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2 years ago

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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A 300-lb gorilla climbs a tree to a height of 30 ft. Find the work done if the gorilla reaches that height in the following time

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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:

$W = Fd$

In which F is the force needed and d is the distance. The time does not interfere on the work.

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

This means that $F = 300, h = 30$

Thus

$W = Fd = 300(30) = 9000$

The work done in both cases is of 9000 ft-lb, as the work is not a function of time.

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2 years ago

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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3 years ago

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Ghella [55]

Answer:

Step-by-step explanation:

A = 1 plus negatives out be minusing positive

B= -8 minus negatives would be nega tive plus negatives

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If you love to love your a lover

if you hate to hate your a lover

if you love to have ur a hater

if you hate to love ur a hater

this all relates to positive and negative addition and subtraction

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1 year ago

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