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3 years ago

The table lists the number of male and female students at Lakeside High school.

Mathematics

1 answer:

AysviL [449]3 years ago

4 0

Answer:

C) 161/163

Step-by-step explanation:

First, you add up all the students for both genders.

Female- 193+ 214 +197+ 201 = 805

Male- 216+ 207+ 198 + 194 = 815

That gives you a fraction on 805/815 which you simplify to 161/163

So your answer would be C, 161/163 :)

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How many configurations are there if we roll a die (6-face) four times?

algol [13]

Each time there are 6 possible outcomes.

If you roll four times, the number of possible outcomes would be 6^4, or 1296.

Final answer: 1296 configurations.

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3 years ago

Can somebody help me? anybody know the answer

tigry1 [53]

Answer:

25 = c

Step-by-step explanation:

You had a good start on filling in the values in the Pythagorean Theorem. The work needed to be finished.

$a^2 +b^2 = c^2\\\\20^2 +15^2=c^2\\\\400+225=c^2\\\\625=c^2\\\\\sqrt{625}=\sqrt{c^2}\\\\\boxed{25=c}$

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2 years ago

(7.8×10^7) + (9.9×10^7)

Hatshy [7]

(7.8×10⁷)+(9.9×10⁷)
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177000000

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3 years ago

Given f(x) = 9x - 10, find f(8) *

Rama09 [41]

Answer:

$\huge\boxed{f(8) = 62}$

Step-by-step explanation:

$\sf f(x) = 9x-10\\\\Put\ x = 8\\\\f(8) = 9(8) -10\\\\f(8) = 72-10\\\\f(8) = 62\\\\\rule[225]{225}{2}$

Hope this helped!

<h2>~AnonymousHelper1807</h2>

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3 years ago

Read 2 more answers

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for <em>S</em>.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for <em>C</em> :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick <em>P</em> such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago