Eso

Mathematics

1 answer:

Fiesta28 [93]3 years ago

4 0

Answer:

Step-by-step explanation:

182

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What is the probability of getting either a sum of 5 or at least one 5 in the roll of a pair of dice?

maria [59]

In a throw of 2 fair dice, there are 6*6=36 equiprobability outcomes.

To get a sum of 5, there are 4 ways, (1,4),(2,3),(3,2),(4,1) with probability of 4/36=1/9

To get at least one 5, there are 6+6-1=11 outcomes (note (5,5) has been counted in both, so subtracted from sum). The probability is 11/36

Since the two events are mutually exclusive (once we have a five, the sum can no longer be 5), we can add the probabilities to get the probability of one event or the other.

P(sum of 5 OR at least one 5)=1/9+11/36=4/36+11/36=15/36=5/9

4 0

4 years ago

28 < 4v solve for v pls help

DedPeter [7]

Answer:

v = 7

Step-by-step explanation:

Divide both sides by 4 to isolate v

28/4 = 4v/4

v = 7

5 0

3 years ago

Will someone help with this ! I was allowed a retry for it and I don’t know how to do it

pashok25 [27]

Answer:

$\frac{dN}{dt}=k(725-N)\\separating the variables and integrating\\\int \frac{dN}{725-N}=\int kdt+c\\-log (725-N)=kt+c\\log (725-N)=-kt-c\\(725-N)=e^{-kt-c} =e^{-kt} *e^{-c} =Ce^{-kt} \\when t=0,N=400\\725-400=Ce^{0} =C\\C=325\\when t=3,N=650\\725-650=325e^{-3k} \\\frac{75}{325}=(e^{-k})^3\\\\e^{-k} =(\frac{3}{13}) ^{\frac{1}{3} } \\when t=5\\725-N=325(\frac{3}{13}) ^{\frac{5}{3} } =325*\frac{3}{13}*(\frac{3}{13} )^{\frac{2}{3}}\\N=725-75*(\frac{3}{13} )^{\frac{2}{3} }=725-28=697$