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3 years ago

Hi, I need help with this question and I need an explanation pls. Please don't copy the answer from the internet.

Mathematics

1 answer:

wlad13 [49]3 years ago

7 0

Answer:

−16(2−−5)

Step-by-step explanation:

hay un factor común entre estos números, por lo que simplemente −16t2+16t+80 divida y reutilice y luego reciba−162+16+80 llegar

−16(2−−5)

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A car rents for $40 per day plus 18¢ per mile. You are on a daily budge of $67. What mileage will allow you to stay within your

IrinaVladis [17]

Out of your $67 for the day, $40 of it goes to the rental before you even drive it out of the lot. That leaves you $27 a day for mileage.

$27 / $0.18 per mile = <u>150 miles</u> per day, tops.

4 0

3 years ago

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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

$f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0 } \atop {0}; Otherwise} \right.$

The value of constant C can be obtained as:

$\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1$

$\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}] } \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1$

$50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1$

$50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1$

$50C[0+\frac{1}{0.5} ] =1$

$100C = 1$ ⇒ $C = \frac{1}{100}$

C = 0.01

3 0

3 years ago

What is the mean median and mode of 5.6 6.8 1.2 6.5 7.9 6.5 ???

rjkz [21]

Median - 6.5
mode- 6.5
mean-5.8

8 0

3 years ago

Which is the correct input-output table for the function ?

alekssr [168]

Do you have any options?

3 0

3 years ago

Help !!!!!!!!!!!!!!!!

geniusboy [140]

Answer:

Hey there!

5(v+3)-10v>-10

5v+15-10v>-10

-5v+15>-10

-5v>-25

v<5

Let me know if this helps :)

5 0

3 years ago

Read 2 more answers