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Bess [88]
3 years ago
13

A company has determined that its weekly profit is a function of the number of items that it sells. Which equation could represe

nt the weekly profit in thousands of dollars, y, when the company sells x items? y squared = 4 x squared minus 100 y = negative x squared + 50 x minus 300 x = negative y squared minus 400 x squared = negative 6 y squared + 200
Mathematics
2 answers:
Mrac [35]3 years ago
7 0

Answer:

B. y= -x^2+50x-300

Step-by-step explanation:

A. y^2=4x^2-100

B. y= -x^2+50x-300

C. x=-y^2-400

D. x^2=-6y^2+200

we are to find profits (y) when the company sells x items

Option A can be used to calculate the profit (y) squared

Option B can be used to calculate profits (y)

Option C can be used to calculate items sold(x)

Option D can be used to calculate items sold squared(x^2)

We are asked to find the weekly profit (y) function which eliminate options A, C and D leaving us with option B

Therefore, the weekly profits (y) function in thousands of dollars when the company sells x items is

B. y= -x^2+50x-300

anygoal [31]3 years ago
7 0

Answer:

B. y= -x^2+50x-300

Step-by-step explanation:

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Police use a radar unit is used to measure speeds of cars on a freeway. The speeds are normally distributed with a mean of 90 km
vagabundo [1.1K]

Answer:

A. P(x≥100)=0.1587

B. P(x≤0)≈0

Step-by-step explanation:

A. Cause we know the distribution of the data, the method used to solve it is called "Normalization" and we need to have the Mean and the Standard deviation of the data. The method consist in the following equation

P(x≤a)=P( z=((x-μ)/σ) ≤ b=((a-μ)/σ) )

Considering <u>μ as the Mean</u> and <u>σ as the Standard deviation</u>. At first, we had a probability in the normal distribution with Mean=90 and STD=10 but <u>that kind of exercises is not meant to find that probability directly but by using this process</u>.

After we normalize the probability, now <u>we have a probability in a specific normal distribution that has Mean=0 and STD=1 and the difference with what we had before is that now we are able to use tools to find probabilities in a normal standard distribution</u>. My favorite of them is a chart that show the approximate values of a lot of probabilities (i attached it to this answer). I´m going to explain point A as an example:

We look for the probability that P(x≥100), but we don´t have an easy method to use there, so we normalize:

P(x≥100)=P( (x-μ)/σ ≥ (100-μ)/σ )

P(x≥100)=P( z ≥ (100-90)/10 )

P(x≥100)=P( z ≥ 1 )

And now we are able to use the chart, let me explain: First, the chart only works with P(z ≤ b), so we have to change it with properties of probabilities before using the table.

P(z≥1)=1-P(z≤1)

And finally we use the chart:

<u>the value of P(z≤1) is in the table, we look for the row with +1 and the column with the decimal part (in this case 0) and with coordinates (1,0) there´s the value</u>:

P(z≤1)=0.8413

But we need P(z≥1) so we use the previous equality

P(z≥1)=1-P(z≤1)

P(z≥1)=1-0.8413

P(z≥1)=0.1587

Because P(x≥100)=P(z≥1), our final answer is 0.1587

B. We use the same process to try to understand what the probability of P(x≤0) represents.

P(x≤0)=P(z≤ (0-90)/10)

P(x≤0)=P( z ≤ -9 )

But when we try to look for its value in the chart It isn´t even there, what could it mean?

<u>A normal distribution function is always increasing</u>, that means that "a≤b if and only if P(x≤a) ≤ P(x≤b)". so we conclude:

P(z≤-9) ≤ P(z≤-3) (The lowest probability in the chart)

P(z≤-9) ≤ 0.0013

P(z≤-9) is way lower than 0.0013 (they aren´t even close) but we know that probability is always positive,  and because of that:

P(x≤0)=P(z≤-9)≈0

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