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o-na [289]
3 years ago
10

Is 730.921 less than, greater than, or equal to 730,921?

Mathematics
1 answer:
Anna35 [415]3 years ago
8 0
730.921 is equal to 730,921
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SOMEONE HELP PLEASE!!!
Oliga [24]

Answer:

x = 2 1/3

y = 14 / (2 1/3) = 14 / (7/3) = 14*3/7 = 42/7 = 6

y = 6

x = 4 1/5

y = 14 / (4 1/5) = 14 / (21/5) = 14*5/21 = 70/21 = 3 7/21 = 3 1/3

y = 3 1/3

x = 7/6

y = 14 / (7/6) = 14*6 / 7 = 12

y = 12

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
An anchor is released from a boat and begins sinking. Its
Andrew [12]

you multiply -10 and nine and you get -90 hope this helps.

Step-by-step explanation:

6 0
3 years ago
Solve for x.<br> 3x + 4x = 14<br> X =<br><br> please explain how you got the answer too
Komok [63]
X=2
You add 3x and 4x which is 7x then divide 14 by 7 which is 2
5 0
3 years ago
From a random sample of 58 businesses, it is found that the mean time the owner spends on administrative issues each week is 21.
zzz [600]

Answer: (20.86, 22.52)

Step-by-step explanation:

Formula to find the confidence interval for population mean :-

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

, where \overline{x} = sample mean.

z*= critical z-value

n= sample size.

\sigma = Population standard deviation.

By considering the given question , we have

\overline{x}= 21.69

\sigma=3.23

n= 58

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, 95% confidence interval for the amount of time spent on administrative issues will be :

21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}

=21.69\pm (1.96)\dfrac{1.7}{7.61577}

=21.69\pm (1.96)(0.223221)

\approx21.69\pm0.83

=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)

Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)

7 0
3 years ago
a total of 1500 raffle tickets were collected during the school fair. if you have 30 raffle tickets, what is the probability tha
hammer [34]

Answer: 1/50 or 0.02

Step-by-step explanation:

30 / 1500 = 1/50

1/50 or 0.02

3 0
3 years ago
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