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3 years ago

You have two spheres. Sphere a has radius ra and sphere b has radius rb = kra. What is b's volumevb intermsofva andk? Vb = va

1 answer:

5 0

$V_a=\dfrac{4}{3}\pi r_a^3\\\\V_b=\dfrac{4}{3}\pi r_b^3\\\\V_b=\dfrac{4}{3}\pi(kr_a)^3\qquad\text{substitute for $r_b$}\\\\V_b=k^3V_a$

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The probability that Brian buys a sandwich is 0.1. The probability that Brian gets the bus is 0.6. Assuming the events are indep

omeli [17]

Answer:

6% chance

Step-by-step explanation:

I think it would be 0.1 * 0.6 which is 0.06

5 0

3 years ago

Read 2 more answers

If f(x) = 1/3x^2 + 5, find f(-9). A. 248 B. -22 C. 11 D. 32

melamori03 [73]

Answer:

D. 32

Step-by-step explanation:

For this problem, we simply plug -9 in for x in f(x):

$f(x)=\frac{1}{3} x^2+5$ ⇒ $f(9)=\frac{1}{3} *(9^2)+5=\frac{1}{3} *81+5=27+5=32$

Thus, the answer is D. 32.

Hope this helps!

7 0

3 years ago

Read 2 more answers

8 times 2/5 = ??? I'm very confused and I do not know how to simplify. I also need this in simplest form.

sergij07 [2.7K]

Make it 8/1 and multiply the top number and the bottom number then fone a common factor and divide by that number

6 0

3 years ago

Read 2 more answers

The probability that a randomly selected 2 2-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0

Mnenie [13.5K]

Answer:

a. $Probability = 0.97735$

b. $Probability = 0.92294$

c. $P(At\ Least\ One) = 1$

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

$P(Live) = 0.98861$

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)\ and\ P(Live)$

$Probability = 0.98861 * 0.98861$

$Probability = 0.9773497321$

$Probability = 0.97735$ --- Approximated

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)^n$

Where $n = 7$

$Probability = 0.98861^7$

$Probability = 0.92294324145$

$Probability = 0.92294$ --- Approximated

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

$P(At\ Least\ One) = 1 - P(None).$

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

$P(Not\ Live) = 0.01139$

The probability that none of the 7 lives up to 3 is:

$P(None) = P(Not\ Live)^7$

$P(None) = 0.01139^7$

Substitute this value for P(None) in

$P(At\ Least\ One) = 1 - P(None).$

$P(At\ Least\ One) = 1 - 0.01139^7$

$P(At\ Least\ One) = 0.99999999999997513055642436060443621$

$P(At\ Least\ One) = 1$ ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0

3 years ago

A lawyer charges a $75 consultation fee and then $120 per hour thereafter.

Ainat [17]

Answer:

(a) The dependant veriable is the $120 per hour and the independant veriable is the $75 consultaion fee.
(b) The table of vaules for this would look like this, $\frac{C}{T} \frac{75}{0} \frac{195}{1} \frac{315}{2} \frac{435}{3} \frac{555}{4}$ , where C = cost and T = time.
(c) The relationship between C and T is linear, beacause with the more hours you take the cost increases.
(d) Yes, it is sensible because the points would be in line due to the steady increase of the hourly rate.

(e) For every hour spent the cost increases 120 dollars.
(f) The fixed cost is 75 dollars for consultation and the veriable cost is 120 per hour.

3 0

2 years ago