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Varvara68 [4.7K]
2 years ago
6

Using the numbers, write two powers that have the same value. 2, 3, 6, 8

Mathematics
2 answers:
kicyunya [14]2 years ago
7 0

Answer:

8^2 = 2^6

Step-by-step explanation:

8^2=64

2^6=64

storchak [24]2 years ago
3 0

Answer:

8^2 = 2^6

8^2=64

2^6=64

Step-by-step explanation:

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Answer:

a) 9000 ft-lb

b) 9000 ft-lb

Step-by-step explanation:

Equation for the work:

The equation for the work done is given by:

W = Fd

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A 300-lb gorilla climbs a tree to a height of 30 ft

This means that F = 300, h = 30

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W = Fd = 300(30) = 9000

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Two events E1 and E2 are called independent if p(E1 â© E2) = p(E1)p(E2). For each of the following pairs of events, which are su
posledela

Answer:  a) Independent

b) Independent

c) Dependent

Step-by-step explanation:

Since, If a coin is tossed three times,

Then, total number of outcomes, n(S) = 8

a)  E_1 : tails comes up with the coin is tossed the first time;

E_1 = { TTT, THH, THT, TTH }

E_2 :  heads comes up when the coin is tossed the second time.

E_2 = { THT, HHH, THH, HHT }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{2}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}

Since,   E_1\cap E_2 = { THH, THT }

n(E_1\cap E_2) = 2

⇒ P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{2}{8}=\frac{1}{4}

Thus,  P(E_1\cap E_2)=P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are independent events.

B)  E_1 :  the first coin comes up tails

E_1 = { TTT, THH, THT, TTH }

E_2 :   two, and not three, heads come up in a row

E_2 = { HHT, THH }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{4}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}

Since,   E_1\cap E_2 = { THH }

n(E_1\cap E_2) = 1

⇒ P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{1}{8}

Thus,  P(E_1\cap E_2)=P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are independent events.

C)  E_1 :  the second coin comes up tails;

E_1 = { HTH, HTT, TTT, TTH }

E_2 :   two, and not three, heads come up in a row

E_2 = { HHT, THH }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{4}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}

Since,   E_1\cap E_2 = \phi

n(E_1\cap E_2) = 0

⇒ P(E_1\cap E_2) = 0

Thus,  P(E_1\cap E_2)\neq P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are dependent events.


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Step-by-step explanation:

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if you love to have ur a hater

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1 year ago
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