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diamong [38]
3 years ago
14

Jonathan and Tim want to predict how many quarters are in a large jar containing 120 coins. Jonathan randomly selects a sample o

f 10 coins from the jar, while Tim randomly selects a sample of 30 coins. Then they count the number of quarters in each sample. Whose sample should more accurately predict the number of quarters in the jar? sample should give a more accurate prediction of the number of quarters in the jar.
Mathematics
2 answers:
Anastaziya [24]3 years ago
8 0


 the larger the sample the more accurate the prediction

 so since Tim looked at a larger quantity of coins his would be more accurate

kozerog [31]3 years ago
7 0
Tim would give a sample that should more accurately predict the numbers in the jar. He has taken out 1/4 of the coins from the jar.
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Hatshy [7]

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5 0
3 years ago
Solve y^3+x^3=(y+x)(y^2-yx+x^2) if x=2 and y=8
Hitman42 [59]

Answer:

Step-by-step explanation:

8³+2³=(8+2)(8²-8(2)+2²)

512+8=(10)(64-16+4)

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512+8=(10)(44)

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5 0
3 years ago
Which table shows a function that is decreasing only over the interval
joja [24]

Answer:

Table 4th is showing the function that is decreasing only over the interval

Step-by-step explanation:

We have been given four tables below:

To determine the function that is decreasing only in the given interval

(-1,\infty)

We will check the values after -1 in all four tables

The table in which values is only  decreasing after -1 will be the table which shows a function is decreasing only.

In table 1 the value of x is increasing

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8 0
3 years ago
Read 2 more answers
HELP ME!!! Answer in Scientific Notation and explain and show work.
umka21 [38]
Multiplying means you add the exponents and dividing means you subtract the exponents. Treat the decimals as you normally would.
(4.3 \times  {10}^{8} ) \times (2.0 \times  {10}^{6} )
4.3 \times 2.0 = 8.6
8 + 6 = 14
8.6 \times  {10}^{14}

(6 \times  {10}^{3} ) \times (1.5 \times  {10}^{ - 2} )
6 \times 1.5 = 9
3 +  - 2 = 1
9 \times  {10}^{1}


(1.5 \times  {10}^{ - 2} ) \times (8.0 \times  {10}^{ - 1} )
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12 \times  {10}^{ - 3}


\frac{(7.8 \times  {10}^{3}) }{(1.2 \times  {10}^{4}) }
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6.5 \times  {10}^{ - 1}


\frac{(8.1 \times  {10}^{ - 2} )}{(9.0 \times  {10}^{2} )}
\frac{8.1}{9.0}  = 0.9
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0.9 \times  {10}^{ - 4}


\frac{6.48 \times  {10}^{5} }{(2.4 \times  {10}^{4} )(1.8 \times  {10}^{ - 2}) }
2.4 \times 1.8 = 4.32
4 +  - 2 = 2
\frac{6.48 \times  {10}^{5} }{4.32 \times  {10}^{2} }
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1.5 \times  {10}^{3}
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