A scuba diver is exploring 200 feet below sea level. What is the diver’s elevation after ascending 200 feet?

{1,3,5,7,9,11} find the subsets and proper subsets

vaieri [72.5K]

Answer:

4

Step-by-step explanation:

now gimme a brainliest

6 0

3 years ago

Solve the equation.

Zina [86]

Answer:

x=5

Step-by-step explanation:

esa es la respuesta correcta

7 0

2 years ago

Find f(-1) if f(x) = -2x^2 -x

kykrilka [37]

I hope it was helpful to you

if u like follow me

3 0

2 years ago

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an lg sm

GuDViN [60]

Answer: 0.951%

Explanation:

Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.

Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by

$P(x = k) = \frac{n!}{k!(n-k)!}p^k (1-p)^{n-k}$

Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones).

Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by

$P(x \ \textless \ 3) = P(x = 0) + P(x = 1) + P(x = 2)
\\ \indent = \frac{12!}{0!(12-0)!}(0.54)^0 (1-0.54)^{12-0} + \frac{12!}{1!(12-1)!}(0.54)^1 (1-0.54)^{12-1} + \\ \indent \frac{12!}{2!(12-2)!}(0.54)^2 (1-0.54)^{12-2}
\\
\\ \indent = \frac{12!}{(1)(12!)}(0.46)^{12} + \frac{12(11!)}{(1)(11!)}(0.54)(0.46)^{11}+ \frac{12(11)(10!)}{(2)(10!)}(0.54)^2(0.46)^{10}
\\
\\ \indent = (1)(0.46)^{12} + (12)(0.54)(0.46)^{11}+ (66)(0.54)^2(0.46)^{10}
\\ \indent \boxed{P(x \ \textless \ 3) \approx 0.00951836732 }
$

Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.

5 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