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Wittaler [7]
3 years ago
5

Baby jacci, who is 2 days old, is shown four drawings: a blue square, a white oval, a drawing of a face, and a bright red circle

. based on fantz's work, which will she probably prefer to look at?
Mathematics
1 answer:
DENIUS [597]3 years ago
4 0
Hello!
.
The answer to your question is "bright red circle".
.
Baby Jacci, who is 2 days old, will probably prefer to look at the bright red circle because it stands out from the other drawings because of its bright color.
:)
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75400 because you times is
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A survey of shoppers is planned to determine what percentage use credit cards. prior surveys suggest​ 63% of shoppers use credit
Alona [7]
We would need a sample size of 560.

We first calculate the z-score associated. with this level of confidence:
Convert 95% to a decimal:  95% = 95/100 = 0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this is associated with a z-score of 1.96.

The margin of error, ME, is given by:
ME=z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

We want ME to be 4%; 4% = 4/100 = 0.04.  Substituting this into our equation, as well as our proportion and z-score,
0.04=1.96\sqrt{\frac{0.63(1-0.63)}{n}}
\\
\\\text{Dividing both sides by 1.96,}
\\
\\\frac{0.04}{1.96}=\sqrt{\frac{0.63(0.37)}{n}}
\\
\\\text{Squaring both sides,}
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\\(\frac{0.04}{1.96})^2=\frac{0.63(0.37)}{n}
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\\\text{Multiplying both sides by n,}
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\\n(\frac{0.04}{1.96})^2=0.63(0.37)
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\\n(\frac{0.04}{1.96})^2=0.2331
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3 0
3 years ago
Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Sea-hawks when they won Su
weeeeeb [17]

Answer:

No. These measures show a thinner team of NFL players according to the mean, variance, standard deviation, and quartiles.

Step-by-step explanation:

1) The measures of variation, namely The Range, Variance, Quartiles, Interquartiles, Sum of Squares, etc. shows us how the data are dispersed.

The Range Δ is calculated:

\Delta =305-189=116 Maximum value - Minimum value for weight

Mean:

\bar{x}=\frac{189+254+235+225+190+305+202+190+252+305}{11}\approx 231.09 \:pounds

Variance:

s^{2}=\frac{\sum_{i=1}^{n} (x_i-\bar{x})^{2}}{n-1}\Rightarrow s=1923.69\\

The Standard Deviation of the sample

s=\sqrt{s^{^{2}}}\Rightarrow s\approx 43.86\\

2) Since there is no preceding exercise, the comparison was made to a recent study in which a NFL player average weight is about 245 pounds (average),

Since 25% of this list are player whose weight is 192.5 lbs and 50% (2nd Quartile) =225 lbs , finally only at the 3rd Quartile we have players above the regular NFL average with 253. This, added with the other data, allow us to say that this list is not a typical of all NFL players.

4 0
3 years ago
What is the area of FGH
Dmitrij [34]
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8 0
3 years ago
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Inessa [10]

Answer:

$153

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Now, we are able to calculate the ultimate cost of coating:

If the cost is $10 per cm^2, then we have to find how much it costs to coat 15.3 cm^2.

15.3 x 10 = $153.

Hope this helps

4 0
3 years ago
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