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Fiesta28 [93]
3 years ago
7

I am pretty sure I know the answer but I need to be certain. Please answer and give explanation if possible.

Mathematics
1 answer:
creativ13 [48]3 years ago
8 0

Answer:

Vertical opposite angles are the same so the answer is D.

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2. A student had taken six tests and received scores of 88, 73, 81, 83, 79, 94. The seventh test
Veronika [31]

Answer:

a) 83

b) mean and mode are both 83

Step-by-step explanation:

a) mean is the average. so add all the numbers and divide by the number of numbers.

<u>88 + 73 + 81 + 83 + 79 + 94 + T</u> = 83

7

b) median - put the numbers in numerical order, smallest to largest, what is the middle number (4th number) 73, 79, 81, 83, 83, 88, 94

mode - which number shows up most frequently?

8 0
3 years ago
Container a and b hold 11,875 liters of water altogether. Container b holds 2,391 liters more than container a holds. How much w
Viefleur [7K]
Container A holds x liters of water.
Container B holds 2391 liters of water more than container A, so container B holds x+2391 liters of water.
Container A and B hold 11875 liters of water altogether.

x+x+2391=11875 \\&#10;2x+2391=11875 \\&#10;2x=11875-2391 \\&#10;2x=9484 \\&#10;x=\frac{9484}{2} \\&#10;x=4742

Container A holds 4742 liters of water.
5 0
3 years ago
g An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single b
Fed [463]

Answer:

Therefore the of blue in the second urn is 4.

Step-by-step explanation:

Let second urn contain x number of blue ball .

Urn            Red Ball          Blue Ball         Total Ball

1                       4                       6                      10

2                      16                       x                    16+x

Getting a red ball from first urn is P(R_1)=\frac{\textrm {Number of red ball}}{\textrm {Total ball}}    =\frac{4}{10}

Getting a blue ball from first urn is P(B_1)=\frac{\textrm {Number of blue ball}}{\textrm {Total ball}} =\frac {6}{10}

Getting a red ball from second urn is P(R_2)=\frac{\textrm {Number of red ball}}{\textrm {Total ball}}    =\frac{16}{16+x}

Getting a blue ball from second urn is P(B_2)=\frac{\textrm {Number of blue ball}}{\textrm {Total ball}} =\frac {x}{16+x}

Getting two red balls from first and second urn is =\frac{4}{10}\times \frac{16}{16+x}

                                                                                  =\frac{32}{5(16+x)}

Getting two blue balls from first and second urn is =\frac{6}{10}\times \frac{x}{16+x}

                                                                                  =\frac{3x}{5(16+x)}

The probability that both balls are the same in color is =\frac{32}{5(16+x)}+\frac{3x}{5(16+x)}

Given that the probability that both balls are the same in color is 0.44.

According to the problem,

\frac{32}{5(16+x)}+\frac{3x}{5(16+x)}=0.44

\Rightarrow \frac{32+3x}{5(16+x)} =0.44

\Rightarrow \frac {32+3x}{(80+5x)} =0.44

\Rightarrow 32+3x =0.44(80+5x)

\Rightarrow 32+3x =35.2 +2.2x

\Rightarrow 3x -2.2 x= 35.2 -32

\Rightarrow 0.8x= 3.2

\Rightarrow x = 4

Therefore the of blue in the second urn is 4.

               

5 0
3 years ago
mrs tart takes a survey she discovers 60% of her students prefer using pens over pencils in class. if 12 students in Mrs. Tart c
mojhsa [17]

Answer:

20

Hope this helps :D

           

5 0
3 years ago
Read 2 more answers
Idk how to do this someone help me
Vlad [161]
I'm thinking its 4 to the 12 power but I am not completely sure on this
4 0
3 years ago
Read 2 more answers
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