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Alex73 [517]
3 years ago
6

Simplify 5^3/5^7 leaving your anwser in index form

Mathematics
1 answer:
Wewaii [24]3 years ago
6 0

Steps:

Simplify:

5³/5⁷

= 5³/5⁷

= 5*5*5/ 5*5*5*5*5*5*5

=1/5⁴

=1/625 (Decimal: 0.0016)

Steps by Step:

5³/5⁷

=125/5⁷

=125/78125

=1/625

Answer:   =1/625       (Decimal: 0.0016)

Please mark brainliest

<em><u>Hope this helps.</u></em>

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Answer:

50

Step-by-step explanation:

6 0
2 years ago
On a flight from New York to London, an airplane travels at a constant speed. An equation relating the distance traveled in mile
lions [1.4K]

Complete question :

On a flight New York to London an airplane travels at a constant speed. An equation relating the distance traveled in miles d to the number of hours flying t is t= 1/500d. How long will it take the airplane to travel 800 miles?

Answer:

1.6 hours

Step-by-step explanation:

Given the function :

Time t taken for flying distance d is given by:

t = 1/500 * d

Tine taken to fly 800 miles;

d = 800 miles

t = 1/500 * (800)

t = 800 / 500

t = 1.6

Hence, time t = 1.6 hours

3 0
3 years ago
Kashif bought a bicycle for rs 10000 and sold it for rs 8000 his loss percentage is
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Answer:

20%

Step-by-step explanation:

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  • Loss=2000
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7 0
2 years ago
(2.05) Choose the missing step in the given solution to the inequality −x − 10 &gt; 14 + 2x. (1 point) −x − 10 &gt; 14 + 2x −3x
sleet_krkn [62]
-x-10>14+2x
Add x for both side
-x+x-10>14+2x+x
-10>14+3x
Subtract 14 for both side
-10-14>14+3x-14
-24>3x
Divided 3 for both side
-24/3>3x/3
-8>x
x<-8
Or
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x<-8.
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4 0
3 years ago
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an
solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. ​(a) What is the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information X\sim N(\mu=268, \sigma^{2}=12^{2}).

(a)

The minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ <em>z</em> = 1.23

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283

Thus, the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248

Thus, the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths is 248 days.

8 0
2 years ago
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