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harina [27]
3 years ago
8

https://www.google.com/search?safe=active&q=Elvira+made+the+trip+on+Tuesday+and+David+made+the+same+trip+on+Wednesday+Elvira

+traveled+at+14+mph.+David+traveled+at+21+mph,+so+his+time+was+3+hours+less+than+Elvira%27s+time.+How+far+did+they+travel?
Mathematics
1 answer:
Neporo4naja [7]3 years ago
8 0

Answer:Let x and (x-3) represent Elvira's and David's times respectively


D = r*t  


Distances traveled are equal in this example***







14mph*x = 21mph(x-3)


solving for x


14x = 21x - 63


 63 = 7x


x = 9hr, the time that Elvira traveled.  David traveled 6hrs   (9hr-3hr = 6hr)




CHECKING our Answer***


14mph*9hr = 126mi = 21mph*6hr = 126mi



Step-by-step explanation:


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Step-by-step explanation:looked it up... have a good day!

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What is the area to this shape ?
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249.42 units²

Step-by-step explanation:

Given only the apothem of an n-sided regular polygon, the area can be computed as ...

... A = n·a²·tan(180°/n)

For n=3 and a=4√3, this is ...

... A = 3·(4√3)²·tan(60°)

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3 years ago
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Read 2 more answers
The value of y varies directly with x. If x =3, then y=21. What is the value of x when y=105?
Goshia [24]

Answer:

x = 15

Step-by-step explanation:

given that y varies directly with x

mathematically we can express this as

y = k x , where k is a constant

step 1: find k

given x = 3 and y = 21

y = k x

21 = k (3)

k = 7

hence the equation becomes

y = 7 x

when y = 105,

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x = 105 / 7 = 15

3 0
3 years ago
The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 . A dotplot s
Ilya [14]

Answer:

The value of the test statistic and degrees of freedom is 2.148 and 11 respectively.

Step-by-step explanation:

Consider the provided information.

The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 .

Thus, n = 12, \bar x=36800 σ = 5000

degrees of freedom = n-1 = 12-1 = 11

H_0: \mu = 33700\ and\ H_a: \mu \neq 33700

Formula to find the value of z is: z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}

Where \bar x is mean of sample, μ is the mean of population, σ is the standard deviation of population and n is number of observations.

z=\frac{36800-33700}{\frac{5000}{\sqrt{12}}}

z=2.148

Hence, the value of the test statistic and degrees of freedom is 2.148 and 11 respectively.

6 0
3 years ago
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