has a pvalue of 0.92. So X when Z = 1.405.
A person must get an IQ score of at least 138.885 to qualify.
Answer:
Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:
a=√{b²+c²−2bc×cos40°}
a=√{149.645−149.645cos40°}
Area Nonagon = (9/4)a²cos40°
=9/4[149.645−149.645cos40°]cot20°
=336.70125[1−cos(40°)]cot(20°)
Applying an identity for the cos(40°) does not get us very far…
= 336.70125[1−(cos2(20°)−1)]cot(20°)
= 336.70125[2−cos2(20°)]cot(20°)
= 336.70125[2−(1−sin2(20°))]cot(20°)
= 336.70125[1+sin2(20°)]cos(20°)sin(20°)
= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²
