This is what I got,

At the city museum, child admission is $5.60 and adult admission is $9.30. On Sunday, 141 tickets were sold for a total sales of

IrinaK [193]

x = child tickets

y = adult tickets

x + y = 141

5.60x + 9.30y = 974.60

y = 141 - x

5.60x + 9.30 ( 141 - x ) = 974.60

5.60x + 1311.3 - 9.30x = 974.60

-3.70x + 1311.3 = 974.60

-3.70x = 974.60 - 1311.3

-3.70x = -336.7

x = 91

91 child tickets were sold that day

5 0

2 years ago

Read 2 more answers

Write an equation for your trend line?

Soloha48 [4]

Find two points on the graph that the line crosses through almost perfectly. It looks like (1,10) and (9,1) will do.

Use them to compute the slope:
m = (1 - 10) / (9 - 1)
= -9/8

Then set up the "point-slope form":
y - y0 = m * (x - x0)

You choose some point (x0, y0) that the line crosses through. We already know the line passes through (1,10) pretty well, so let's use that.

x0 = 1
y0 = 10

Now finish plugging into the equation:
y - 10 = -9/8 * (x - 1)

The above equation will work fine for an answer, but let's go a step further and solve for y.

y - 10 = -9/8x + 9/8
y = -9/8x + 9/8 + 10
y = -9/8x + 9/8 + 80/8
y = -9/8x + (9 + 80)/8
y = -9/8x + 89/8

4 0

4 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means: