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3 years ago

Why is the constant of proportionality a useful concept?

2 answers:

5 0

It implies that there exist a unit-whatever that means to u. Take basis vectors which are linearly independent, they are maximal ideals in a ring R which generates a Field. similar to idea of Hilbert space

alexdok [17]3 years ago

4 0

❓❓❓Well think of real life problems?

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How do you solve 9x+2=8x+9

avanturin [10]

Try solving for the variable and get everything to one side

7 0

3 years ago

Read 2 more answers

What brand is better ? Brand A: $2.50 Brand B: 35 for $3.25 Brand C: 50 for $5.25 Brand D: 100 for $10.00

arlik [135]

I belive that the answer should be D) 100 for $10.00 because I had this exact same question one time.

Hope my answer helps!

Thank you for letting me answer!

Thank you,

Florida Mermaid

3 0

3 years ago

Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha

GalinKa [24]

Answer:

$\theta_1 \ n\ \theta_2 = 120, 240$

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

$Nonagon = 9\ sides$

$Hexagon = 6\ sides$

Divide 360 degrees by the number of sides in each angle, then find the multiples.

Nonagon

$\theta = \frac{360}{9} =40$

List the multiples of 40

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

Hexagon

$\theta = \frac{360}{6} =60$

List the multiples of 60

$\theta_2 = 60, 120, 180, 240, 300$

List out the common angles

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

$\theta_2 = 60, 120, 180, 240, 300$

$\theta_1 \ n\ \theta_2 = 120, 240$

This means that, only a rotation of $120, 240$ will lift both shapes onto themselves, when applied to both shapes.

The other angles will only work on one of the shapes, but not both at the same time.

7 0

3 years ago

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

Step-by-step explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$