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3 years ago

1.3( x + 2 )-( y - 6) Help please

Mathematics

1 answer:

mestny [16]3 years ago

6 0

Basically, we need to plug in the given values for the variables $x$ and $y$ into the given expression $1.3(x+2)-(y-6)$ . First off, we can plug in all the given values into the expression, giving us $1.3(4+2)-(2-6)$ . Now, perform the operations on the inside of the parentheses. Doing this, we get $1.3(6)-(-4)$ . Now, we use the distributive property to simplify. This gives us $7.8+4$ . Finally, when we add the two numbers, we get $\boxed{11.8}$ . Hope this helped!

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What properties does a square have in common with a quadrilateral?

Rus_ich [418]

Both figures always have four sides

8 0

3 years ago

Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know

USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

$P(A/B)= \frac{P(AnB)}{P(B)}$

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

$P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33$

$P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33$

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0

3 years ago

Nine less than a number

olga2289 [7]

Let us say that this number is n

Then we could say:

n - 9

3 0

3 years ago

Read 2 more answers

Find the slope of the line (Image attached) A) 5 B) -5 C) 1/5 D) -1/5

Inessa [10]

The slope is 1/5 as [rise/run]

7 0

3 years ago

Read 2 more answers

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