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2 years ago

Need help on this ASAP quickly please.

Mathematics

1 answer:

makvit [3.9K]2 years ago

3 0

Well if we’re adding them all together 1.2+1.2+1.2+1.2+1.2= 6kg

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On a biased dice the probability of getting 1 is 0.3. The ice is rolled 150 times. How many ones would you expect to roll?

Montano1993 [528]

Answer: 45

=====================================================

Work Shown:

n = number of rolls = 150

p = probability of rolling a "1" = 0.3

n*p = expected number of "1"s that show up out of the 150 rolls

n*p = 150*0.3

n*p = 45

We expect about 45 ones will show up.

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2 years ago

Consider the initial value problem:

Inessa [10]

Answer:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

Step-by-step explanation:

The given initial value problem is;

$y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2$

Let

$u(t)=y(t)----(2)\\v(t)=y'(t)----(3)$

Differentiating both sides of equation (1) with respect to $t$ , we obtain:

$u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)$

Differentiating both sides of equation (2) with respect to $t$ gives:

$v'(t)=y''(t)----(6)$

From equation (1),

$y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)$

Putting t=0 into equation (2) yields

$u(0)=y(0)\\\implies u(0)=-5$

Also putting t=0 into equation (3)

$u'(0)=y'(0)\\u'(0)=2$

The system of first order equations is:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

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