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Pepsi [2]
3 years ago
14

Round 668294.018492 to the nearest hundred.

Mathematics
1 answer:
AveGali [126]3 years ago
8 0
=668294.02
here this is the answer u think
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Explain the steps you would take to write 36/10 as a decimal
Feliz [49]
Just divide 36÷10 and that equals 3.6. 
      
3 0
3 years ago
Read 2 more answers
Please help me please pleaseee
MatroZZZ [7]

Answer:

2.57

Step-by-step explanation:

10*10*10 = 1000

2570/1000 = 2.57

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3 years ago
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Anyone knows the answers with their explanations
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4 years ago
1. A zoo has an Asian elephant that weighs 6 tons. The elephant got sick and lost 1450 pounds. What is the elephants new weight
Marianna [84]

Answer:

1. The new weight of elephant is 5.342 tones.

3. The number of red flowers are 280.

Step-by-step explanation:

1. weight of elephant = 6 tons

weight lost = 1450 pounds

As , 1 pound = 0.000454 ton

So, 1450 pounds = 0.658 ton

The new weight of elephant is

= 6 - 0.658 = 5.342 tones

3.

red flower to the yellow flower is 5 : 7

Number of yellow flowers = 392

Let the red flowers are R.

So,

\frac {5}{7}=\frac{R}{392}\\\\R = 280

3 0
3 years ago
A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were
Rufina [12.5K]

Answer:

a) ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

b) 0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

\hat p=\frac{372}{900}=0.413 estimated proportion of students were pursuing liberal arts degrees

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

The margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

If we replace we have:

ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

And replacing into the confidence interval formula we got:

0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

7 0
3 years ago
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