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Serjik [45]
2 years ago
7

Shawn said that in standard notation,

Mathematics
1 answer:
lawyer [7]2 years ago
7 0

Answer:

This is incorrect.  

Step-by-step explanation:

You move the decimal 7 places to the right after the 7.  Since 2 is the third digit, you're adding 4 more zeros.

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Drag expressions to complete each equation.
Ainat [17]
A^-b is the same as 1/a^b.

When there is a negative power, place the number and power over 1.

a^b/a^c = a^(b-c).

c is a negative power, because it is being divided, and is underneath b, which is a positive (and so it stays in the numerator).

a^c/b^c = (a/b)^c

Inside this one, the power of c is distributed to all numbers inside the parenthesis, in this case a and b.


hope this helps
3 0
3 years ago
Read 2 more answers
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
Tell whether the following function is linear. If so; graph the function: 2x + 3y = 5.
Brums [2.3K]
Here’s the correct answer, how I got it, how to help for next time, and the equation graphed :)

4 0
2 years ago
Find the value of x. (Hint: Use a trigonometric ratio or what you know about the lengths of sides in special right triangles.)
Leona [35]
Tan 45 = perpendicular/base 

1= 9 / x

x= 9 units
7 0
3 years ago
Read 2 more answers
HELP PLEASE!! THIS IS DUE IN AN HOUR!!
Neko [114]

Answer:

Your answer is 30.

Step-by-step explanation:

You plug in b=12, and 5=h

12*5/2

60/2

--->30<---

7 0
3 years ago
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