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mojhsa [17]
3 years ago
12

Trevon’s school is selling tickets to a fall musical. On the first day of ticket sales the school sold 13 adult tickets and 12 s

tudent tickets for a total of $211. The school took in $65 on the second day by selling 5 adult tickets and 3 student tickets. What is the price of one adult ticket and one student ticket?
Write two separate equations and what is the price of one adult ticket and one student ticket.
Mathematics
1 answer:
julsineya [31]3 years ago
7 0

9514 1404 393

Answer:

  • adult: $7
  • student: $10

Step-by-step explanation:

Let a and s represent the prices of adult and student tickets, respectively.

  13a +12s = 211 . . . . . . ticket sales the first day

  5a +3s = 65 . . . . . . . ticket sales the second day

Subtracting the first equation from 4 times the second gives ...

  4(5a +3s) -(13a +12s) = 4(65) -(211)

  7a = 49 . . . . . . . simplify

  a = 7 . . . . . . . divide by 7

  5(7) +3s = 65 . . . . substitute into the second equation

  3s = 30 . . . . . . . subtract 35

  s = 10 . . . . . . . divide by 3

The price of one adult ticket is $7; the price of one student ticket is $10.

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mariarad [96]

Answer:

Hay 200 botellas de 5 litros y 1000 botellas de 2 litros.

Step-by-step explanation:

Un sistema de ecuaciones lineales es un conjunto de dos o más ecuaciones de primer grado, en el cual se relacionan dos o más incógnitas.

Resolver un sistema de ecuaciones consiste en encontrar el valor de cada incógnita para que se cumplan todas las ecuaciones del sistema.

En este caso, las variables a calcular son:

  • x= cantidad de botellas de 2 litros.
  • y= cantidad de botellas de 5 litros.

Una empresa aceitera ha envasado 3000 litros de aceite en 1200 botellas de dos y de cinco litros. Entonces es posible plantear el siguiente sistema de ecuaciones:

\left \{ {{2*x+5*y=3000} \atop {x+y=1200}} \right.

Existen varios métodos para resolver un sistema de ecuaciones. Resolviendo por el método de sustitución, que consiste en despejar o aislar una de las incógnitas y sustituir su expresión en la otra ecuación, despejas x de la segunda ecuación:

x= 1200 - y

Sustituyendo la expresión en la primer ecuación:

2*(1200 - y) + 5*y=3000

Resolviendo se obtiene:

2*1200 - 2*y + 5*y= 3000

2400 +3*y= 3000

3*y= 3000 - 2400

3*y= 600

y= 600÷3

y= 200

Reemplazando en la expresión x= 1200 - y:

x= 1200 - y

x=1200 -200

x= 1000

<u><em>Hay 200 botellas de 5 litros y 1000 botellas de 2 litros.</em></u>

7 0
2 years ago
One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla
Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0
3 years ago
Need help figuring this one out
Dennis_Churaev [7]

Answer:

4.426020408 × 10∧19

Step-by-step explanation:

First, I found what 540 multiplied by 0.0000000028 was. Next, I took 66921428571428.6 and divided it by the product I got on top.

7 0
3 years ago
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Andreyy89
I doubt the options you've mentioned here.
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2 years ago
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marysya [2.9K]
<u><em>Hey There!</em></u>
<u><em>Perimeter=L+W×2</em></u>
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8 0
3 years ago
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