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Tanzania [10]
3 years ago
10

Urgenttttt please I need it fast​

Mathematics
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

Below in bold.

Step-by-step explanation:

c) 3 x (9^2)^3/4 x ((81^3)^5/6

= 3 x 81^3/4 x 81^15/6

= 3 x 81^(3/4 + 15/6)

= 3 x 81^13/4

= 3 x 3^13

= 3^14

= 4,782,969.

f) (5x^-1y^2)^-2 / (25 x^2 y - 1)^2

=  5^-2 x^2y^-4 / 625 x^4y^-2

=  5^-2 x^-2 y^-2 / 5^4

= 5^-6 x^-2y^-2

= 0.000064x^-2y^-2.

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When Brad was born, his Grandma put, $1,500 in a new account for him. Since then, the balance of the account, has grown by 6.5%
marin [14]

Answer:

$969.04

Step-by-step explanation:

now at age of 18 Brad has 1500 * 1,065^18

at age of 21 he will have 1500 * 1,065^21

if he waits the difference will be 1500 * 1,065^21 - 1500 * 1,065^18 = $969.04

6 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
Which line has the smallest slope?<br>y= 3x<br>v=9x<br>y= .4x<br>y= .25x​
pentagon [3]
Y=.25x is the answer
3 0
3 years ago
Read 2 more answers
Question Help
mina [271]

The answer is 44 and if he deposits it is 33 dollars:

11x4

1x4=4

1x4=4

So 11x4=44

And if he deposits 11 he will have 33

3 0
3 years ago
What is 4n squared plus n squared?
Nutka1998 [239]

Answer:

Step-by-step explanation:

Itś 5n²

8 0
3 years ago
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