All categories

3 years ago

Can someone please help me

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

6 0

Answer: D. (2,0) and (2,5) have the same x-value by different y-values

In other words, the input x = 2 leads to multiple outputs y = 0 and y = 5 at the same time. A function is where any input leads to exactly one output only (assuming that input is in the domain). So whenever x repeats itself like this is when we don't have a function.

You might be interested in

Hay sorry I forgot the picture last time but can some one help

FrozenT [24]

Answer:

Marina's design uses more cardboard.

in Kristen's design, it uses:

3*4/2*2+10*5+3*10+4*10=132 sq inches.

in Marina's design, it uses:

6*3*2+3*7*2+7*6*2=162 sq inches.

132<162

Therefore Marina's design uses more cardboard.

The difference between 2 designs are:

162-132

=30 sq inches

7 0

3 years ago

Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 4 grams per liter is added at

Alla [95]

Answer:

a) dx/dt = 600 - 6x

b) x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F = 6 L/min

a) Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dx/dt = Fᵢxᵢ - Fx

Fᵢ = 6 L/min, C = 4 g/L, F = 6 L/min

dC/dt = 24 - 6C

dx/dt = 25 (dC/dt), (dC/dt) = (1/25) (dx/dt) and C = x/25

(1/25)(dx/dt) = 24 - (6/25)x

dx/dt = 600 - 6x

b) dC/dt = 24 - 6C

dC/(24 - 6C) = dt

∫ dC/(24 - 6C) = ∫ dt

(-1/6) In (24 - C) = t + k (k = constant of integration)

In (24 - 6C) = -6t - 6k

-6k = K

In (24 - 6C) = K - 6t

At t = 0, C = 15 g/25 L = 0.6 g/L

In (24 - 6(0.6)) = K

In 20.4 = K

K = 3.02

So, the equation describing concentration of salt at anytime in the tank is

In (24 - 6C) = K - 6t

In (24 - 6C) = 3.02 - 6t

24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾

6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)

C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)/6

C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

But C = x/25

x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) when x = 20 g

20 = 100 - 4.12(e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

- (6t - 3.02) = In 80

- (6t - 3.02) = 4.382

(6t - 3.02) = -4.382

6t = -4.382 + 3.02

t = 1.362/6 = 0.227 min = 13.62 s

4 0

4 years ago

Find value of x from the given figure

Marizza181 [45]

I an not sure but i hope it will help you.
please mark me as brainlest.

7 0

3 years ago

Order the following from least to greatest: 14 -6, 0, 8, 3.4, -8, 2

Leto [7]

Answer:

-6. -8 0 3.4 6 8 14

Step-by-step explanation:

I think I am correct

5 0

3 years ago

Read 2 more answers

Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting

slega [8]

Answer:

(a) λ = 2.

(b) P (X = 0) = 0.1353; P (X = 1) = 0.2706;

P (X = 2) = 0.2706; P (X = 3) = 0.1804

Step-by-step explanation:

Let X = number of arrivals at the drive-up teller window.

The average number of arrivals at the drive-up teller window per minute is,

p = 0.4 customers/ minute.

(1)

Compute the expected number of customers at the drive-up teller window in n = 5 minutes as follows:

$E(X)=\lambda\\=np\\=5\times 0.4\\=2$

Thus, the mean number of customers that will arrive in a five-minute period is λ = 2.

(2)

The random variable X follows a Poisson distribution with parameter λ = 2.

The probability mass function of X is:

$P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...$

Compute the probability of exactly 0 arrivals in 5 minutes as follows:

$P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.1353\times 1}{1}=0.1353$

Compute the probability of exactly 1 arrivals in 5 minutes as follows:

$P(X=1)=\frac{e^{-2}2^{1}}{1!}=\frac{0.1353\times 2}{1}=0.2706$

Compute the probability of exactly 2 arrivals in 5 minutes as follows:

$P(X=2)=\frac{e^{-2}2^{2}}{2!}=\frac{0.1353\times 4}{2}=0.2706$

Compute the probability of exactly 3 arrivals in 5 minutes as follows:

$P(X=3)=\frac{e^{-2}2^{3}}{3!}=\frac{0.1353\times 8}{6}=0.1804$

Thus, the values are:

P (X = 0) = 0.1353

P (X = 1) = 0.2706

P (X = 2) = 0.2706

P (X = 3) = 0.1804

(3)

Delays occur in the service time if there are more than three customers arrive during any five-minute period.

Compute the probability that there are more than 3 customers as follows:

P (X > 3) = 1 - P (X ≤ 3)

$=1-\sum\limits^{3}_{x=0}{\frac{e^{-2}2^{x}}{x!}}\\=1-(0.1353+0.2706+0.2706+0.1804)\\=1-0.8569\\=0.1431$

Thus, the probability that delays will occur is 0.1431.

3 0

4 years ago