Answer:
3) girls = 24% = 456
let the total number of students be = x
the total number of students is 1900
4) increase = 20%
new salary = 24000 rupees
let his original salary be = y
his original salary is 20,000 rupees
<span>For given hyperbola:
center: (0,0)
a=7 (distance from center to vertices)
a^2=49
c=9 (distance from center to vertices)
c^2=81
c^2=a^2+b^2
b^2=c^2-a^2=81-49=32
Equation of given hyperbola:
..
2: vertices (0,+/-3) foci (0,+/-6)
hyperbola has a vertical transverse axis
Its standard form of equation: , (h,k)=(x,y) coordinates of center
For given hyperbola:
center: (0,0)
a=3 (distance from center to vertices)
a^2=9
c=6 (distance from center to vertices)
c^2=36 a^2+b^2
b^2=c^2-a^2=36-9=25
Equation of given hyperbola:
</span>
Answer:
(a)23 (b)90 (c)3
Step-by-step explanation:
The equation for the line of best fit for this situation is given as
where x=average temperature in degrees
y=average number of hot dogs she sold,
(a) The expected number of hot dogs sold when the temperature is 50° would be___hot dogs.
When x=50°
When the temperature is 50°, the expected number of hot dogs sold would be 23.
(b)If the vendor sold 35 hot dogs, the temperature is expected to be ___degrees.
If y=35
Multiply both sides by 10/3
If the vendor sold 35 hot dogs, the temperature is expected to be 90 degrees.
(c) Based on the line of best fit, for every 10-degree increase in temperature, she should sell 3 more hot dogs.
Answer:
the answer is -8 or 7
Step-by-step explanation:
Sorry im not good w negatives...
<em>Answer: h = 120 ft; w = 80 ft </em>
<em></em>
<em>A = 9600 ft^2</em>
<em />
<em>Step-by-step explanation: Let h and w be the dimensions of the playground. The area is given by:</em>
<em></em>
<em>A = h*w (eq1)</em>
<em></em>
<em>The total amount of fence used is:</em>
<em></em>
<em>L = 2*h + 2*w + w (eq2) (an extra distance w beacuse of the division)</em>
<em></em>
<em>Solving for w:</em>
<em></em>
<em>w = L - 2/3*h = 480 - 2/3*h (eq3) Replacing this into the area eq:</em>
<em></em>
<em></em>
<em></em>
<em>We derive this and equal zero to find its maximum:</em>
<em></em>
<em> Solving for h:</em>
<em></em>
<em>h = 120 ft. Replacing this into eq3:</em>
<em></em>
<em>w = 80ft</em>
<em></em>
<em>Therefore the maximum area is:</em>
<em></em>
<em>A = 9600 ft^2</em>
<em />
