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Makovka662 [10]

3 years ago

7

The expression 10a+6c gives the cost (in dollars) for a adults and c children to eat at a buffet restaurant. A. The cost for 1 a

dult is $ . The cost for 1 child is $ .

1 answer:

Liono4ka [1.6K]3 years ago

4 0

Answer: The total cost is $\$62$ assuming the cost for 1 adult is $\$5$ and the cost for 1 child is $\$2$

Step-by-step explanation:

Assuming the cost for 1 adult is $\$5$ and the cost for 1 child is $\$2$ :

$a=\$5$ and $c=\$2$

Then the expression that gives the total cost $10 a+6c$ is solved as:

$10(\$5)+6(\$2)=\$62$

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Use interval notation to describe.<br><br> John is at least 26 years old.

andrezito [222]

John= greater than or equal to 26 (a >26 with a line under the symbol)

5 0

3 years ago

As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur

Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math. It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

$y=Ce^{kt}$

Filling in our formula with the 2 conditions we are given:

$65=Ce^{10k}$ and $85=Ce^{15k}$

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k. If

$65=Ce^{10k}$ then

$\frac{65}{e^{10k}}=C$ which, by exponential rules is the same as

$C=65e^{-10k}$

Since that value of C is the same as the value of C in the other equation, we sub it in:

$85=65e^{-10k}(e^{15k})$

Divide both sides by 65 and use the rules of exponents again to get

$\frac{85}{65}=e^{-10k+15k}$ which simplifies down to

$\frac{85}{65}=e^{5k}$

Take the natural log of both sides to get

$ln(\frac{85}{65})=5k$

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

$65=Ce^{10(.0536527973)}$

which simplifies down to

$65=Ce^{.536527973}$

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

$y=38.01038064e^{(.0536527973)(23)}$ which simplifies a bit to

$y=38.01038064e^{1.234014338}$

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565

6 0

3 years ago

Round 389,935 to the nearest hundred thousand

Ivenika [448]

Answer:

400 thousand or 400,000

Step-by-step explanation:

just round 389 to the nearest hundred ------ 400

3 0

3 years ago

Read 2 more answers

La media de una población es 683.5g y su desviación estándar 52.1g. Determina la puntuación estándar para el valor 652.1

Ket [755]

Respuesta:

- 0,60

Explicación paso a paso:

Dado que:

Media poblacional, μ = 683,5

Desviación estándar, σ = 52,1

Puntuación, x = 652,1

La puntuación estandarizada, Z = (x - μ) / σ

Puntuación Z = (652,1 - 683,5) / 52,1

Puntuación Z = - 31,4 / 52,1

Zscore = −0.602687

Puntaje estandarizado = - 0.60

7 0

3 years ago

A large mixing tank currently contains 350 gallons of water, into which 10 pounds of sugar have been mixed. A tap will open, pou

Marysya12 [62]

Answer

Let t be number of minutes since tap is opened

water is increased by 20 gallons of water per minute.

sugar increase rate of 3 pounds per minute

water :- W(t) = 350 + 10 t

sugar :- S(t) = 10 + 3 t

concentration

C(t) = $\dfrac{S(t)}{W(t)}$

at t = 15 minutes

C(15) = $\dfrac{S(15)}{W(15)}$

C(15) = $\dfrac{10 + 3\times 15 }{350 + 10\times 15}$

C(15) = 0.11

at beginning t = 0 minutes

C(0) = $\dfrac{S(0)}{W(0)}$

C(0) = $\dfrac{10 + 3\times 0}{350 + 10\times 0}$

C(0) = 0.0286

C(15) > C(0)

hence, concentration is increasing as time is passing.

3 0

3 years ago