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3 years ago

What common base can be used to rewrite each side of the resulting equation? 2 4 16 32

Mathematics

2 answers:

Scorpion4ik [409]3 years ago

6 0

Answer:2

Step-by-step explanation:

Sauron [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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Laketown had 215 inches of snowfall last year, but only 155 inches this year. What was the percent decrease in snowfall for Lake

skelet666 [1.2K]

Okay, so FIRST you would subtract 215 - 155, which is 60, and THEN divide it by 215, so 60 divided by 215 = 0.27, and rounded that is 0.28, then you move the decimal two places to the right to make it a percentage so your answer would be A., 28%
I hope I helped! =D

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3 years ago

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Pls answer this! I need to pass!

KengaRu [80]

Answer:

Step-by-step explanation:

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3 years ago

Romeo will roll a 6-sided number cube, numbered 1-6, twice. What is the probability of rolling an odd number on the 1st roll and

elena-s [515]

1st rol= 3/6
2nd roll= 1/6

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2 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

Help me please sos!!

Anvisha [2.4K]

Answer:

75 ≤ length < 85

Step-by-step explanation:

You know a number 5 or greater will round up to the next higher number in the number place to its left. Similarly, numbers less than 5 round down (get replaced by zero).

This means that 75 will round up to 80. and 84.9999... will round down to 80.

a) The lower bound on length is 75. The length must be at least 75.

b) The upper bound on length is 85. The length must be less than 85.

4 0

2 years ago