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3 years ago

What is equivalent to 5/1,215^x?

Mathematics

1 answer:

Inessa05 [86]3 years ago

5 0

5th quare root is the same as power with 1/5
So the solution is 1,215^x/5 , the 2nd option

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The midpoint of gh is m(4,-3). One endpoint is G(-2,2). Find the coordinates of endpoint H

Mars2501 [29]

Answer:

Step-by-step explanation:

(-2 + x)/2= 4

-2 + x = 8

x = 10

(2 + y)/2 = -3

2 + y = -6

y = -8

(10, -8) for endpoint H

4 0

3 years ago

35 of 39

natali 33 [55]

Step-by-step explanation:

5x 3 1/2 = 3 x 5 1/2 is false

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3 years ago

Rewrite the product as a power:<br> <img src="https://tex.z-dn.net/?f=%28-a%29%5E%7B3%7D%20b%5E%7B3%7D" id="TexFormula1" title="

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${ - a}^{3} . {b}^{3} = ( { - a \times b})^{3} = ( { - ab})^{3} \\$

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4 0

3 years ago

An acorn is falling from 13 feet in the air. If it descends 4.8 feet every second, write and solve an equation to determine how

podryga [215]

Answer:

2.5

Step-by-step explanation:

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5 0

4 years ago

A box contains 20 lightbulbs, of which 5 are defective. If 4 lightbulbs are picked from the box randomly, the probability that a

madam [21]

Answer:

Option D. 938/969

Step-by-step explanation:

At most 2 defective means 2 or less than 2 bulbs are defective

So, we have 3 cases:

a) No defective bulb. b) 1 defective bulb. c) 2 defective bulbs

Case a) No defective bulb

Total number of bulbs = 20

Number of defective bulbs = 5

Number of non-defective bulbs = 15

Total number of ways to select 0 defective bulb = 15C4 = 1365

Case b) 1 defective bulb

Total number of ways to select 1 defective bulb = 15C3 x 5C1 = 2275

Case c) 2 defective bulbs

Total number of ways to select 2 defective bulbs = 15C2 x 5C2 = 1050

Therefore, total number of ways to select at most 2 defective bulbs = 1365 + 2275 + 1050 = 4690

Total number of ways to select 4 bulbs from 20 = 20C4 = 4845

Therefore, probability of selecting at most 2 defective bulbs = $\frac{4690}{4845}=\frac{938}{969}$

Therefore, option D gives the correct answer.

6 0

3 years ago