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3 years ago

Write 10,000 in scientific notation

Mathematics

1 answer:

Semenov [28]3 years ago

3 0

1*10^4
Or
10^4
Either way

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Coach Kevin brought 18 gallons of water to the football game.

VLD [36.1K]

Answer:

288

Step-by-step explanation:

16 cups per gallon so 16x18=288

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3 years ago

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Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units ass

FrozenT [24]

Answer:

H0 is rejected.There difference in the means for the three assembly methods.

Step-by-step explanation:

Calculating gives the following results.

ANOVA Table

Source DF Sum Mean F Statistic

of Square Square

Treatments 2 4560 2280 9.865

(between groups)

<u>Error 27 6240 231.111 </u>

Total 29 10,800

Error SS= SST -SSTR = -10,800-4560=6240

d.f for SSTr = r-1= 3-1

D.f for SST= n-r= 30-3= 27

Total D.f= d.f for SSTr+D.f for SST= 27+29

MSTR= SSTR/d.f= 4560/2= 2280

MSError= Error SS/ d.f= 6240/27=231.111

F test= MSTR/ MsError= 2280/231.11= 9.865

The P- value < 0.01 which is less than 0.05 therefore H0 is rejected.

There is sufficient evidence to reject H0.

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3 years ago

What is the answer to thus problem

Dominik [7]

The numerator is 16 and the denominator is 3.

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4 years ago

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What is the median of the following data values?

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The median of the following data values is 100

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3 years ago

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"Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning." These words begin

krok68 [10]

Answer:

The answer is "(0.461 , 7.206)"

Step-by-step explanation:

In the given question some information is missing, that is data. so, the correct answer to this question can be defined as follows:

UnLogged:

$x_1=17.500\\s_1=3.529\\n_1=12$

Logged:

$x_2=13.667\\ s_2=4.500\\n_2=9\\$

formula of std error:

$\ std \ error =\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

$=\sqrt{\frac{3.529^2}{12}+\frac{4.500^2}{9}}\\\\=\sqrt{\frac{3.529^2\times 3+4.500^2\times 4}{36}}\\\\=1.8133$

Point differential estimation = $x_1-x _2$

$= 17.500-13.667\\=3.833$

For 90 percent t= 1.860 Cl & 8 df

error of margin E = $t\times \ std \ error$

$= 1.860 \times 1.8133\\= 3.373$

Lower bound =mean difference -E=0.461

Upper bound = mean difference+ E=7.206

In the above 90% confidence interval were the population mean= (0.461 , 7.206)

7 0

3 years ago