The probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline = P(x > 4.3) = 0.00621

Step-by-step explanation:

This is a normal distribution problem

The mean of the sample = The population mean

μₓ = μ = 4 ounces

But the standard deviation of the sample is related to the standard deviation of the population through the relation

σₓ = σ/√n

where n = Sample size = 100

σₓ = 1.2/√100

σₓ = 0.12

The probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline = P(x > 4.3)

To do this, we first normalize/standardize the 4.3 ounces

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (4.3 - 4)/0.12 = 2.5

To determine the probability of finding an average in excess of 4.3 ounces of this ingredient from 100 randomly inspected 1-gallon samples of regular unleaded gasoline = P(x > 4.3) = P(z > 2.5)

We'll use data from the normal probability table for these probabilities

P(x > 4.3) = P(z > 2.5) = 1 - P(z ≤ 2.5) = 1 - 0.99379 = 0.00621

5 0

3 years ago

Please helppp meee I need it done noww

bearhunter [10]

Answer:

7/2

Step-by-step explanation:

3 wholes is equal to 6 halves plus a half equals 7 halves

6 0

4 years ago

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

3 0

2 years ago