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3 years ago

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In a food manufacturing company, the decision to reorder raw materials when inventories drop to a certain level is an example of

a(n) _____. a. programmed decision b. aberrant decision c. nonprogrammed decision d. unique decision

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

a. programmed decision

Step-by-step explanation:

When a decision is routinely made based on rules that look at specific measurable criteria, it is a <em>programmed decision</em>.

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in 1990 there were about 5.4 billion people in the world. if the population has been growing at 1.95% per year, estimate the yea

kompoz [17]

Answer:

2011

Step-by-step explanation:

Here is the formula.

$A=P(1+r)^{n}$

A = Final Value

P = Starting Value

R = Rate

N = Time in Years

A = 8,000,000,000

P = 5,400,000,000

R = 0.0195

N = Time in Years

Solve for n.

20.35 Years from 1990; so sometime in 2011 the population would be 8 billion.

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2 years ago

Solve the following inequality for the variable and graph the solution on the number line.

Likurg_2 [28]

The solved inequality is t > 12.

Divide both sides by 9 to isolate the variable.

9t becomes t, and 108 becomes 12.

<u>We do not flip the sign because we are dividing by a </u><u>positive</u><u>.</u>

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3 years ago

A sign in the the store window advertises 3-t shirts for $16.50 which coronet pierced as a unit rate of the T-shirt sale sc $16.

Kruka [31]

You would have to divide 16.50 into 3 then put your answer down hope this helps

8 0

3 years ago

If f(x) = x + 7 and g(x) = 1 / x - 13, whaf is the domain of (f°g)(x)?

ZanzabumX [31]

Answer: Choice D) x can be anything but 13

========================================================

Explanation:

The domain of $(f \circ g)(x) = f(g(x))$ is the same as the domain of g(x)

The domain for g(x) is $\{x|x \ne 13\}$ saying we can plug in any number we want as long as it's not 13. This is to avoid dividing by zero. The same domain applies for the composite function because

$f(x) = x+7\\\\\\f(g(x)) = g(x)+7\\\\\\f(g(x)) = \frac{1}{x-13}+7\\\\\\(f \circ g)(x) = \frac{1}{x-13}+7\\\\\\$

and we can see that we still need to kick out x = 13 from the domain to avoid the division by zero issue.

6 0

3 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

3 years ago