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never [62]
3 years ago
14

675,360.41 in expanded notation

Mathematics
2 answers:
denis-greek [22]3 years ago
8 0
{600,000,00} {70,000,00} {300,00} {6000} {40} {1}
Alisiya [41]3 years ago
7 0
I got Osteoporosis
I got Osteoporosis
I got Osteoporosis 
I got Osteoporosis

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What is the best estimate for the product of 289 and 7
Katarina [22]
I would estimate it as (300 x 7) = 2100 

Then I would subtract (11 x 7) + 77 from that to get the exact answer of 2023 

~~~~~~~~~~ 
I didn't use a calculator or paper. The 'trick' is to notice that 289 = (300 - 11)
7 0
3 years ago
Bart ran 5000 from the cop and an average speed of 6 meters/ seconds before he got caught. how long did he run?
Masja [62]

Answer:

13.8 minutes

Step-by-step explanation:

5000/6= 833/60

3 0
3 years ago
What are the names of three planes that contain point G?
liubo4ka [24]

Answer:

B is the correct answer to your question

4 0
2 years ago
Question #18: Three times as many boys as girls visited the mall in Deptford. A
romanna [79]

(boys)  x=3(y)

(girls)   y=x/3

Step-by-step explanation:

4 0
3 years ago
Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars
GuDViN [60]

Answer:

Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

Step-by-step explanation:

We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.

<em>Let X = incomes for the industry</em>

So, X ~ N(\mu=95,\sigma^{2}=5^{2})

Now, the z score probability distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean income of firms in the industry = 95 million dollars

            \sigma = standard deviation = 5 million dollars

So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)

    P(X < 100) = P( \frac{X-\mu}{\sigma} < \frac{100-95}{5} ) = P(Z < 1) = 0.8413   {using z table]

                                                     

Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

5 0
3 years ago
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