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Mice21 [21]
3 years ago
15

there are approximately 3 million births in the unitad states each year .find the birth rate per minute

Mathematics
2 answers:
juin [17]3 years ago
3 0

Answer:

Approx. 6 b/m

Step-by-step explanation:

Find out How many minutes in a year which is 525600 then divide that by 3,000,000 which makes 5.7077 so rounded is 6.

Dvinal [7]3 years ago
3 0

Answer:

5.7/min

Step-by-step explanation:

<u>Divide 3 million by 365 to represent 1 day:</u>

3,000,000 / 365 = approximately 8,219.

<u>Divide 8,219 by 24 to represent 1 hour:</u>

8,219 / 24 = approximately 342.

<u>Divide 342 by 60 to represent one minute:</u>

342 / 60 = 5.7.

The birth rate in United States is approximately 5.7/min

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Find the GCF of the terms of the polynomial. 18x4 – 27x3 – 36x2
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Answer:

9 x^ 2

Step-by-step explanation:

18x4 – 27x3 – 36x2

<u>GCF :</u>

<em>Find the prime factors of each term in order to find the greatest common factor (GCF)</em>

= 9 x^ 2

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4 years ago
The sum of three intergers is 242. The 2nd number is 3 more than twice the first and the 3rd number is 9 less than 5 times the f
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Answer:

31, 65, 146

Step-by-step explanation:

math

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3 years ago
Kathy has ribbons that have lengths of 7 inches, 10 inches, and 12 inches. Explain how she can use these ribbons to measure a le
Snowcat [4.5K]
Very   easy:
put the   10 and 12 end to end  then    back off the 7 inches
22-7 = 15

7 0
4 years ago
How many solutions does the equation −4y + 4y + 2 = 2 have? (1 point) a Two b None c One d Infinitely many
elena55 [62]

Answer:

Infinity

Step-by-step explanation:

-4y + 4y + 2 = 2

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6 0
3 years ago
Read 2 more answers
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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