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3 years ago

How do I find this answer

Mathematics

1 answer:

aleksklad [387]3 years ago

6 0

Answer:

i think its 35

Step-by-step explanation:

ab/bc = 7/3

this means ac = 10

ab/ac= 7/10

7/10= ab/50

50*7= 350

350/10= 35

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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 is the domain and range and function ??? Then can u explain how u did it please and thank you

enot [183]

The domain is
$- 10 \leqslant x \leqslant 10$
and the range is
$- 5 \leqslant y \leqslant 5$
Domain is x values (horizontal axis), while range is y values (vertical axis). On a line like this one, domain is calculated from the left most point to the right most, while range is calculated from the down most to the upmost.
Because this line is a solid line rather than a dotted line, less than or equal to
$\leqslant$
signs are used rather than just
$<$
less than signs.

6 0

3 years ago

Brian works 40 hours a week. He spends 40% of his time helping customers, 13% doing paperwork, and the remainder performing othe

MissTica

40% helping customers + 13% doing paper work = 53% of his time.

This means 100% - 53% = 47% of his time he is doing other tasks.

7 0

3 years ago

What would be the measure of segment FH?

Alisiya [41]

60 degrees because it is an equilateral triangle. Hope this helps and pls give thanks!!

4 0