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3 years ago

10

heyyy! i haven't posted a question in a long time lolz. anyways, i really needed help with this!! if you can answer my question

i'll give brainliest, thank you

1 answer:

Andre45 [30]3 years ago

7 0

2 ones and 8 hundredths

00.00
the first zero is tens, ones, and after the decimal, to the right is the tenths, hundredths and so on

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What is the value of the 3 in 63?

Rina8888 [55]

Answer:

ones

Step-by-step explanation:

63

6=60

3=3

so it is in the ones place

6 0

3 years ago

Read 2 more answers

What is (x+2)(x+5) in polynomial standard form

Harlamova29_29 [7]

Answer:

=x^2+7x+10

Step-by-step explanation:

Simplify: xx+5x+2x+2.5: x+7x+10

Add similar elements: 5x+2x= 7x

=xx+7x+2.5

xx=x2

2.5=10

=x+7x+10

Answer: =x^2+7x+10

<em><u>Hope this helps.</u></em>

3 0

3 years ago

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

Solve for X.<br> X = [?]<br> 5x - 16 X + 10

Aleks [24]

$\boxed{ \sf{ \color{purple}\huge Answer :}}$

$\: \:$

$\large \tt \purple↪ \: \: \: \: \: 5x - 16 = x + 10$

$\: \:$

$\large \tt \purple↪ \: \: \: \: \:5x - x = 10 + 16$

$\:$

$\large \tt \purple↪ \: \: \: \: \: 4x = 26$

$\:$

$\large \tt \purple↪ \: \: \: \: \: x = \cancel\frac{26}{4}$

$\: \:$

$\underline{ \boxed{\large \tt \purple↪ \red{ \: \: \: \: x = \frac{13}{2} }}}{ \large✓}$

$\: \:$

hope helps ~

4 0

2 years ago

Someone please help with parts a b c and d. I will mark brainlest!!!

maria [59]

Answer:

A. 49 feet

B. 66 feet (round to the nearest foot)

C. 4 seg

Step-by-step explanation:

A. What is the height of the ball after 3 seconds?

For $t=3 seg$

$h(3)=-16(3)^{2} +63(3)+4$

$h(3)=-144+189+4\\h(3)=-144+193\\h(3)=49 feet$

B. What is the maximum height of the ball? round to the nearest foot

$h(t)=-16(t)^{2} +63(t)+4\\h'(t)=-32(t) +63\\$

then

$-32(t)+63=0\\t=\frac{-63}{-32}\\t=1.969seg$

For $t=1.969seg$

$h(1.969)=-16(1.969)^{2}+63(1.969)+4\\h(1.969)= 66.016\\h(1.969)=66 feet$

C. When will the ball hit the ground?

The ball will hit the ground when $h(t)=0$

so, $-16t^{2} +63(t)+4=0$

Using the quadratic equation

$t=4seg$

6 0

3 years ago