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3 years ago

Round 2,461,612,242 to the nearest billion

Mathematics

2 answers:

beks73 [17]3 years ago

7 0

Answer:

2 billion

Step-by-step explanation:

if the # is 4 or brow you round down

Gre4nikov [31]3 years ago

6 0

Answer:

2,000,000,000.

Step-by-step explanation:

We know that when rounding, we take a look at the digit next to the digit we are rounding. So if I were to ask for you to round 122 to the nearest hundred, then the answer would be 100.

Remember that we round down if the digit next to it is 0-4. If the digit is 5-9, then we round up.

In this case, the digit next to the 2 is a 4. We know that we round down with a 4, so we go to the nearest whole billion, in this case 2 billion.

I hope this helps!

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According to a medical institution, about 44% of people from a certain country have type A blood. Suppose 80 people show up at a

vesna_86 [32]

Do 80 divided by 44%

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3 years ago

the expression 4(1.5g+ 3.5) is equal to 6g plus a constant term what is the value of the constant term

riadik2000 [5.3K]

4*3.5 =14.0
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3 0

3 years ago

Consider the sequence {an}={3n+13n−3n3n+1}. Graph this sequence and use your graph to help you answer the following questions.

Fantom [35]

Part 1: You can simplify $a_n$ to

$\dfrac{3n+1}{3n}-\dfrac{3n}{3n+1} = \dfrac1{3n}+\dfrac1{3n+1}$

Presumably, the sequence starts at n = 1. It's easy to see that the sequence is strictly decreasing, since larger values of n make either fraction smaller.

(a) So, the sequence is bounded above by its first value,

$|a_n| \le a_1 = \dfrac13+\dfrac14 = \boxed{\dfrac7{12}}$

(b) And because both fractions in $a_n$ converge to 0, while remaining positive for any natural number n, the sequence is bounded below by 0,

$|a_n| \ge \boxed{0}$

Part 2: Yes, $a_n$ is monotonic and strictly decreasing.

Part 3:

(a) I assume the choices are between convergent and divergent. Any monotonic and bounded sequence is convergent.

(b) Since $a_n$ is decreasing and bounded below by 0, its limit as n goes to infinity is 0.

Part 4:

(a) We have

$\displaystyle \lim_{n\to\infty} \frac{10n^2+1}{n^2+n} = \lim_{n\to\infty}10+\frac1{n^2}}{1+\frac1n} = 10$

and the (-1)ⁿ makes this limit alternate between -10 and 10. So the sequence is bounded but clearly not monotonic, and hence divergent.

(b) Taking the limit gives

$\displaystyle\lim_{n\to\infty}\frac{10n^3+1}{n^2+n} = \lim_{n\to\infty}\frac{10+\frac1{n^3}}{\frac1n+\frac1{n^2}} = \infty$

so the sequence is unbounded and divergent. It should also be easy to see or establish that the sequence is strictly increasing and thus monotonic.

For the next three, I'm guessing the options here are something to the effect of "does", "may", or "does not".

(d) does not : a monotonic sequence has to be bounded in order to converge, otherwise it grows to ± infinity.

(e) does : this is true and is known as the monotone convergence theorem.

5 0

3 years ago

Please help me please

hram777 [196]

Answer:

detailed explanation is in image

Step-by-step explanation:

ans is 60 unit²

4 0

2 years ago

Read 2 more answers

A thermometer is taken from a room where the temperature is 24oc to the outdoors, where the temperature is −15oc. After one minu

kap26 [50]

Solution:

Use Newton's Law of Cooling.

T = T_s + (T_0 - T_s)*e^(-kt)

where

T = temperature at any instant

T_s = temperature of surroundings

T_0 = original temperature

t = elapsed time

k = constant

Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.

T = 14, T_0 = 24, T_s = -15, t = 1, k = ?

T = T_s + (T_0 - T_s)*e^(-kt)

==> 14 = -15 + (24 - 7)*e^(-k)

==> 14 = 7 + 17*e^(-k)

==> 7 = 17*e^(-k)

==> 7/13 = e^(-k)

==> -k = ln(7/17)

==> k = -ln(7/17) ≈ 0.774

Now,

Let's calculate temperatures!

T = ?, T_0 = 24, T_s = -15, k = 0.773, t = 3

T = T_s + (T_0 - T_s)*e^(-kt)

==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]

==> T = -15 + 39*e^(-1.548)

==> T ≈ 15.72° C

This the required answer.

7 0

3 years ago