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Afina-wow [57]
4 years ago
5

HELP!!!!! Please!!!!!

Mathematics
1 answer:
Mekhanik [1.2K]4 years ago
5 0
What are you confused by?
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On the first season of "The Biggest Loser",
amid [387]

Answer:

33%

Step-by-step explanation:

311-208=103 ponds lost

103/311=33%

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The digit to the right of the hundreds place is? <br><br> The number is 3.2906
Tcecarenko [31]

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3.2906, The 9

<em>hope this helps :)</em>

8 0
3 years ago
Hi! I really suck at math so please help :)
MArishka [77]

Answer:

(-2,-4), (-1, -1), (0,2), (1,5)(2, 8)

Step-by-step explanation:

you plug x into the equation to get y

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3 years ago
Read 2 more answers
Find the other endpoint of the line segment with the given endpoint and midpoint: endpoint(1.7, -4.6) midpoint (-7.5, 0.3).. (ne
Sergio [31]

Answer:

(- 16.7, 5.2 )

Step-by-step explanation:

let the coordinates of the endpoint be (x , y )

using the midpoint formula

consider the x- coordinate

\frac{1}{2}(x + 1.7) = - 7.5 ( multiply both sides by 2 )

x + 1. 7 = - 7.5 ( subtract 1.7 from both sides )

x = - 16.7

consider the y-coordinate

\frac{1}{2}(y - 4.6 ) = 0.3 ( multiply both sides by 2 )

y - 4.6 = 0.6 ( add 4.6 to both sides )

y = 5.2

endpoint = (- 16.7, 5.2 )


3 0
4 years ago
A veterinarian collects data on the number of times race horses are raced during their careers. The veterinarian finds that the
SpyIntel [72]

Answer:

c. (12.12, 18.48)

Step-by-step explanation:

Hello!

The study variable is X: number of times a racehorse is raced during its career.

The average number is X[bar]= 15.3 and the standard deviation is S= 6.8 obtained from a sample of n=20 horses.

To estimate the population mean you need that the variable has a normal distribution, in this case, we have no information about its distribution so I'll assume that it has a normal distribution. With n=20 the most accurate statistic to use for the estimation is a Students-t for one sample, the formula for the interval is:

X[bar] ± t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{20} }

t_{n-1;1-\alpha /2} = t_{19;0.975}= 2.093

[15.3 ± 2.093 * \frac{6.8}{\sqrt{20} }]

[12.12; 18.48]

Using a significance level of 95% you'd expect that the true average of times racehorses are raced during their career is included in the interval [12.12; 18.48].

I hope it helps!

8 0
4 years ago
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