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irga5000 [103]
3 years ago
10

Proportions in Triangles (10)

Mathematics
1 answer:
jolli1 [7]3 years ago
4 0
X would equal 8 because 8 is 2/3 of 12
And 24 is 2/3 of 36
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Find the measure of x and round to the nearest tenth!!
Jet001 [13]

Answer:24

Step-by-step explanation:

the number of the right triangle is to subtract the number

4 0
3 years ago
Complete the sentence: similar figures have the
umka21 [38]
I'm assuming it would be the same shape? That's also assuming that it's referring to congruent shapes and angles. 
7 0
3 years ago
The product of -3 and -7 is positive.<br><br><br> True False
dmitriy555 [2]
True!! If you multiply -3 x -7 you get 21!!
3 0
2 years ago
Read 2 more answers
In how many distinct ways can the letters of the word mathematics be arranged? (first, does the order matter?)
TEA [102]
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).

Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
4 0
3 years ago
There are 20 machines in a factory. 7 of the machines are defective.
adelina 88 [10]

Answer:

0.1225

Step-by-step explanation:

Given

Number of Machines = 20

Defective Machines = 7

Required

Probability that two selected (with replacement) are defective.

The first step is to define an event that a machine will be defective.

Let M represent the selected machine sis defective.

P(M) = 7/20

Provided that the two selected machines are replaced;

The probability is calculated as thus

P(Both) = P(First Defect) * P(Second Defect)

From tge question, we understand that each selection is replaced before another selection is made.

This means that the probability of first selection and the probability of second selection are independent.

And as such;

P(First Defect) = P (Second Defect) = P(M) = 7/20

So;

P(Both) = P(First Defect) * P(Second Defect)

PBoth) = 7/20 * 7/20

P(Both) = 49/400

P(Both) = 0.1225

Hence, the probability that both choices will be defective machines is 0.1225

4 0
3 years ago
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