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3 years ago

Which of the following has a slope of -3 and a y intercept of -5

1 answer:

5 0

The answer is 2, y=-3x-5
Slope-intercept form is y=mx+b, where m is slope and b is the y-intercept.
So if m=-3 and b=-5, the equation is y = -3x -5

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The average age at which adolescent girls reach their adult height is 16 years. Suppose you have a sample of 29 adolescent girls

kumpel [21]

Answer:

standard error = 1.63

Step-by-step explanation:

To calculate the standard error we need to know the standard deviation (σ) and the sample size (n) (see the attached formula).

To obtain the standard deviation we use the given sample variance of 77.4 years:

σ²= variance

Therefore:

σ = √77.4 = 8.8

Now we can calcuate the estimated standard error:

standard error = σ /√n = 8.8/√29 = 1.63

The standard error gives us an estimation about how far the mean of the sample is from the mean of the entire population, and in this case is 1.63.

7 0

2 years ago

Which of the following is equivalent to (x^2 − 9)(x + 2) − (x^2 + 4x + 4)(x − 5)?

Aleonysh [2.5K]

The correct answer is D.

7 0

2 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

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

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

2 years ago

What compound inequality describes this graph

nadezda [96]

Answer:

0≥x<4

Step-by-step explanation:

first, let's look at this number line.

there is a closed circle at 0 and an open circle at 4. this means that 0 is included (≤ or ≥) and that 4 is not included (< or >).

these are the endpoints, meaning that in this compound inequality, the numbers next to the symbols are 0 and 4.

x is in the middle of this compound inequality.

0 x 4

now, we have to figure out the symbols in between. i wrote out our choices above for each number. the highlighted portion is greater than or equal to 0 and less than 4, so we can write this compound inequality as the following:

0≥x<4

x is greater than or equal to 0, but less than 4

4 0

3 years ago

A six-week fitness program was designed to decrease the time it takes retired individuals to walk one mile. At the beginning of

valkas [14]

Answer:

c. The sampling distribution of the sample means can be assumed to be approximately normal because the distribution of the sample data is not skewed

Step-by-step explanation:

From the given data, we have;

The category of the sample = Retired individuals

The number of participants in the sample = 20

The duration of program = six-weeks

The improvement seen by most participants = Little to no improvement

The improvement seen by few participants = Drastic improvement

Therefore, given that the participants are randomly selected and the majority of the participants make the same observation of improvement in the time to walk a mile, we have that, the majority of the outcomes show little difference in walk times after the program, therefore, the distribution of the sample data is not skewed and can be assumed to be approximately normal

7 0

2 years ago