-(1+7 x) – 61–7

(c) Evaluate P P(0). What might this stand for?

Naily [24]

I dont know but i think today you need a break being yourslef is awesome

3 0

2 years ago

Please help me i only have two tries! (slopes) (brainliest)

svet-max [94.6K]

The answer should be y=-1/7x

8 0

2 years ago

Read 2 more answers

Algebra 1: Factoring Polynomials Factor the common factor out of each expression: 40n^4m^2+40n^3m^3-32n^3

Nina [5.8K]

Taking out the greatest common factor

GCF of 40, 40 and 32 = 8
GCF of n^4, n^3 and n^3 is n^3

so the factors are

8n^3(5nm^2 +5m^3 - 4)

7 0

3 years ago

Suppose xy=−4 and dy/dt=−3. Find dx/dt when x=−1.

user100 [1]

When x=-1: $\quad (-1)y=-4\qquad\to\qquad y=4\)$

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

$x'y+xy'=0$

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.

From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4

and solve for x'.

Hope that helps!

4 0

3 years ago

. Lightning Strikes It has been said that the probability of being struck by lightning is about 1 in 750,000, but under what cir

mrs_skeptik [129]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$P(U |D ) = 0.198$

b

$P(O\ n \ B) = 0.188$

c

$P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

$T = 16 + 23 + \cdots + 16$

$T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

$T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

$T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

$T_d = 182$

Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

=>

Also

$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

=>

So

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

$P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001 and the value is 117

So

$P(O \ n \ B) = \frac{117}{623}$

=> $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

So

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

5 0

3 years ago

-(1+7 x) – 61–7 – x)= 36