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Lyrx [107]
3 years ago
14

. Select ALL the true statements. There is more than one answer. *

Mathematics
1 answer:
Dafna1 [17]3 years ago
8 0
I think the answer is D
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liraira [26]

poopy head ate da pilot

4 0
3 years ago
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were o
ad-work [718]

The question is incomplete. Here is the complete qeustion.

In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15

(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.

(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?

Answer: (a) 0.089 ≤ μ ≤ 0.171

(b) No

Step-by-step explanation:

(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample

X = \frac{0.1+0.13+0.16+0.15+0.14+0.08+0.15}{7}

X = 0.13

s = \sqrt{\frac{(0.1-0.13)^{2} + (0.13 - 0.13)^{2} + ... + (0.15 - 0.13)^{2}}{7-1} }

s = 0.029

The degrees of freedom is

N - 1 = 7 - 1 = 6

And since the confidence is of 99%:

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

The t-test statistics for t_{6,0.005} is 3.707

(Value found in the t-distribution table)

Now, calculate Error:

E = t_{6,0.005} . \frac{s}{\sqrt{N} }

E = 3.707. \frac{0.029}{\sqrt{7} }

E = 0.041

The interval will be:  

0.13 - 0.041 ≤ μ ≤ 0.13+0.041

0.089 ≤ μ ≤ 0.171

(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.

7 0
3 years ago
Decompose x^2+4x-3/x-2​
mariarad [96]

Answer:

\dfrac{x^3-2x^2+4x-3}{x-2}

Step-by-step explanation:

The given equation is :

x^2+\dfrac{4x-3}{x-2}

We need to decompose this equation.

We have,

x^2+\dfrac{4x-3}{x-2}\\\\=\dfrac{x^2(x-2)+4x-3}{x-2}\\\\=\dfrac{x^3-2x^2+4x-3}{x-2}

So, \dfrac{x^3-2x^2+4x-3}{x-2} is the decomposed form of the given expression.

5 0
2 years ago
A cube is packed with decorative pebbles. If the cube has a side length of 2 inches, and each pebble weighs on average 0.5 lb pe
kow [346]

We can solve this problem by first solving for the total volume of the cube. The formula for volume of cube is given as:

V cube = s^3

Where s is the length of one side which is equivalent to 2 inches. Therefore:

V cube = (2 inches)^3

V cube = 8 cubic inch

 

Assuming that all the pebble fills the cube without any spaces, then the total weight of the pebbles in the cube would simply be:

Total weight = 0.5 lb per cubic inch * 8 cubic inch

Total weight = 4 lb

 

Therefore, the total weight of the pebbles in the cube is <u>4 lbs</u>.

5 0
3 years ago
The circle below is centered at the origin and has a radius of 4. What is its
BlackZzzverrR [31]

Answer:

Step-by-step explanation:

The standard form of a circle's equation is

(x-h)^2+(y-k)^2=r^2 where h and k are the circle's center and r is the radius. We are told, and we can see, that the center of the circle is (0, 0) so h = 0 and k = 0. The radius, given as 4, goes into the equation squared, so the equation is

(x-0)^2+(y-0)^2=16 which simplifies down to

x^2+y^2=16, choice C.

8 0
3 years ago
Read 2 more answers
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