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FromTheMoon [43]
3 years ago
14

4m^2×4m^3. how do i put this as a single power

Mathematics
1 answer:
Gala2k [10]3 years ago
6 0

You add exponents so 4m^5

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Simplify: -1 | 2/3 - 4 | / 5/6
stealth61 [152]

Answer:

<h3>4</h3>

Step-by-step explanation:

- 1 | \frac{2}{3}  - 4|  \div  \frac{5}{6}  \\  - 1 | \frac{2}{3} -  \frac{4 \times 3}{1 \times 3}  |  \div  \frac{5}{6}  \\  - 1 | \frac{2 - 12}{3} |  \div  \frac{5}{6}  \\  - 1 | \frac{ - 10}{3} |  \div  \frac{5}{6}  \\  \frac{10}{3}  \div  \frac{5}{6}  \\  \frac{10}{3}  \times  \frac{6}{5}  \\  \frac{2}{1}  \times  \frac{2}{1} = 4

3 0
3 years ago
In the souvenir shop, Gary finds a shirt and a key chain he would like to buy. The shirt costs $13.75 and the key chain costs $2
Firlakuza [10]

Answer:

$16.08

Step-by-step explanation:

13.75+2.33=16.08

3 0
3 years ago
A company has two manufacturing plants with daily production levels of 5x+11 items and 2x-3 ​items, respectively. The first plan
Elan Coil [88]

Answer:

alr 5x+11 and 2x-3

5x+11=16x

2x-3=-1x

Step-by-step explanation:

df

f

ff

f

f

g

g

t

t

t

t

t

t

y

y

y

y

4 0
3 years ago
This year the CDC reported that 30% of adults received their flu shot. Of those adults who received their flu shot,
Vlad [161]

Using conditional probability, it is found that there is a 0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Person has the flu.
  • Event B: Person got the flu shot.

The percentages associated with getting the flu are:

  • 20% of 30%(got the shot).
  • 65% of 70%(did not get the shot).

Hence:

P(A) = 0.2(0.3) + 0.65(0.7) = 0.515

The probability of both having the flu and getting the shot is:

P(A \cap B) = 0.2(0.3) = 0.06

Hence, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.515} = 0.1165

0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

To learn more about conditional probability, you can take a look at brainly.com/question/14398287

7 0
2 years ago
Can someone please help me with this question, ill give brainiest.
Dmitry_Shevchenko [17]

Answer:

uhm... what question? I don't see one

4 0
3 years ago
Read 2 more answers
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