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bulgar [2K]
2 years ago
3

Use the graph representing bacteria decay to estimate the domain of the function and solve for the average rate of change across

the domain.
An exponential function titled Bacteria Decay with x axis labeled Time, in Minutes, and y axis labeled Amount of Bacteria, in Thousands, decreasing to the right with a y intercept of 0 comma 90 and an x intercept of 30 comma 0.

0 ≤ y ≤ 90, −0.33
0 ≤ y ≤ 90, −3
0 ≤ x ≤ 30, −0.33
0 ≤ x ≤ 30, −3

Mathematics
2 answers:
Alona [7]2 years ago
7 0

Answer:

D

Step-by-step explanation:

Took the same test and got that question

Tanya [424]2 years ago
5 0

Answer:

0 ≤ x ≤ 30, −3 is the answer.

Step-by-step explanation:

The domain of the function is all the x values between the highest and the lowest values of x.

The graph shows that the function starts at x=0 and ends at x=30; and it is not defined for x>30 and x; therefore the domain is

\boxed{0\leq x\leq 30.}

The average rate of change of the function is the slope of the line that best approximates the function, and we see that a line from (0,90) to (30,0) best approximates the function, and its slope is:

\frac{90-0}{0-30} =-3

Therefore the average slope is -3.

So the correct choice is the third one: 0\leq x\leq 30, -3

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A weatherman stated that the average temperature during July in Chattanooga is 80 degrees or less. A sample of 32 Julys is taken
baherus [9]

Answer:

The correct set of hypotheses is

H_{0}: d=80 degrees

H_{0}: d>80 degrees

Step-by-step explanation:

The weatherman claims that the average temperature during July in Chattanooga is 80 degrees or less. To test this claim, following hypotheses need to be taken:

Let d be the the average temperature during July in Chattanooga, then

H_{0}: d=80 degrees

H_{0}: d>80 degrees

7 0
2 years ago
A copper smelting process is supposed to reduce the arsenic content of the copper to less than 1000 ppm. let μ denote the mean a
Hitman42 [59]
The rejection region is give by 

|z_{test}|\ \textgreater \ z_{\alpha/2}

where the test statistics is given by

\frac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \frac{980-1000}{100/\sqrt{75}} \\ \\ = \frac{-20}{100/8.6603} = \frac{-20}{11.5470} =-1.73

i.e. |z_{test}|=|-1.73|=1.73

Thus, z_{\alpha/2}=1.73

Using the statistical table, the level of the test is 0.04.
6 0
2 years ago
Set up a double integral for calculating the flux of the vector field F⃗ (x,y,z)=xi⃗ +yj⃗ through the open-ended circular cylind
katrin [286]

Parameterize the cylinder (call it S) by

\vec s(\theta,z)=\sqrt8\cos\theta\,\vec\imath+\sqrt8\sin\theta\,\vec\jmath+z\,\vec k

with 0\le\theta\le2\pi and 0\le z\le9. Then

\vec F(x(\theta,z),y(\theta,z),z(\theta,z))=\sqrt8\cos\theta\,\vec\imath+\sqrt8\sin\theta\,\vec\jmath

Take the normal vector to S to be

\vec s_\theta\times\vec s_z=\sqrt8\cos\theta\,\vec\imath+\sqrt8\sin\theta\,\vec\jmath

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(\theta,z),y(\theta,z),z(\theta,z))\cdot(\vec s_\theta\times\vec s_z)\,\mathrm dz\,\mathrm d\theta

=\displaystyle\int_0^{2\pi}\int_0^98(\cos^2\theta+\sin^2\theta)\,\mathrm dz\,\mathrm d\theta=\boxed{144\pi}

7 0
2 years ago
Solve the equation by completing the square. Round to the nearest hundredth if necessary. x2 – 4x = 5
Ksenya-84 [330]
Solve for x over the real numbers:
x^2 - 4 x = 5

Subtract 5 from both sides:
x^2 - 4 x - 5 = 0

x = (4 ± sqrt((-4)^2 - 4 (-5)))/2 = (4 ± sqrt(16 + 20))/2 = (4 ± sqrt(36))/2:
x = (4 + sqrt(36))/2 or x = (4 - sqrt(36))/2

sqrt(36) = sqrt(4×9) = sqrt(2^2×3^2) = 2×3 = 6:
x = (4 + 6)/2 or x = (4 - 6)/2

(4 + 6)/2 = 10/2 = 5:
x = 5 or x = (4 - 6)/2

(4 - 6)/2 = -2/2 = -1:

Answer:  x = 5 or x = -1
8 0
2 years ago
Read 2 more answers
Why are you allowed to move the decimal points before dividing with decimals? Explain your reasoning. I will give the brainliest
oksano4ka [1.4K]

Answer:

so you can make both your dividend and your divisor equal so you can divide

Step-by-step explanation:

4 0
1 year ago
Read 2 more answers
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