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Afina-wow [57]
3 years ago
7

I need Help with THIS LIKE...NOWWW!!!!!!

Mathematics
2 answers:
stellarik [79]3 years ago
7 0
You would add the exponents. So you would just add 5 and 2. keep the 3. So it would be 3 to the power of 7
Mrac [35]3 years ago
4 0

The correct answer is 3 to the 7th power. When multiplying exponents, you are supposed to keep the base (which is 3) and multiply the exponent (5 x 2 = 7). Therefore, the answer is...

3^7

Hope this helps!

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The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand
Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})

P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})

P(X > 1560) = P(Z > \dfrac{110}{220})

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean \mu_x = np = 52 × 0.3085 = 16.042

The standard deviation =  \sqrt {n \time p (1-p)}

The standard deviation = \sqrt {52 \times 0.3085 (1-0.3085)}

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})

Pr ( Y > 20) = P(Z >1 .338)

From z tables

P(Y > 20) \simeq 0.0903

7 0
3 years ago
Need some help with this​
sergij07 [2.7K]

Answer:

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Step-by-step exp

4 0
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Step-by-step explanation:

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