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Arlecino [84]
3 years ago
14

Describe a situation where it is easier to use decimals than friction,and explain why.

Mathematics
1 answer:
Bess [88]3 years ago
3 0
<span>Describe a situation where it is easier to use decimals than fractions and explain why:
</span>When you must multiply numbers that are NOT whole, decimals are easier to change in the sense that setting up a multiplication problem with decimals rather than fractions is much simpler and easier to calculate.
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Two students compared the scores on their math tests. Their results on 9 tests are shown in the box plots below.
Pani-rosa [81]

The statements which are true regarding the students whose results on 9 tests are as shown are; The median of Nadia's data is equal to the median of Ben's data.

Nadia had the highest score on a test.

<h3>Which statements are true regarding the scores on the tests?</h3>

According to the task content, it follows that the results of the students in discuss are indicated by means of a box plot as in the attached image.

Consequently, it follows from observation that the median of Nadia and Ben as indicated in the attached box plot is 92 in both cases as indicated by the vertical line in both boxes.

Additionally, the highest score by Nadia is 100 while that for Ben is; 99.

Hence, Nadia had the highest score on a test.

Read more on median in box plots;

brainly.com/question/14277132

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4 0
2 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

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3 0
1 year ago
PLEASE HELP?! <br><br> Polygon LMNOP ~ Polygon QRSTU What is TU?
ycow [4]

Answer:

The answer is B.

Step-by-step explanation:

9 : 12 = 3 : 4

12 : 16 = 3 : 4

6 0
3 years ago
Read 2 more answers
Solve 3 x + k equals c for x
olga nikolaevna [1]

Answer:

Step-by-step explanation:

3 0
3 years ago
Someone pls help me !
oksian1 [2.3K]
The correct anser is 1 and only 1
3 0
3 years ago
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