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mafiozo [28]
2 years ago
8

Which statement compares the two numbers correctly?

Mathematics
2 answers:
Otrada [13]2 years ago
8 0

Answer: Its A

Step-by-step explanation: Because six hundred seventy thousandths has a greater number then 0.8 so the answer is A tell me if this is right

Alex_Xolod [135]2 years ago
7 0
Six hundred seventy-two thousandths <0.8
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point symmetry if no symmetry isn't an option as it looks grayed out and not clickable

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1 year ago
1oo divided into ratio 3:7
masha68 [24]
 <span>100 is not divisible by 3 when u divide 100 by 3 u get 33.333333...... not 3.33333..... 
the good explanation of your answer is that 0.999999....is equal to 1 
we can prove it by the following way: 
let x=0.9999.... 
then 10x = 9.999999..... 
subtract we will get 10x - x = 9.9999....- 0.99999...... 
which gives us 9x = 9 and therefore x =1 
i hope this will help u with your confusion.......  :)</span><span>
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8 0
3 years ago
Read 2 more answers
Kam
I am Lyosha [343]

Answer:Im pretty sure it's 25% percent loss

Step-by-step explanation:

sorry if im wrong

6 0
2 years ago
Suppose X has an exponential distribution with mean equal to 23. Determine the following:
e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

Mean equal to 23.

This means that m = 23, \mu = \frac{1}{23} = 0.0435

(a) P(X >10)

P(X > 10) = e^{-0.0435*10} = 0.6473

So

P(X > 10) = 0.6473

(b) P(X >20)

P(X > 20) = e^{-0.0435*20} = 0.4190

So

P(X > 20) = 0.4190

(c) P(X <30)

P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

P(X > x) = e^{-\mu x}

0.05 = e^{-0.0435x}

\ln{e^{-0.0435x}} = \ln{0.05}

-0.0435x = \ln{0.05}

x = -\frac{\ln{0.05}}{0.0435}

x = 68.87

5 0
3 years ago
Write twelve and five hundred ninety nine thousandths in standard form
stellarik [79]
12.599
Hope this helps!
7 0
3 years ago
Read 2 more answers
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