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Paha777 [63]
3 years ago
8

The school that Stefan goes to is selling tickets to a choral performance. On the first day of ticket sales the school sold 3 se

nior tickets and 2 child tickets. Find the price of a senior citizen ticket and the price of a child ticket.
Mathematics
1 answer:
Dominik [7]3 years ago
6 0

Answer: the price of a senior citizen's ticket is $8.

the price of a child's ticket is $14

Step-by-step explanation:

Let x represent the price of a senior citizen's ticket.

Let y represent the price of a child's ticket.

On the first day of ticket sales, the school sold 3 senior citizen tickets and 1 child ticket for a total of $38. It means that

3x + y = 38- - - - - - - - - - - -1

The school took in $52 on the second day by selling 3 senior citizen and 2 child tickets. It means that

3x + 2y = 52- - - - - - - - - - - -2

Subtracting equation 2 from equation 1, it becomes

- y = - 14

y = 14

Substituting y = 14 into equation 1, it becomes

3x + 14 = 38

3x = 38 - 14 = 24

x = 24/3

x = 8

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Which of the following statements must be true about this diagram?
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Amelia competed in a race in which she ran at a constant speed. A graph that represents the relationship between the number of s
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Answer:

B (3, 13.5)

Step-by-step explanation:

Using point (0,0) and (15, 67.5) we can find the slope(gradient).

(y - y¹) / (x - x¹) = (67.5 - 0) / (15 - 0)

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slope = 4.5

using the point given in option A (0, 4.5) with point (15, 67.5) to calculate the slope it gives 4.2 which is not equal to what we calculated.

using the option B (3, 13.5) with (15, 67.5) gives a slope of 4.5 which is equal to the slope of the line.

3 0
3 years ago
3. A rare species of aquatic insect was discovered in the Amazon rainforest. To protect the species, environmentalists declared
navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

8 0
2 years ago
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