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ycow [4]
3 years ago
9

Round 344,501 to the nearest 10,000

Mathematics
1 answer:
Luda [366]3 years ago
3 0

Answer:

Step-by-step explanation:

Round 344,501 to the nearest ten thousand

340000 to the nearest ten thousand

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Most analysts focus on the cost of tuition as the way to measure the cost of a college education. But incidentals, such as textb
Charra [1.4K]

Answer:

The population of interest to the researcher were the 250 first-year students that were monitored.

Step-by-step explanation:

In descriptive statistics, the portion of the cost of college education to be determined and has been selected for analysis is calle d "sample", the sample the researcher is interested in, considers the textbooks cost of first-year students, therefore the 250 first-year students is the researcher´s population of interest. This method involved the collection, presentation, and characterization.

8 0
2 years ago
David has $100.00 saved in the bank. a payment is taken out of his account for $150.00. he makes a deposit for $50.00 but at the
Dmitry_Shevchenko [17]

Gimme a few mins, imma edit this later... ;-;

6 0
3 years ago
Read 2 more answers
Help on both please ! Im a new student at this school and didn't learn this
larisa86 [58]
For the first one, you have to convert the fractions to an improper fraction. To do that you need to multiply the bottom denominator number (3) by the whole number (1) then you need to add the numinator, so 3x1+2= 5. You have to keep the denominator so 1 2/3 is equal to 5/3. Then do the same to 2 1/5, and you get 11/5. Now you have to find a common denominator, that's basically the smallest number that both numbers can go Into, the lowest common denominator for 3 and 5 is 15. So 3x5= 15, so we have to multiply the top number by 5 which is 25. So 5/3 is equal to 25/15, then 5x3= 15, so you need to multiply 11 by 3 which is 33. So 11/5 is equal to 33/15. Then you add them. Add the numinators (25+33=58. Then you keep the denominator 15. So when u add it it's 58/15 then you need to simplify that and you get 3 13/15.

The second one you turn them into improper fractions like I told you how to before (multiply the bottom number by the whole number then add he top number, then add he same denominator.) do that for both. Then you line them right next to each other and multiply across. (I just realized that they were the same number so they are equal to 5/3 and 11/5)
Then you do 5x11 and you get 55 then do 3x5 and you get 15. 55/15 is your answer, but you need to simplify it, you need to divide 55 by 15, (not all the way just the first number) so you do 15x3 and that's 45, then you subtract that from 55, and you get 10, so then you take your denominator (15) and you answer is 3 10/15. But when you simplify it it's 3 and 2/3


Hope I helped sorry it's so long and sorry for any typos it's so long I didn't go back and check
5 0
3 years ago
Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

7 0
2 years ago
Simplify. Assume that letters represent any real number.
TiliK225 [7]
\sqrt[5]{x^5}=x\\\\32^\frac{3}{5}=\left(32^\frac{1}{5}\right)^3=\left(\sqrt[5]{32}\right)^3=2^3=8
7 0
3 years ago
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