All categories

3 years ago

PLEASE HELP ASAP WILL GIVE 30 POINTS

Mathematics

1 answer:

kykrilka [37]3 years ago

3 0

Answer:

C. $\displaystyle (−∞, 5) ∪ (5, ∞)$

Step-by-step explanation:

This is obvious the correct answer, considering that the denominator can NEVER equal zero.

I am joyous to assist you anytime.

You might be interested in

The F ratio increases as _________. Group of answer choices the variability between means increases relative to the variability

Alex Ar [27]

Answer:

The F-Ratio increases as, the variability between means increases relative to the variability within groups

Step-by-step explanation:

The F-Ratio is given as follows;

$F-Ratio = \dfrac{MS_{between}}{MS_{within}}$

Where;

$MS_{between}$ = The variance between groups

$MS_{within}$ = The variance within groups

Therefore, as the F-Ratio increases, the variability between groups (means) increases relation to the variability within groups

3 0

2 years ago

Points Which values should you plot to show a linear relationship?

Alexus [3.1K]

Answer: where’s the picture?

Step-by-step explanation:

4 0

3 years ago

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) Large and small drains are opened

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) Only the small drain is opened

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$