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kodGreya [7K]
3 years ago
13

What is standard form of 104,47

Mathematics
2 answers:
Ymorist [56]3 years ago
4 0
It's already in standard form. 
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Nady [450]3 years ago
3 0
104.47 is already in standard form :)
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What are the characteristics of a quadrilateral?
gregori [183]

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Most known one: 4 sides

Step-by-step explanation:

They are closed shapes, their sides should be straight, they are strictly 2 dimensional, and finally, all quadrilaterals have 4 sides.

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2 years ago
Use the following algorithm:
andriy [413]

Answer:

First it says that you have to multiply the number x times 2, this will give you 2x as a result. Next you have to build your equation which is 2x=10, when you have an equation like this, you divide the result with the number before x (x=10÷2) your result will be x=5. Now you have to square the result (which is x) to get y, when squaring a number you have to multiply it by itself (depending of how many times it's asking to do so when there's a small number at the upper right corner of your number, that's how many times you multiply it) so if you have to square 5, it should be 5×5 which gives you the result of y being 25.

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Step-by-step explanation:

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3 years ago
1. Find the mean, median, mode, and range of the number of chapters per book of the Old Testament. Be sure to show how you arriv
Savatey [412]

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3 years ago
The 2003 Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United Sta
soldi70 [24.7K]

Answer:

a. P(x=0)=0.2967

b. P(x=1)=0.4444

c. P(x=2)=0.2219

d. P(x=3)=0.0369

Step-by-step explanation:

The variable X: "number of meals that exceed $50" can be modeled as a binomial random variable, with n=3 (the total number of meals) and p=0.333 (the probability that the chosen restaurant charges mor thena $50).

The probabilty p can be calculated dividing the amount of restaurants that are expected to charge more than $50 (5 restaurants)  by the total amount of restaurants from where we can pick (15 restaurants):

p=\dfrac{5}{15}=0.333

Then, we can model the probability that k meals cost more than $50 as:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{3}{k} 0.333^{k} 0.667^{3-k}\\\\\\

a. We have to calculate P(x=0)

P(x=0) = \dbinom{3}{0} p^{0}(1-p)^{3}=1*1*0.2967=0.2967\\\\\\

b. We have to calculate P(x=1)

P(x=1) = \dbinom{3}{1} p^{1}(1-p)^{2}=3*0.333*0.4449=0.4444\\\\\\

c. We have to calcualte P(x=2)

P(x=2) = \dbinom{3}{2} p^{2}(1-p)^{1}=3*0.1109*0.667=0.2219\\\\\\

d. We have to calculate P(x=3)

P(x=3) = \dbinom{3}{3} p^{3}(1-p)^{0}=1*0.0369*1=0.0369\\\\\\

6 0
2 years ago
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