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MrRa [10]
3 years ago
9

Question in above picture explanation please not jusr answer

Mathematics
2 answers:
Lena [83]3 years ago
7 0

Answer:

x+16

Step-by-step explanation:

For this one the negative signs become positive because, if you multiply a negative number*a negative number it becomes positive. That is a rule.

-(-x-16)\\-1(-1x)-1(-6)\\x+16

son4ous [18]3 years ago
4 0

Answer:

C.) x+16

  1. Distribute parentheses -(-x) - (-16)
  2. Apply minus-plus rules

x+16

<h2><em>:) Hope This Helps You :)</em></h2>

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geniusboy [140]

The circumference of a circle is equal to the diameter times pi. So, let's look at what happens if you double the radius of a circle -- say from 2 to 4. The area will go from 12.56 to 50.24. This means that it has quadrupled.

4 0
3 years ago
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
What lump sum do parents need to deposit in an account earning 9%, compounded monthly, so that it will grow to $100,000 for thei
gogolik [260]

Answer:

They need to deposit $31,172.49

Step-by-step explanation:

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Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time the money is invested or borrowed for, in years.

In this problem

We want to find P for which A = 100000 when r = 0.09, n = 12, t = 13

So

A = P(1 + \frac{r}{n})^{nt}

100000 = P(1 + \frac{0.09}{12})^{12*13}

3.208P = 100000

P = \frac{100000}{3.208}

P = 31172.49

They need to deposit $31,172.49

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Multiply. Write in simplest form. Identify the two whole numbers between which the product lies.
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The answer is:
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~Hope this Helped~
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yawa3891 [41]

Answer:

ASA, ASA, SAS, SSS

Step-by-step explanation:

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