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EastWind [94]
3 years ago
7

What is the mean of 90, 55, 75, 60, 80, and 90?

Mathematics
2 answers:
ElenaW [278]3 years ago
8 0
90+55+75+60+80+90=450
450/6=75
Nataly [62]3 years ago
7 0
(90+55+75+60+80+90)/7=64.2857143
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if Sarah repaired her computer her bill was 1444.50 dollars and she included a 15 percent gratuity whats her new total cost
tiny-mole [99]
The new cost would be 1661.175 if u want it rounded it would be 1661.18 because you would need to do 1444.5 x .15 and then add that to 1444.5
7 0
3 years ago
If 4x^2-100=0 the roots of the equation are
Marrrta [24]

Your equation:                   4x² - 100           =  0

Factor the left side:      (2x + 10) · (2x - 10) = 0

This equation is true if either factor is zero.

-- One factor:                     2x + 10 = 0
  
Divide each side by 2:            x + 5 = 0
Subtract 5 from each side:      x      = -5 .


-- The other factor:               2x - 10  =  0

Divide each side by 2:            x + 5 = 0
Add  5  to each side:              x      =  5 .

8 0
3 years ago
Juanita wants to buy 12.5 pounds of stones to decorate her garden. The stone normally sells for $1.80 a pound. The salesman offe
anastassius [24]

Step-by-step explanation:

$1.80 per pound

The amount she's supposed to buy it is

1.80 × 12.5 = $22.5

She paid $21.25 for the 12.5 pounds, that means she saved

$22.5 - $21.25 = $1.25

the amount she saved per pound is

1.25/12.5 = $0.1 per pound

5 0
3 years ago
use the information provided to write the standard form equation of each hyperbola vertices:(4,14), (4,-10) foci: (4,15), (4,-11
Evgen [1.6K]
So... if you notice the picture below, based on the given vertices, is a hyperbola with a vertical traverse axis

meaning for the equation, the fraction with the "y" variable is the positive fraction, thus \bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1

so... what' the point h,k for the center?  well, just take a peek at the graph, the center is half-way between both vertices, that's h,k

what's the "a" component or the traverse axis... well, is the distance from one vertex to the center, notice the picture, notice how many units from either vertex to the center, that's "a"

now.. what's "b"?  well, "b" comes from the conjugate axis, or the other runnning over the x-axis   hmm the smaller one in this case.... well, we dunno what "b" is

however, we know the distance from either focus, to the center of the hyperbola, and that distance "c", is  \bf c=\sqrt{a^2+b^2}

notice the picture, notice the distance from either focus to the center, that's the distance "c", thus  \bf c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b

that'd be "b"

bearing in mind that   \bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center= ({{ h}},{{ k}})\\\\ b=\sqrt{c^2-a^2}
\end{cases}

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=y%20%2B%205%20%3D%20x%20%7B%3F%7D%5E%7B2%7D%20" id="TexFormula1" title="y + 5 = x {?}^{2} " al
zimovet [89]
What is your question?
3 0
3 years ago
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