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3 years ago

The class sizes of the Introductory Psychology courses at a college are shown below.

Mathematics

2 answers:

Ganezh [65]3 years ago

8 0

Answer:

can u send a lic of the acc question

yawa3891 [41]3 years ago

8 0

Answer:

pic of question

Step-by-step explanation:

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What is the exact volume of the cone in terms of pi?

kenny6666 [7]

Answer:

D) 3456 m

Step-by-step explanation:

Diameter of the cone = 48 m

Hence radius of the cone = $\frac{48}{2}$ = 24 m

Height of the cone = 18 m

Volume of a cone is given as $\[\frac{1}{3}*pi*radius^{2}*height\]
$

= $\[\frac{1}{3}*pi*24^{2}*18\]
$

= $\[\frac{1}{3}*pi*10368\]
$

= $\[pi*3456\]
$

Hence among the given options, option D is the correct one.

7 0

3 years ago

In the table, what are the domain and range of the function relating the pressure and volume of an ideal gas at a constant tempe

jasenka [17]

In the table what are the donain and range of the function relating the pressure and volume of an idea

7 0

2 years ago

Kelly has to read a 15 page poem she has read 10 pages so far,what fraction of the poem did she read?

Free_Kalibri [48]

Answer:

Kelly has read 2/3 of the poem.

Step-by-step explanation:

If you divide the 10 pages she has read by the total 15 pages, you get 0.66. Another way to get this is to put 10 over 15. Divide both the numerator and denominator by 5 to get 2/3.

5 0

3 years ago

Read 2 more answers

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago

Solve for x .<br><br> 18 ÷ x = -2 <br> x =

Ratling [72]

X would equal -9
Because -9 times -2 equals 18

5 0

2 years ago

Read 2 more answers