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Scorpion4ik [409]

3 years ago

15

The ABC Automobile Company assembles cars, but buys it’s parts from another company. The company buys tires from a wholesaler th

at ships the tires in containers that hold “t” tires. The company buys “b” brake pads in packages that contains 4 times as many brakes as the tire containers contain tires and purchase steering wheels shipped in pallets that contain half as many steering wheels “w” as the brake pad packages contain brakes. If the company orders 6 packages of brake pads and 2 pallets of steering wheels, what is the total number of brake pads and steering wheels it orders in terms of “t”? A) 2.5t B)4t C)20t D)28t

1 answer:

Vinvika [58]3 years ago

6 0

Answer: i'm pretty sure the answer is C

Step-by-step explanation:

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SMART PPL PLEASE HELP MEEEE

Neko [114]

Answer:

it's 80% of 125

the value of x = 125

Step-by-step explanation:

6 0

2 years ago

Please help Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation, then find the domain. (F-g)(x)

Vesna [10]

F(x) = 2x² + x - 3
g(x) = x - 1
(f - g)(x) = (2x² + x - 3) - (x - 1)
(f - g)(x) = 2x² + (x - x) + (-3 + 1)
(f - g)(x) = 2x² - 2

The answer is D.

4 0

3 years ago

let X represent the amount of time till the next student will arriv ein the library partking lot at the university. If we know t

AlekseyPX

Answer:

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean of 4 minutes

This means that $m = 4, \mu = \frac{1}{4} = 0.25$

Find the probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot:

This is:

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2)$

In which

$P(X \leq 132) = 1 - e^{-0.25*132} = 1$

$P(X \leq 2) = 1 - e^{-0.25*2} = 0.393469$

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2) = 1 - 0.393469 = 0.606531$

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

4 0

2 years ago

What is the standard form of the line with x-intercept of 6 and y-intercept of 5?

Vanyuwa [196]

Answer:

It should be 5x-6y=30

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

2 years ago

Read 2 more answers