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Mashutka [201]
3 years ago
11

The width of a rectangular computer screen is 2 inches more than its height. If the area of the screen is 323 square inches, fin

d its dimensions.
Mathematics
1 answer:
Lana71 [14]3 years ago
5 0
Width= 19 in
Height= 17 in

WxH=323

19x17=323
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Elena-2011 [213]
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1. Find P (phone has internet or phone plays music) using this Venn diagram. 
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Step-by-step explanation:

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the venn diagram represents the results of a survey that asked participants whether they would want a rabbit or a snake as a pet
rosijanka [135]
For this case we will complete the table part by part.
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The odds of winning a contest are 3:7. What is the probability winning the contest?
DiKsa [7]

Answer:

0.3

Step-by-step explanation:

Given: The odds of winning a contest are 3:7

To find: probability of winning the contest

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3 years ago
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Dvinal [7]

                                             Question 11)

Answer:

\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80

Step-by-step explanation:

Given the expression

\log _2\left(63\right)-\log _2\left(9\right)

\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)

\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(\frac{63}{9}\right)

=\log _2\left(\frac{63}{9}\right)

\mathrm{Divide\:the\:numbers:}\:\frac{63}{9}=7

=\log _2\left(7\right)

=2.80

Therefore,

\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80

                                            Question 12)

Answer:

\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32    

Step-by-step explanation:

Given the expression

\log _2\left(3\right)+\log _2\left(15\right)-\log _2\left(9\right)

\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)

\log _2\left(3\right)+\log _2\left(15\right)=\log _2\left(3\cdot \:15\right)

=\log _2\left(3\cdot \:15\right)-\log _2\left(9\right)

\mathrm{Multiply\:the\:numbers:}\:3\cdot \:15=45

=\log _2\left(45\right)-\log _2\left(9\right)

\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)

\log _2\left(45\right)-\log _2\left(9\right)=\log _2\left(\frac{45}{9}\right)

=\log _2\left(\frac{45}{9}\right)

\mathrm{Divide\:the\:numbers:}\:\frac{45}{9}=5

=\log _2\left(5\right)

=2.32

Therefore,

\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32                            

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3 years ago
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