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Rufina [12.5K]
3 years ago
7

Round to the nearest Ten-thousand: 849,708​

Mathematics
2 answers:
Andre45 [30]3 years ago
8 0

Answer: 850,000

Concept:

Here, we need to know the order and name of each place value.

Please refer to the attachment below for the specified names.

Solve:

8 = Hundred thousands

4 = Ten thousands

9 = One thousands

7 = Hundreds

0 = Tens

9 = Ones

Since the values before the ten thousands place, which would be the one thousands place, is greater than 5, then we should round up.

Therefore, the rounded value would be \boxed{850,000}

Hope this helps!! :)

Please let me know if you have any questions

jarptica [38.1K]3 years ago
5 0

Answer:

850,000

Step-by-step explanation:

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Let x be the amount of weeks.

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3 years ago
A n =−1+3(n−1) an=−1+3(n−1) what is the 55 term in the sequence
Sindrei [870]
ANSWER

a_{55}=161


EXPLANATION

The general term for the sequence is


a_n=-1+3(n-1)


To find the 55th term, we have to substitute
n = 55
in to the general term and simplify.



This implies that,

a_{55}=-1+3(55-1)



a_{55}=-1+3(54)




a_{55}=-1+162



a_{55}=161


Therefore the 55th term is 161.
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2 years ago
Difference between the smallest term in the data set and the largest term in the data set
tamaranim1 [39]

Range definition:

The difference between the smallest term and the largest term in the data set is called the range.

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6 0
3 years ago
Please Help Best answer gets brainliest first answer gets five stars
Brrunno [24]

Hi,

(1/4)^−2−(5^0)(2)(1^−1)

=16−(5^0)(2)(1^−1)

=16−(1)(2)(1^−1)

=16−2(1^−1)

=16−(2)(1)

=16−2

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Have a great day!

4 0
3 years ago
Read 2 more answers
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

3 0
3 years ago
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