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A home security company plans to open a calling center to respond to customer calls and

Answer:

$x+y\geqslant 12$

$x+2y\leqslant 20$

Step-by-step explanation:

<u>Inequalities </u>

The home security company will choose between two options for cubicles: the small cubicle for only one operator and the large cubicle to hold 2 operators. Let x be the number of small cubicles and y the number of large cubicles. Since each small cubicle has 49 square feet, the total area occupied by them will be 49x. Each large cubicle occupies 80 square feet, so the y cubicles use 80y square feet. The total area used by them is

Area=49x+80y

Assuming there is no operator without cubicle and all cubicles are fully occupied, the number of operators will be

Operators=1x+2y=x+2y

The number of cubicles is

# cubicles=x+y

The conditions of the problem state that

$x+y\geqslant 12$

$x+2y\leqslant 20$

7 0

3 years ago

I need serious help with these 2 questions. I already used the first attempt and have 1 left so please help me

Westkost [7]

Answer:

c and full truat i nwver wrong

6 0

3 years ago

Read 2 more answers

Use the graph of f below. Assume the entire function is graphed below.<br> Find the domain

enot [183]

ANSWER

The domain of the function is

$- 3 \leqslant x \leqslant 6$

or

[-3,6]

EXPLANATION

The domain of a function refers to the values of x for which the function is defined.

From the graph, we can see that the function begins at (-3,6) and ends at (6,-4).

Therefore the function is defined for x=-3 to x=6.

Hence the domain of the function is

$- 3 \leqslant x \leqslant 6$

or

[-3,6]

5 0

3 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

5 0

3 years ago

The accountant charged $35 for the first hour of work and $23 for each hour after that.He earned a total of$127 how many hours d

dexar [7]

Answer:

He worked a total of 5 hours.

Step-by-step explanation:

The first hour the man makes $35 dollars for the first hour then for the other hours he earns $23 dollars, right? So we start off with 35.

35 = one hour.

35 + 23 = 2 hours . = 58.

35 + 23 + 23 = 3 hours. = 81.

35 + 23 + 23 + 23 = 4 hours. = 104.

35 + 23 + 23 + 23 + 23 = 5 hours. = 127.

We were looking for if the man earned $127 dollars how many hours would be be working.

He'd be working 5 hours.

Hope this helps (:

5 0

3 years ago