Ask question

Ask question

All categories

1 year ago

9

A pet store has 15 birds and 75 fish.

2 answers:

Ludmilka [50]1 year ago

6 0

Answer:

1/5

Step-by-step explanation:

Alex Ar [27]1 year ago

6 0

Answer: b

Step-by-step explanation: 1/5

You might be interested in

Which expression is equivalent to 12r + 8p –34r – 2p?

ELEN [110]

Answer:

It is C -14r+6p

Step-by-step explanation:

If you want the explanation just tell me

3 0

1 year ago

Read 2 more answers

Every $4,000 spent on cable TV will reach 160 fans while every $2,750 spent on radio ad will reach 125 fans. A stadium has estab

laila [671]

Answer:

Neither of the two advertising options meet the goal.

Step-by-step explanation:

Every $4,000 spent on cable TV will reach 160 fans.

Therefore, the goal of reaching 200000 fans will be fulfilled by $\frac{4000 \times 200000}{160} = 5000000$ dollars i.e. $5 million which is greater than the budget of $3 million.

Now, every $2,750 spent on radio ads will reach 125 fans.

Therefore, the goal of reaching 200000 fans will be fulfilled by $\frac{2750 \times 200000}{125} = 4500000$ dollars i.e. $4.5 million which is again greater than the budget of $3 million.

Therefore, neither of the two advertising options meet the goal. (Answer)

8 0

1 year ago

The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)

Lemur [1.5K]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

$\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt$

Do, integration by parts

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}$

Substitute the value in 1

Cardiac output: $F=\frac{3}{54.49}$

Cardiac output: $F=0.055 L\s$

Hence Cardiac output: $F=0.055 L\s$

4 0

1 year ago

What is (0.5x + 15) + (8.2x- 16.6)

nadya68 [22]

1.6 I think I’m sorry if it’s incorrect

3 0

1 year ago

Which polynomial is prime? x3 + 3x2 – 2x – 6 x3 – 2x2 + 3x – 6 4x4 + 4x3 – 2x – 2 2x4 + x3 – x + 2

alexandr1967 [171]

D. 2x4+x3–x+2 <span>is prime</span>

4 0

1 year ago

Read 2 more answers