Need help with statistics exercise?
Hi, With 39,000 gallons being 9,000 gallons above the mean with a standard deviation of 4000, then the Z score for this is 9,000/4,000 = 2.25. A z score of 2.25 is .4878 above the mean so that only leaves .5 - .4878 or .0122 left above 39,000. .0122 is a probability of 1.22% <==ANSWER I hope that helps!! :-)
I need some help with this statistics exercise?
A craft store has 25 assorted grab bags on sale for $3.00 each. 15 of the bags contain $3.00 worth of merchandise, six contain $2.00 worth, two contain $5.00 worth of merchandise,and there are one each containing $10.00 and $20.00 worth of merchandise.Suppose that you purchase one bag; What is your expected, gain or lose?
I need help with this calculus exercise?
. ( a ) Q(t) = 78/( 5 + 72*e^-t ) = lim t→∞ { 78/( 5 + 72*e^-t ) } = lim t→∞ { 78/( 5 + 72/e^t ) } = 78/( 5 + 0 ) = 15.6 thousand = 15600 people ━━━━━━━ ( b ) Q(t) = 78/( 5 + 72*e^-t ) Q(t) = 78 ( 5 + 72*e^-t )⁻¹ Q’(t) = -(1)(78) ( 5 + 72*e^-t )⁻² * (-72e^-t ) Q’(t) = 5616 e^-t / ( 5 + 72*e^-t )² Q’(t) = 5616 e^t / ( 72 + 5*e^t )² ━━━━━━━━━━━━━━ Q’(0) = 5616 e⁰ / ( 72 + 5*e⁰ )² Q’(0) = 5616 / 5929 thousand people / day Q’(0) = 947 people / day ━━━━━━━━━━━ ( c ) Q’(t) = 5616 e^t / ( 72 + 5*e^t )² Q’(5) = 5616 e⁵ / ( 72 + 5*e⁵ )² thousand people / day Q’(5) = 1258 people / day ━━━━━━━━━━━━ ( d ) Q’(t) = 5616 e^t / ( 72 + 5*e^t )² ( 72 + 5*e^t )² Q’(t) = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) Q’(t) + ( 72 + 5*e^t )² Q'’(t) = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) Q’(t) + ( 72 + 5*e^t )² * 0 = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) Q’(t) = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) * 5616 e^t / ( 72 + 5*e^t )² = 5616 e^t 56160e^t = 5616 ( 72 + 5*e^t ) 56160e^t = 404352 + 28080e^t e^t = 14.4 t = 2.67 days ━━━━━━
I need help with this exercise!!!?
I think your answers are correct! (Either that, or we're both wrong in the same way!) The first thing you would want to do is take the derivative dd/dh: dd/dh = (3.53)(1/2)h^(-1/2) = (1.765) / h^(1/2) = (1.765) / √h If you wanted to, you could rationalize the denominator, like this: dd/dh = (1.765)√h / h It's not strictly necessary, but some instructors prefer it. Anyway, now just plug in 200 for h, and that will give you the rate of change of d when h = 200. dd/dh = (1.765) / √200 = 0.1248 This agrees with your answer. If you need to know the distance to the horizon at a height of 200 meters, simply plug 200 into the original equation, like so: d = (3.53)(200)^(1/2) = 49.92 km This also agrees with your answer, so I'd say you're spot on. If you're still not certain, we can run a quick check. We know that at a height of 200 meters, we should gain 0.1248 km of distance for every meter risen. So let's just plug in 201 meters and see if we get approximately 49.92 km + 0.1248 km = 50.04 km d = (3.53)(201)^(1/2) = 50.05 km ≈ 50.04 km <== CHECK! I hope that helps. Good luck!
Chemestry exercises, i need help !?
Hi Im having a hard time doing these exercises : 1/ FE03 + ...CO --> ... CO2 b) Iron (III) oxide is reduced to iron in the blast furnace according to the equation completed in part (a) . How much iron (III) is needed to produce 112 g of iron ? c) How much iron can be made from 320 tonnes of iron (III) oxide? d) A certain iron ore is impure iron (III) oxide. 320 tonnes of this ore will make 202 tonnes of iron . What is the percentage purity of the iron ore? And this exercice: It was found that 25 cm3 of 0.1mol/dm3 sodium hydroxide solution was neutralised by exactly 20cm3of hydrochloric acid. The concentration of the acid was not known. a) how many moles of sodium hydroxide were there in the experiment? b) HCl+ NAOH --> NACL + H20 How many moles of hydrochloric were needed to neutralise the sodium hydroxide? c) What was the concentration of the hydrochloric acid? Thanks!
(2/2) I need your help to complete this exercise, i'm not sure with my own answer. Please share your opinion.
2. Send messages (preposition) ____(9)___ fax or email. A long-distance fax cost only about 35 cents, (conjunction)____(10)____ local message cost nothing. Email messages are equally inexpensive. 3. Use the FedEx or UPS account number (preposition)____(11)____ the recipient whenever possible. 4. Plan ahead so that (pronoun)____(12)_____ can use FedEx or UPS ground service. These ground (noun) _____(13)___ take about 3-5 days. Some overnight shipments, of course, (verb)____(14)____ critical. However, to retain our budget for those essential shipment, we must (verb)_____(15)____ our overall use by one half before April 1. If you can think of (adjective)_____(16)_____ ways to reduce to overnight shipment, please call me ext. 213. I appreciate your ideas and your (noun)____(17)____ in solving this problem. (thanks for your help)
I need help with this exercises of chemistry?
You are asking too many questions all at once. To help you answer them, you have to keep in mind the following: 1. 1 mL = 1cm3 1000 mg = 1 g density = mass / volume volume = mass / density 2. 1 mL = 1cm3 1000 mg = 1 g density = mass / volume mass = density x volume 3. volume = length x width x height density = mass / volume 4. If the density of an object is less than the density of water, it will float. If the density of an object is more than the density of water, it will sink. 5. volume = length x width x height density = mass / volume 6. density = mass / volume 7. volume = pi x radius squared x height radius = 1/2 diameter volume = [pi x diameter squared x height] / 4 density = mass / volume 8. mass = density x volume 9. volume = mass / density 10. What is the question? Since equal masses of water at two different temperatures have different volumes, it looks like water at the higher temperature is less dense than water at lower temperature because its volume is larger. 11. volume = mass / density Using the formulas I gave you, you should be able to figure out all the answers.
¿Integral : Please I need help with this exercise... ∫t sen 3t dt?
...∫t sen 3t dt Integral by parts ... uv - ∫ v du ... u = t => du = dt ... dv = Sen t dt => v = (-1/3)Cos3t .. ∫t sen 3t dt = (-1/3)tCos3t + (1/3)∫ Cos3t dt ..... = (-1/3)tCos3t + (1/3)(1/3)Sen 3t + C ..... = - tCos3t/3 + Sen 3t/9 + C OK!! bye...