Express x in term of y:x/7+2y=6​

Answer:

As stated in the question, two tractors are pulling two identical stone blocks the same distance across similar surfacesAdditionally, block A moves at double the speed of block B when completing the race

This implies both blocks suffer from comparable friction loss

Moreover, we understand that

Energy loss from friction is

Thus, the friction loss should be identical for both blocks

therefore, option d is the accurate choice

Answer:

Step-by-step explanation:

a. The variable X represents the height of a Goomba, which follows a normal distribution with a mean of μ= 12 inches and a standard deviation of δ= 6 inches.

To find the probability that a Goomba picked at random has a height between 13 and 15 inches, you express it as:

P(13≤X≤15)

Considering that standard normal probability tables provide cumulative values, you can express this range as the cumulative probability up to 15 minus the cumulative probability up to 13. You'll first need to standardize these variable heights to obtain corresponding Z values:

P(X≤15) - P(X≤13)

P(Z≤(15-12)/6) - P(Z≤(13-12)/6)

P(Z≤0.33) - P(Z≤0.17)= 0.62930 - 0.56749= 0.06181

b. Now we have Y as the variable indicating the height of a Koopa Troopa. This variable also follows a normal distribution, with a mean μ= 15 inches and a standard deviation δ=3 inches.

The query concerns the probability that a Koopa Troopa stands taller than 75% of Goombas.

First step:

You need to determine the height of a randomly chosen Koopa Troopa that exceeds 75% of the Goomba population.

This entails determining the value of X corresponding to the limit below which 75% of the population falls, denoted by:

P(X ≤ b)= 0.75

Step 2:

Search the standard normal distribution for the Z value that has 0.75 beneath it:

Next, you will reverse the standardization to solve for "b"

Z= (b - μ)/δ

b= (Z*δ)+μ

b= (0.674*6)+12

b= 16.044 inches

Step 3:

With the height that identifies a Koopa Troopa taller than 75% of the Goomba population determined, compute the probability of selecting that Koopa Troopa:

P(Y≤16.044)

This time, utilize the Koopa’s average height and standard deviation to find the probability:

P(Z≤(16.044-15)/3)

P(Z≤0.348)= 0.636

The likelihood of randomly selecting a Koopa Troopa that is taller than 75% of Goombas is 63.6%

I hope this information is useful!

The equation representing the circle centered at (-27, 120) that passes through the origin is:

Solution:

The general equation of a circle is expressed as:

Where,

(a, b) denotes the center of the circle

r signifies the radius

Given the center as (-27, 120)

Thus;

a = -27

b = 120

Considering it intersects the origin, meaning (x, y) = (0, 0)

Substituting (a, b) = (-27, 120) and (x, y) = (0, 0) into the equation

Input = 15129 and (a, b) = (-27, 120) into the equation

Hence, the equation characterizing the circle is determined

Answer:

Carmen's estimate is too low, since rolling 200 times results in a total of 50.

Step-by-step explanation:

Initially, we will define the sample space for this scenario.

The sample space is Ω = {1,2,3,4,5,6,7,8}

For the event A: "Rolling an 8-sided die to get a multiple of 4"

The probability for event A is

This is because there are two multiples of 4 (4 and 8) out of a total of eight numbers (1 through 8).

Next, considering the random variable X: "Total count of multiples of 4 if she rolls an 8-sided die 200 times"

X can be described as following a Binomial distribution.

X ~ Bi (n,p)

X ~ Bi (200,)

Where n is the number of rolls and p is the probability of success, defined as rolling a multiple of 4.

The mean for this variable is

Thus, we conclude that Carmen's prediction is low, as rolling 200 times yields 50.