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2 years ago

The value of an algebraic expression depends on the value of each variable in the expression.

1 answer:

3 0

The value of an algebraic expression depends on the value of each variable in the expression in other words, the value of an algebraic expression varies for different values of the variables.

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Considere a função de oferta como sendo y= 5x - 1/2 e a função de demanda como sendo y= -7x + 47/2 , onde x é a quantidade e y é

swat32

A) Temos que igualar isso a 2,50, que é o valor de y, ou seja, o valor do preço. Façamos isso, então, na função de demanda.

$y = -7x + \frac{47}{2}$

Substituir.

$2,50 = - 7x + \frac{47}{2}$

Passar o que é x de um lado e o que é número do outro lado.

$-7x = - \frac{47}{2} + 2,50$

A fração 47/2 corresponde a 23,5. É mais interessante para nós, nesse caso, utilizar o valor decimal.

$-7x = - 23,5 + 2,50$

Somar logo.

$-7x = -21$

Podemos multiplicar por -1 para tirar os sinais negativos.

$7x = 21$

Passar o x dividindo

$x = \frac{21}{7}$

Dividir

$\boxed{x = 3}$

Para um preço de R$ 2,50, a quantidade demandada é de 3 unidades.

B) Basta igualar as duas formulinhas.

$5x - \frac{1}{2} = -7x + \frac{47}{2}$

Passar o que é x de um lado e o que é número do outro lado.

$5x + 7x = \frac{47}{2} + \frac{1}{2}$

Somar tudo. Como as frações têm o mesmo denominador, nada de mmc! Basta somar os numeradores.

$12x = \frac{48}{2}$

Podemos dividir lá, olha:

$12x = 24$

Passar o x dividindo

$x = \frac{24}{12}$

Dividir

$\boxed{x = 2}$

O ponto de equilíbrio é o valor 2.

C) Basta, então, igualar as fórmulas da oferta e da demanda a 10.

Fórmula da oferta

$10 = 5x - \frac{1}{2}$

Passar o que é x de um lado e o que é número do outro lado.

$5x = 10 - \frac{1}{2}$

Se 1/2 = 0,5, dá pra resolver sem fazer mmc:

$5x = 10 - 0,5$

Subtrair

$5x = 9,5$

Passar dividindo

$x = \frac{9,5}{5}$

Dividir.

$\boxed{x = 1,9}$

Fórmula da demanda

$10 = -7x + \frac{47}{2}$

Passar o que é x de um lado e o que é número do outro lado.

$-7x = \frac{47}{2} - 10$

Divide 47/2, vai.

$-7x = 23,5 - 10$

Subtrair.

$-7x = 13,5$

Passar o 7 dividindo.

$x = - \frac{13,5}{7}$

Vamos dividir e transformar em decimal.

$\boxed{x = -1,9285714285714285714285714285714}$

Subtrair a oferta pela demanda.

$x = 1,9 - (-1,9285714285714285714285714285714)$

Subtrair.

$\boxed{x = 3,8285714285714285714285714285714}$

Arredondando, o preço será de R$ 3,83.

8 0

3 years ago

15 out of 25 perfer skittles to M&MS. what percent of students prefer skittles

likoan [24]

Answer:

60%

Step-by-step explanation:

Just divide 15 by 25 and then move the decimal point 2 places to the right. Then turn that number into a percentage.

6 0

3 years ago

Read 2 more answers

16. A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours

Reika [66]

Answer:

a) 23.11% probability of making exactly four sales.

b) 1.38% probability of making no sales.

c) 16.78% probability of making exactly two sales.

d) The mean number of sales in the two-hour period is 3.6.

Step-by-step explanation:

For each phone call, there are only two possible outcomes. Either a sale is made, or it is not. The probability of a sale being made in a call is independent from other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours, find:

Six calls per hour, 2 hours. So

$n = 2*6 = 12$

Sale on 30% of these calls, so $p = 0.3$

a. The probability of making exactly four sales.

This is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{12,4}.(0.3)^{4}.(0.7)^{8} = 0.2311$

23.11% probability of making exactly four sales.

b. The probability of making no sales.

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.3)^{0}.(0.7)^{12} = 0.0138$

1.38% probability of making no sales.

c. The probability of making exactly two sales.

This is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.3)^{2}.(0.7)^{10} = 0.1678$

16.78% probability of making exactly two sales.

d. The mean number of sales in the two-hour period.

The mean of the binomia distribution is

$E(X) = np$

$E(X) = 12*0.3 = 3.6$

The mean number of sales in the two-hour period is 3.6.

4 0

3 years ago

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp

Katena32 [7]

Let's move like a crab, backwards some.

after 2 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2$

after 3 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3$

is that enough to pay the $1100?

now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000 <---- standard form.

8 0

3 years ago

if you’re good with set theory in math 30 please help with questions 31 and 32!! real answers only !!

GREYUIT [131]

Answer: 31) d 32) c

Step-by-step explanation:

31)

A = {1, 3, 5, 15}

B = {2, 3, 5, 7}

A ∪ B = {1, 2, 3, 5, 7, 15}

C = {2, 4, 6, 8}

(A ∪ B) ∩ C = {1, 2, 3, 5, 7, 15} ∩ {2, 4, 6, 8} = {2}

2 is the only number in both sets

32)

The pattern is: reflection, add symbol, ..., reflection, add symbol.

The last image shown is: add symbol

So the next image will be: reflection (flipped to the left)

7 0

3 years ago