Using it's concept, it is found that the prevalence of frequent multivitamin use is of 0.4523 = 45.23%.
<h3>What is the prevalence of a sample?</h3>
The prevalence of a sample is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.
In this problem, 588 out of 1300 women reported taking a multivitamin at least 4 times a week during the month before pregnant, hence the prevalence is given by:
p = 588/1300 = 0.4523 = 45.23%.
More can be learned about prevalence at brainly.com/question/14398287
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Answer:
29
Step-by-step explanation:
x=4. 4-1=3. 7x3=21. 21+8=29. Viola
Answer:
Question 1 ) Difference of Volume = 112.25 cm³
Question 2) Volume = 6400π feet³
Step-by-step explanation:
<h3>As part of a new advertising campaign, a beverage company wants to increase the dimensions of their cans by a multiple of 1.10. If the cans are currently 12 cm tall, 6 cm in diameter, and have a volume of 339.12 cm3, how much more will the new cans hold? Use 3.14 for π and round your answer to the nearest hundredth.</h3>
Diameter = 6 cm
Radius = 3 cm
Height = 12 cm
If we increase the dimension by 1.10, new dimesnions are:
Radius = 3 · 1.1 = 3.3 cm
Height = 12 · 1.1 = 13.2 cm
Volume = (Area)(Height) = (πr²)(Height)
Volume = (π)(3.3²)(13.2)
Volume = 451.37 cm³
Difference of Volume = 451.37 cm³ - 339.12 cm³
Difference of Volume = 112.25 cm³
<h3>
The circumference of a redwood tree trunk is 16π ft, and it is 100 ft tall. What is the approximate volume of the redwood tree trunk? </h3>
Circumference = 2πr = 16π = 2π(8) feet
Radius = 8 feet
Volume = Volume = (Area)(Height) = (πr²)(Height)
Volume = (π)((8²)(100)
Volume = 6400π feet³
1. = 5 animals
2. = 22 animals
3. = 32 animals
Hope this helped
Answer:
See Explanation
Step-by-step explanation:
The question is incomplete as the matrix is not given.
The general form of a matrix is:
![b[m*n] =\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}........b{1n}}\\b_{21}&b_{22}&b_{23}........b{2n}\\b_{31}&b_{32}&b_{33}........b{3n}\\ \\ \\b_{m1}&b_{m2}&b_{m3}........b{mn}\end{array}\right]](https://tex.z-dn.net/?f=b%5Bm%2An%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db_%7B11%7D%26b_%7B12%7D%26b_%7B13%7D........b%7B1n%7D%7D%5C%5Cb_%7B21%7D%26b_%7B22%7D%26b_%7B23%7D........b%7B2n%7D%5C%5Cb_%7B31%7D%26b_%7B32%7D%26b_%7B33%7D........b%7B3n%7D%5C%5C%20%5C%5C%20%5C%5Cb_%7Bm1%7D%26b_%7Bm2%7D%26b_%7Bm3%7D........b%7Bmn%7D%5Cend%7Barray%7D%5Cright%5D)
Where m and n represent the 
respectively.
Notice the position of b11
This means that b11 is at the first column and first row.
Take for instance matrix b is:
![b = \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]](https://tex.z-dn.net/?f=b%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D)
The element at the first row and first column is 1.
Hence
