You need to add, (9+2X)+(2+3c)

The chief chemist for a major oil and gasoline production company claims that the regular unleaded gasoline produced by the comp

Svetradugi [14.3K]

Answer:

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

4 years ago

One type of lubricating oil weighs approximately 52 pounds per cubic foot, or 7 pounds per gallon. a 42-gallon barrel will hold

irina [24]

42 gallons weigh 7(42) or 294 pounds

<span>each cubic foot weighs 52 pounds </span>
<span>so the number of cubic feet = 294/52 = 5.654 ft^3</span>

4 0

3 years ago

Read 2 more answers

The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied y

ioda

Answer:

$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

$s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7$

$s= \sqrt{856849.7}=925.662$

Step-by-step explanation:

For this case we can create the following table

Interval Frequency (f) Midpoint(xi) xi *f xi^2* f

0-499 9 249.5 2245.5 560252.3

500-999 13 749.5 9743.5 7302753

1000-1499 33 1249.5 41233.5 51521258.25

1500-1999 115 1749.5 201193.5 361986278.8

2000-2499 125 2249.5 281187.5 632531281.3

2500-2999 81 2749.5 222709.5 612339770.3

3000-3499 47 3249.5 152726.5 496284761.8

3500-3999 45 3749.5 168727.5 632643761.3

4000-4499 22 4249.5 93489 397281505.5

4500-4999 10 4749.5 47495 225577502.5

Total 500 1220750 3408029125

$\sum f_i = 500 , \sum x_i f_i = 1220750, \sum x^2_i f_i = 3408029125$

For this case we can calculate the sample mean with this formula:

$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

And for the sample variance we can use the following formula:

$s^2= \frac{\sum x^2_i f_i - \frac{(\sum x_i f_i)^2}{n}}{n-1}$

And if we replace we got:

$s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7$

And the deviation is just the square root of the sample variance and for this case is:

$s= \sqrt{856849.7}=925.662$

4 0

3 years ago

What is the solution to the system of equations graphed below

aliina [53]

The point where the lines intersect is the point that satisfies both equations.

It is (1, 5), selection C.

3 0

3 years ago

Read 2 more answers

NEED MATH HELP NOW!!! PLEASE HELP!!!

andrezito [222]

Answer:

See explanation

Step-by-step explanation:

Vertex: -b/2a = 8/4 = 2. Plug 2 back into the equation to get (2, -2)

y-intercept: plug 0 for all the x values to get 6. The y intercept is (0,6)

x intercept: factor the equation:

2(x^2 - 4x + 3)

2(x-3)(x-1) <— by splitting the middle variable

so the x intercepts are (x-3) and (x-1) or 3 and 1.

Axis of symmetry : 2 (the x value of the vertex)

Domain: ARN ( all real numbers)

Range: ARN except y cannot be less than -2.

To graph you simply just plug in x values in a table and get the y values. Then plot them on the graph and draw a smooth curve.

4 0

4 years ago