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Airida [17]
3 years ago
6

Mary runs 4 laps in 8 minutes.Nicole runs 12 laps in 18 minutes.Who runs the fastest?

Mathematics
2 answers:
frozen [14]3 years ago
5 0
\frac{4}{8} =  \frac{x}{1}

4/8 or 1/2 laps per minute 

\frac{12}{18} =  \frac{x}{1}

12/18 or 2/3 laps per minute 

Nicole ran the fastest 
Ugo [173]3 years ago
3 0
4/8 is mary and it is 0.5
12/18 is nicole and it is .6666
mary runs faster
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Find the value of two numbers if their sum is 112 and their difference is 66.
snow_lady [41]

Answer:

x+y = 112 ........ (1)

x-y = 62............ (2)

from (1) and (2)

2x = 174

x = 174/2

x = 87

sub x = 87 in (1)

x+y = 112

87+y = 112

y = 112 - 87

y = 25

so the numbers are 87 and 25

Step-by-step explanation:

8 0
2 years ago
How do I solve this I’m dumb
laila [671]
-1/4 + f/8 = 1/2.
The LCD is 8.  Rewrite this equation as    -8/4 + 8f/8 = 8(1/2) and then reduce the result:

-2 + f = =4

Then f = 6.

You should check this result.

8 0
3 years ago
What is the mean median and mode of 99, 69, 96, 69, 78. Round to the nearest tenth if necessary
KiRa [710]
A median is the middle number of a data set.
To find it put the numbers in order.
78, 69, 69, 96, 99
Then find the middle number
78, 69, *69,* 96, 99

The median is 69

A mode is the number that appears the most in a set of numbers.
78, *69, 69,* 96, 99

The mode is 69

To find the mean or average of a set of numbers, add all the numbers together then divide by the number of numbers there are.
99+69+96+69+78 = 411
411/5 = 82.2

The mean is 82.2

Hope this helped! If you have anymore questions or don't understand, please comment on my profile or DM me. :)
6 0
3 years ago
Use the given values of n and p to find the minimum usual value μ−2σ and the maximum usual value μ+2σ. Round to the nearest hund
lbvjy [14]

Answer:

μ−2σ = 1,089.26

μ+2σ = 1,097.62

Step-by-step explanation:

The standard deviation of a sample of size 'n' and proportion 'p' is:

\sigma=\sqrt{\frac{p*(1-p)}{n} }

If n=1139 and p =0.96, the standard deviation is:

\sigma=\sqrt{\frac{p*(1-p)}{n}}\\\sigma = 0.001836

The minimum and maximum usual values are:

\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n

\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62

5 0
3 years ago
Please help question in picture!
eduard

Answer:

b

Step-by-step explanation:

6 0
3 years ago
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