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3 years ago

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Adult tickets to a basketball game cost $5. Student tickets cost $1. A total of $2,699 was collected on the sale of 1,179 ticket

s. How many of each type of ticket were sold?

1 answer:

lana66690 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

The Answer 3182121

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Which property of addition is shown? <br><br> -3 + 5 = 5 + (-3)

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a successful music app tracked the number of downloads each day for 4 artists, represented by lines l, j, m and d over the cours

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3 years ago

Fred placed 3 strands of bacteria in a petri dish. The bacteria strands double every hour. Write a formula to represent how many

miv72 [106K]

Answer:

C. $f(n) = 3(2)^n$ ; 48 strands of bacteria

Step-by-step explanation:

Fred placed 3 strands of bacteria in a petri dish.

The bacteria strands double every hour.

Formula : y = ab^x

Where a is the initial value

b = growth rate

x = hours

So, A formula to represent how many bacteria strands are present at the nth hour : $f(n)=3(2)^n$

Now we are supposed to find how many strands of bacteria are present at 4 hours.

Substitute the value in the formula :

$f(4)=3(2)^{4}$

f(4)=48

Hence 48 strands of bacteria are present at 4 hours.

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3 years ago

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

6 0

3 years ago