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3 years ago

Which is the dependent variable in the following situation?

Mathematics

2 answers:

IrinaVladis [17]3 years ago

8 0

The number of binders
the dependent variable is the one you can change

Leviafan [203]3 years ago

4 0

Answer:

C number of binders because the teacher would have to order binders depending on how many students there are.

Step-by-step explanation:

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A company plans to sell a new type of cleaner for $280 each. The estimated cost, y, of manufacturing the cleaners is a function

Minchanka [31]

Suppose x is sold, $280 each , and y is the total financial cost. The first equation in the system is then certainly y = 280x.

The vertex form y-k = a(x-h)^2 where (h,k) is the vertex, and y is the y-intercept.

So, plug the values in.
11,000-24,000 = a(0-500)^2
-13,000=250,000a
a=-0.052

y-24,000 = -0.052(x-500)^2
y= -0.052(x-500)^2 + 24,000
This is the second equation in the system.

The answer is then A, which contain the system of equations which can be used to determine must be sold for the company to make a profit.

4 0

3 years ago

У – бх — 11 — 2x — Зу = -7

fiasKO [112]

Answer: -2y -8x = 4

Step-by-step explanation:

y - 6x - 11 - 2x - 3y = -7

y -3y -6x -2x = -7 +11

-2y -8x = 4

3 0

3 years ago

"A manufacturer of automobile batteries claims that the average length of life for its grade A battery is 55 months. Suppose the

STALIN [3.7K]

Answer:

$\\ P(z>-2) = 0.97725$ or P(x>49) is about 97.725% (or being less precise 97.5% using the empirical rule).

Step-by-step explanation:

We solve this question using the following information:

We are dealing here with normally distributed data, that is "the frequency distribution of the life length data is known to be mound-shaped".
The normal distribution is defined by two parameters: the population mean ( $\\ \mu$ ) and the population standard deviation ( $\\ \sigma$ ). In this case, we have that $\\ \mu = 55$ months, and $\\ \sigma = 3$ months.
To find the probabilities, we have to use the standard normal distribution, which has $\\ \mu = 0$ and $\\ \sigma = 1$ . The probabilities for this distribution are collected in the standard normal table, available in Statistics books or on the Internet. We can also use statistics programs to find these probabilities.
For most cases, we need to use the cumulative standard normal table, and for this we have to previously "transform" a raw score (x) into a z-score using the next formula: $\\ z = \frac{x - \mu}{\sigma}$ [1]. A z-score tells us the distance from the mean that a raw score is from it in standard deviations units. If this value is negative, the raw score is below the mean. Conversely, a positive value indicates that it is above the mean.
The cumulative standard normal table is made for positive values of z. Since the normal distribution is symmetrical around the mean, we can find the negative values of z using this formula: $\\ P(z$ [2].

Having all this information, we can solve the question.

<h3>The percentage of the manufacturer's grade A batteries that will last more than 49 months</h3>

First Step: Use formula [1] to find the z-score of the raw score x = 49 months.

$\\ z = \frac{49 - 55}{3}$

$\\ z = \frac{-6}{3}$

$\\ z = -2$

This means that the raw score is represented by a z-score of $\\ z = -2$ , which tells us that it is two standard deviations below the population mean.

Second Step: Consult this value in the cumulative standard normal table for z = 2 and apply the formula [2] to find the corresponding probability.

For a z = 2, the probability is 0.97725.

Then

$\\ P(z$

$\\ P(z2)$

But we are not asked for P(z<-2) but for P(z>-2) = P(x>49). This probability is the complement of the previous result, that is

$\\ P(z>-2) = 1 - P(z$

$\\ P(z>-2) = 1 - 0.02275$

$\\ P(z>-2) = 0.97725$

That is, the "percentage of the manufacturer's grade A batteries will last more than 49 months" is

$\\ P(z>-2) = 0.97725$ or about 97.725%

A graph below shows this result.

Notice that if we had used the 68-95-99.7 rule (also known as the empirical rule), that is, in a normal distribution, the interval between one standard deviation below and above the mean contains, approximately, 68% of the observations; the interval between two standard deviations below and above the mean contains, approximately, 95% of the observations; and the interval between three standard deviations below and above the mean contains, approximately, 99.7% of the observations, we could have concluded that 2.5 % of the manufacturer's grade A batteries will last less than 49 months, and, as a result, 1 - 0.025 = 0.975 or 97.5% will last more than 49 months.

We can conclude that with a less precise answer (but faster) because of the symmetry of the normal distribution, that is, 1 - 0.95 = 0.05. At both extremes we have 0.05/2 = 0.025 or 2.5% and we were asked for P(x>49) = P(z>-2) (see the graph below).