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seropon [69]
4 years ago
8

Use the information about budgets on page 352 to explain how inequalities are used in finances. Include an explanation of the re

strictions placed on the domain and range of the inequality that describes the number of times Hannah can buy lunch. Describe three possible solutions of the inequality. ( Info ) Hannah budgets $30 a month for lunch. On most days, she brings her lunch. She can also buy lunch at the cafeteria or at a fast-food restaurant. She spends an average of $3 for lunch at the cafeteria and an average of $4 for lunch at a restaurant. How many times a month can Hannah buy her lunch and remain within her budget? The cost of eating in the cafeteria plus the cost of eating in a restaurant is less than or equal to$30. Let x = the number of days she buys lunch at the cafeteria.Let y = the number of days she buys lunch at a restaurant 3x +4y ≤30 There are many solutions for this inequality. Each solution represents a different combination of lunches bought in the cafeteria and in a restaurant.
Mathematics
1 answer:
mrs_skeptik [129]4 years ago
7 0
*it by 15 and you will get the answer
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En una fiesta hay 40 asistentes de los cuales el número de varones que no bailan es el doble que el número de las mujeres que no
Arte-miy333 [17]

Answer:

50%

Step-by-step explanation:

Para resolver este problema vamos a definir 4 variables:

V= número de varones bailando.

M= número de mujeres bailando.

V' = número de varones que no están bailando.

M' = número de mujeres que no están bailando.

Sabemos que hay 40 personas en la fiesta, por lo que podemos construir la siguiente ecuación:

V+M+V'+M'=40

Vamos a suponer que cada mujer que está bailando, está bailando varón respectivamente. (El problema no da más información, entonces podemos suponer esto.)

Entonces si hay 5 mujeres bailando, entonces también hay 5 hombres bailando, por lo que nuestra ecuación se reescribe de la siguiente manera:

5+5+V'+M'=40

y simplificamos:

10+V'+M'=40

V'+M'=40-10

V'+M'=30

Ahora bien, el problem nos dice que el número de varones que no bailan es el doble del número de mujeres que no bailan en un determinado momento. Entonces con esta información podemos construir la siguiente ecuación:

V'=2M'

Y podemos despejar el número de mujeres que no bailan, lo que nos da:

M'=\frac{V'}{2}

Entonces podemos sustituir esto dentro de nuestra ecuación para obtener:

\frac{V'}{2}+V'=30

y podemos entonces despejar V'

\frac{3V'}{2}=30

V'=\frac{2(30)}{3}

V'=20

Entonces hay 20 varones que no están bailando, por lo que la probabilidad de que el varón que se escoge al azar no esté bailando está dada por la siguiente fórmula:

P=\frac{V'}{total}

P=\frac{20}{40}=\frac{1}{2}

P=50%

6 0
3 years ago
Given: f(x) = 3x+4 and g(x) = -x^2+4x-5<br> find: -5f(x) - 4g(x)
sattari [20]

Answer:

642

Step-by-step explanation:

bc it is trust me

3 0
3 years ago
Determine if y = 5 is a solution for 2y = 15.
Greeley [361]

Answer:

y=5 does not equal 2y=15

Step-by-step explanation:

If you plug y into the equation we get 2(5)=15

The next step is to multiply 2*5, which equals 10

10≠15

4 0
4 years ago
Read 2 more answers
Three friends said their heights were 5 and start fraction 13 over 30 end fraction feet, 5 and one-sixth feet, and 5 and start f
KengaRu [80]

Answer:

5\frac{1}{6} < 5\frac{13}{30} < 5\frac{68}{90}

Step-by-step explanation:

Represent the friends with A, B and C

A = 5\frac{13}{30}

B = 5\frac{1}{6}

C = 5\frac{68}{90}

Required

Order from least to highest

First, we convert the given parameters to decimal

A = 5\frac{13}{30}

A = 5 + \frac{13}{30}

A = 5 + 0.43

A = 5.43

B = 5\frac{1}{6}

B = 5 + \frac{1}{6}

B = 5 + 0.17

B = 5.17

C = 5\frac{68}{90}

C = 5 + \frac{68}{90}

C = 5 + 0.76

C = 5.76

So, we have:

A = 5.43      B = 5.17      C = 5.76

Order from least to greatest

B = 5.17       A = 5.43          C = 5.76

Replace them with the original values

B = 5\frac{1}{6}       A = 5\frac{13}{30}      C = 5\frac{68}{90}

Hence, the correct order is:

5\frac{1}{6} < 5\frac{13}{30} < 5\frac{68}{90}

5 0
3 years ago
Read 2 more answers
Integral of (x^2)/(e^x^3)
Lesechka [4]

I=\displaystyle\int\frac{x^2}{e^{x^3}}\,\mathrm dx

Substitute y=x^3 and \mathrm dy=3x^2\,\mathrm dx. Then

I=\displaystyle\frac13\int\frac{\mathrm dy}{e^y}=-\frac1{3e^y}+C=-\frac1{3e^{x^3}}+C

5 0
3 years ago
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