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Drupady [299]
4 years ago
10

Suppose you have two similar trapezoids with a scale factor of 2. If the angle measures of trapezoid ABCD are 80∘, 80∘, 100∘, an

d 100∘, what are the measures of the angles in trapezoid EFGH?
Mathematics
1 answer:
deff fn [24]4 years ago
7 0

Answer:

160, 160, 200, 200

Step-by-step explanation:

All you have to do is double the numbers.

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Most everyday situations involving chance and likelihood are examples of ______. simple probability permutations conditional pro
Fynjy0 [20]

Answer:

Most everyday situations involving chance and likelihood are examples of simple probability.

Explanation:

The probability is the chance or likelihood of any event happening. In our everyday life, we unintentionally use the probability. For example, we say there is 70% chance that tomorrow will be rain, there is 50% chance that Brazil will win the world cup, there is less likelihood of he arriving today and soon. In all these concepts we are dealing with uncertainty and there is chance factor involved in all these examples. So in most everyday situations which involve chance and likelihood are actually examples of simple probability.

6 0
3 years ago
10-6-(8)???? ???????
Andrej [43]
-4 is the answer. 10-6 is 4. 4-8 is -4
5 0
3 years ago
Read 2 more answers
What is 3.75x-5-8.75+4.5+0.5 I need help ASAP it is a simplify the expression question
miv72 [106K]

Answer:

3.75x-8.75 or

Step-by-step explanation:

3 0
3 years ago
Four different sets of objects contain 2,5,6, and 7 objects, respectively. How many unique combinations can be formed by picking
Ilia_Sergeevich [38]
There are 2 choices for the first set, and 5 choices for the second set. Each of the 2 choices from the first set can be combined with each of the 5 choices from the second set. Therefore there are 2 times 5 combinations from the first and second sets. Continuing this reasoning, the total number of unique combinations of one object from each set is:
2\times5\times\times6\times7=420\ combinations
8 0
3 years ago
PLEASE HELP!
butalik [34]
We know for the problem that the performer earned $120 at a performance where 8 people attend. We also know that he u<span>ses 43% of the money earned to pay the costs involved in putting on each performance, so we need to find the 43% of $120. To do that, we are going to divide 43% by 100%, and then multiply it by $120:
</span>\frac{43}{100}*120=51.6
Now we know that the performer uses $51.6 of $120 to pay the costs involved in putting on each performance. The only thing left to find his profits is subtract $51.6 from $120:
120-51.6=68.4

We can conclude that the performer makes a profit of $68.4 when 8 people attend his performance.
 
3 0
3 years ago
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