Divide partial quotient 308 divide by 4

A coin is tossed and a number cube labelled 1 to 6 is rolled. How many likely outcomes are there? Use a tree diagram

Arte-miy333 [17]

H T
/ / / \ \ \ / / / \ \ \
1 2 3 4 5 6 1 2 3 4 5 6

12 different outcomes
since each side can land with one number

5 0

3 years ago

Which of these is a Euler diagram that represents p->q

Klio2033 [76]

It’s the 3rd one ...

5 0

3 years ago

At one point the average price of regular unleaded gasoline was $3.43 per gallon. Assume that the standard deviation price per

disa [49]

Answer:

(a) The percentage of gasoline stations had prices within 4 standard deviations of the mean is 93.75%.

(b) The percentage of gasoline stations had prices within 1.5 standard deviations of the mean is 55.56%. The prices for this stations goes from $3.325 to $3.535.

(c) The minimum percentage of stations that had prices between $3.22 and $3.64 is 88.88%.

Step-by-step explanation:

The Chebyshev's inequality states that at least 1-(1/K^2) of data from a sample must fall within K standard deviations from the mean, being K any positive real number greater than one.

It can be expressed as

$P(|X-\mu| \geq k\sigma)\leq \frac{1}{k^2}$

In this problem, we have, for the gasoline price, a normal distribution with mean of 3.43 and standar deviation of 0.07.

(a) The percentage of gasoline stations had prices within 4 standard deviations of the mean is equal to <em>one less the percentage of gasoline stations that had prices out of 4 standard deviations of the mean:</em>

$P(|X-\mu| \leq k\sigma)=1-P(|X-\mu| \leq k\sigma) \\\\1-P(|X-\mu| \leq k\sigma) \geq 1-\frac{1}{k^2} \\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-\frac{1}{4^2}\\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-1/16\\\\1-P(|X-\mu| \leq 4\sigma) \geq 0.9375$

The percentage of gasoline stations had prices within 4 standard deviations of the mean is 93.75%.

(b) The percentage of gasoline stations had prices within 1.5 standard deviations of the mean is 55.56%.

$P(|X-\mu| \leq k\sigma)\geq 1-\frac{1}{k^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 1-\frac{1}{1.5^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 0.5556\\$

The prices for this stations goes from $3.325 to $3.535.

$X=\mu\pm 1.5\sigma=3.43\pm 1.5*0.07=3.43 \pm 0.105\\\\X_{upper} =3.43+0.105=3.535\\X_{lower}=3.43-0.105=3.325$

(c) To answer, we have to calculate k for this range of prices:

$x=\mu\pm k\sigma\\\\k=\frac{|x-\mu|}{\sigma} =\frac{|3.64-3.43|}{0.07}=\frac{0.21}{0.07}= 3$

For k=3, the Chebyshev's inequality states:

$P(|X-\mu| \leq 3\sigma)\geq 1-\frac{1}{3^2}\\\\P(|X-\mu| \leq 3\sigma)\geq 0.8889$

So, the minimum percentage of stations that had prices between $3.22 and $3.64 is 88.88%.

7 0

3 years ago

How you do this find the equation.

Tanya [424]

#1 -3/2
#2 undefined
#3 4
#4 0
That is this four questions? Hope that can help you!

8 0

4 years ago

Suppose you’re making a model we are one block represents 5 feet about how many blocks tall is your model of the empire state bu

ASHA 777 [7]

Answer:

Step-by-step explanation:

Scale factor is 1:5

create a proportion: 1/5 = x/1450

cross multiply: 5x = 1450

divide both sides by 5

x = 280 blocks

6 0

3 years ago