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3 years ago

Help please! 15 points!

Mathematics

1 answer:

Ivan3 years ago

5 0

It’s 3!! I checked on a calculator :)

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A travel agent currently has 80 people signed up for a tour. The price of a ticket is $5000 per person. The agency has chartered

Nikolay [14]

So hmm let's take a peek at the cost first

so, they chartered the plane for 150 folks with a fixed cost of 250,000
now, incidental fees are 300 per person, if we use the quantity "x", for how many folks, then if "x" persons are booked, then incidental fees are 300x

so, more than likely an insurance agency is charging them 300x for coverage

anyway, thus the cost C(x) = 250,000 + 300x

now, the Revenue R(x), is simple is jut price * quantity

well, the price, thus far we know is 5000 for 80 folks, but it can be lowered by 30 to get one more person, thus increasing profits

so... let's see what the price say y(x) is $\bf \begin{array}{ccllll}
quantity(x)&price(y)\\
-----&-----\\
80&5000\\
81&4970\\
82&4940\\
83&4910
\end{array}\\\\
-----------------------------\\\\$

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 80}}\quad ,&{{ 5000}})\quad 
% (c,d)
&({{ 83}}\quad ,&{{ 4910}})
\end{array}
\\\quad \\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-90}{3}\implies -30
\\ \quad \\\\
% point-slope intercept
y-{{ 5000}}={{ -30}}(x-{{ 80}})\implies y=-30x+2400+5000\\
\left.\qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=-30x+7400$

so.. now we know y(x) = -30x+7400

now, Revenue is just price * quantity
the price y(x) is -30x+7400, the quantity is "x"

that simply means R(x) = -30x²+7400x

now, for the profit P(x)

the profit is simple, that is just incoming revenue minus costs, whatever is left, is profit
so P(x) = R(x) - C(x)

P(x) = (7400x - 30x²) - (250,000+300x)

P(x) = -30x² + 7100x - 250,000

now, where does it get maximized? namely, where's the maximum for P(x)?

well $\bf \cfrac{dp}{dx}=-60x+7100$

and as you can see, if you zero out the derivative, there's only 1 critical point, run a first-derivative test on it, to see if its a maximum

7 0

3 years ago

Write y=1/6x+7 in standard form using integers

Law Incorporation [45]

The standard form using integers of the equation is 6y - x = 42

<h3>How to determine the standard form</h3>

Note that solving in standard form means writing it as a simple equation

Given y = 1/6x + 7

Find the LCM of the right side, we have

y = $\frac{x + 42}{6}$

Cross multiply

6 × y = x + 42

6y = x + 42

Make the variables move to the left side

6y - x = 42

Thus, the standard form using integers of the equation is 6y - x = 42

Learn more about algebraic expressions here:

brainly.com/question/723406

#SPJ1

8 0

2 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Which set of numbers is arranged in order from least to greatest?

charle [14.2K]

The answer is: [C]: -0.7, ⅕, 0.35, ⅔ .
________________________________________
Explanation:
_________________________________________
<span>
Note that in this correct Answer choice "C" given, we have the following arrangement of numbers:
_____________________________________________________
</span>→ -0.7, ⅕, 0.35, ⅔ ;
______________________________________
We are asked to find the "Answer choice" (or, perhaps, "Answer choices?") given that show a set of numbers arranged in order from "least to greatest"; that is, starting with a value that is the smallest number in the arrangement, and sequentially progressing, in order from least to greatest, with the largest (greatest) number in the arrangement appearing as the last number in the arrangement.
______________________
Note the EACH of the 4 (four) answer choices given consists of an arrangement with ONLY one negative number, "- 0.7". Only TWO of the answer choices—Choices "B" and "C"—have an arrangement beginning with the number, "-0.7 "; So we can "rule out" the "Answer choices: [A] and [D]".
________________________
Let us examine: Answer choice: [B]: <span>-0.7, 0.35, ⅕, ⅔ ;
</span>_________________________
Note: The fraction, "⅕" = "2/10"; or, write as: 0.2 .
________________________________________
The fraction, "⅔" = 0.6666667 (that is 0.6666... repeating; so we often see a "final decimal point" rounded to "7" at some point.
___________________________________________
Through experience, one will be able to automatically look at these 2 (two) fractions and immediately know their "decimal equivalents".
____________________________________________
Otherwise, one can determine the "decimal form" of these values on a calculator by division:
_________________________
→ ⅕ = 1/5 = 1 ÷ 5 = 0.2
_________________________
→ ⅔ = 2/3 = 2 ÷ 3 = 0.6666666666666667
___________________________________
For Answer choice: [B], we have:
______________________________
→ -0.7, 0.35, ⅕, ⅔ ;
_________________________
→ So, we can "rewrite" the arrangement of "Answer choice [B]" as:
___________________________________________
→ -0.7, 0.35, 0.2, 0.666666667 ;
________________________________
→ And we can see that "Answer choice: [B]" is INCORRECT; because
"0.2" (that is, "⅕"), is LESS THAN "0.35". So, "0.35" should not come BEFORE "⅕" in the arrangement that applies correctly to the problem.
_______________________________________
Let us examine: Answer choice: [C]: -0.7, ⅕, 0.35, 0.666666667 .
____________________________________________
→ Remember from our previous— and aforementioned—examination of "Answer Choice: [B]" ; that:
____________________________
→ ⅕ = 0.2 ; and:
→ ⅔ = 0.666666667
_______________________
So, given:
____________
→ Answer choice: [C]: -0.7, ⅕, 0.35, ⅔ ;
______________________
→ We can "rewrite" this given "arrangement", substituting our known "decimal values for the fractions:
______________________________
→ Answer choice: [C]: -0.7, 0.2, 0.35, 0.666666667 ;
_________________________________________
→ As mentioned above, this sequence starts with "-0.7", which is the ONLY negative number in the sequence; as such, the next positive number is correct. Nonetheless, "0.2" (or, "(⅕") is the next number in the sequence, and is greater than "-0.7". The next number is "0.35. "0.35" is greater than "⅕" (or, "0.2"). Then next number is "(⅔)" (or, "0.666666667").
"(⅔)"; (or, "0.666666667") is greater than 0.35.
____________________________
This set of numbers: "-0.7, ⅕, 0.35, ⅔" ; is arranged in order from least to greatest; which is "Answer choice: [C]: -0.7, ⅕, 0.35, ⅔" ; the correct answer.
________________________________________________________

6 0

3 years ago

Dion has a pizza with a diameter of 10 1/3 in . Is the square box shown big enough to fit the pizza inside? The box is 10.38 in

Artemon [7]

Yes, the box is wide enough because 10 1/3 is 10.33 repeating and the box is 10.38 (.05 inches wider than the pizza)

7 0

2 years ago