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3 years ago

10

Can you help me with this problem?

1 answer:

Neko [114]3 years ago

8 0

Answer:

2,795

Step-by-step explanation:

The real answer is 2,795.1807, but because 1 isn't greater than 5 you can't make the 5 in the one's place a 6. So, 2,795 would be your final answer. Hope this helped and please mark brainliest! :^)

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Sam is using an industrial kitchen to make several batches of his famous chocolate chip granola bars. He needs to weight out 78

aleksandr82 [10.1K]

Sam is using an industrial kitchen to make several batches of his famous chocolate chip granola bars. He needs to weight out 78 ounces of chocolate chips, plus or minus 2.5 ounces. The equation that can be used to find the minimum or maximum amount,c, of chocolate chips that he can weigh out is-

Absolute value equation is = |c-2.5| = 78

7 0

3 years ago

Help MOI for the Third time PLS<br> Solve for t.<br><br> –9t − 19t + 9t − –18t − t = 14

Lorico [155]

Answer:

–9t − 19t + 9t − –18t − t = 14

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Step-by-step explanation:

Hope this helps!!

3 0

3 years ago

Whats 9 +10 will give brainlest

shtirl [24]

Answer:

19

Step-by-step explanation:

"You stupid"

"No Im not"

"Whats 9+10"

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4 0

3 years ago

Read 2 more answers

Let X denote the temperature (degree C) and let Y denote thetime in minutes that it takes for the diesel engine on anautomobile

BlackZzzverrR [31]

Answer:

Step-by-step explanation:

Given $f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y$

a)

we know that $\int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1$

therefore $\int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1$

on integrating we get

c=(1/6640)

b)

$P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy$

on doing the integration we get

=0.37349

c)

marginal density of X is

$f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy$

on doing integration we get

f(x)=(4x+3)/3320 ; 0<x<40

marginal density of Y is

$f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx$

on doing integration we get

$f(y)=\frac{(y+40.5)}{83}$

d)

$P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy$

solve the above integration we get the answer

e)

$P(X>20, 0$

solve the above integration we get the answer

f)

Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.

we know f(x,y)

In the (c) bit we got f(x) and f(y)

f(x,y)cramster-equation-2006112927536330036287f(x).f(y)

therefore X and Y are not independent

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3 years ago

As part of a science experiment, Greta measured the amount of water flowing from Container A to Container B. Container B had hal

Morgarella [4.7K]

Answer:

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Step-by-step explanation:

add the flow plus the original number

5 0

3 years ago