All categories

3 years ago

How To Round 17 5/8 To The Nearest Whole Number

1 answer:

5 0

Answer:
18

Explanation:
To round a number to the nearest whole number, we would check the fraction part:
If the fraction is less than 0.5, then we would simply ignore the fraction
If the fraction is equal to or more than 0.5, then we would add 1 to the whole number and eliminate the fraction.

In the given , we have:
 $17 \frac{5}{8}$ .

The fraction part is $\frac{5}{8}$ which is greater than 0.5
This means that we would add one to the whole number (round it up) and eliminate the fraction.
Therefore, the answer would be 18.

Hope this helps :)

You might be interested in

What is the solution to the system of equations? y = 1/3x – 10 2x + y = 4

kolezko [41]

The solution is (6, -8) (solved by graphing with Desmos.com/calculator)

5 0

3 years ago

Read 2 more answers

Charles asked his teammates how many hours they practiced swimming during the week. He recorded his data in the table below and

Kitty [74]

Answer:

The answer is C.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The lemonade stand at the county fair sells the lemonade at a price of two cups for $3.60. Complete the table at right to find w

kakasveta [241]

$3.60 for 2 cups would be $1.80 for one cup.

$3.60 for 2

$5.40 for 3

$7.20 for 4

$9 for 5

$10.80 for 6

$12.60 for 7

$14.40 for 8

4 0

3 years ago

Hi I really need help with my math questions because I am on the edge of failing math class please help me out thank you!

Troyanec [42]

survey every employee who works in the store

Step-by-step explanation:

7 0

3 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

3 years ago