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OlgaM077 [116]
3 years ago
10

What is 74000 times 5000

Mathematics
2 answers:
WINSTONCH [101]3 years ago
7 0

Answer:

370000000

Step-by-step explanation:

The easiest way is probably to multiply 74,000 by 10,000 and divide the product by 2.

svetoff [14.1K]3 years ago
6 0
It is 370,000,000
370000000
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We first find our z-score.  We want a 95% confidence interval:

0.95/2 = 0.475

Looking this up in the z-table, (http://www.statisticshowto.com/tables/z-table/) we see the z-score is 1.96.

The formula we will use is:

p \pm z\sqrt{\frac{p(1-p)}{n}}

In this problem, p = 20/100 = 0.2, and n=100:

0.2 \pm 1.96\sqrt{\frac{(0.2)(0.8)}{100}}
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4 years ago
Choose the definition for the function
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Answer:

In mathematics, a function is a binary relation over two sets that associates to every element of the first set exactly one element of the second set. Typical examples are functions from integers to integers or from the real numbers to real numbers.

Step-by-step explanation:

hope this helps you :)

7 0
3 years ago
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grigory [225]
There is no solution to this
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4 years ago
A line passes through the points (8, –1) and (–4, 2).
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The velocity of a turtle is recorded at 1 minute intervals (in meters per second). Use the right-endpoint approximation to estim
aliina [53]

Answer:

The total distance that the turtle traveled during the 5 seconds recorded is Distance \:traveled\approx 3.838\:\frac{m}{s}

Step-by-step explanation:

To estimate distance traveled of an object moving in a straight line over a period of time, from discrete data on the velocity of the object, we use a Riemann Sum. If we have a table of values

\left\begin{array}{ccccccc}time\:=\:t_i&t_0=0&t_1&t_2&...&t_n\\velocity\:=\:v(t_i)&v(t_0)&v(t_1)&v(t_2)&...&v(t_n)\end{array}\right

where \Delta t=t_i-t_{i-1}, then we can approximate the displacement on the interval [t_{i-1},t_i] by v(t_{i}) \times\Delta t.

Therefore the distance traveled of the object over the time interval [0,t_n] can be approximated by

Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+...+|v(t_n)|\Delta t

This is the right endpoint approximation.

We are given a table of values for <em>v(t)</em>

\left\begin{array}{cccccccc}t(sec)&0&1&2&3&4&5\\v(t)&0.078&0.83&0.75&0.98&0.853&0.425\end{array}\right

Applying the right endpoint approximation formula we get,

\Delta t = 1\sec

Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+|v(t_3)|\Delta t+|v(t_4)|\Delta t+|v(t_5)|\Delta t\\\\Distance \:traveled\approx 0.83(1)+0.75(1)+0.98(1)+0.853(1)+0.425(1)\\\\Distance \:traveled\approx 3.838\:\frac{m}{s}

The total distance that the turtle traveled during the 5 seconds recorded is Distance \:traveled\approx 3.838\:\frac{m}{s}

4 0
3 years ago
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