How do I do this plzzzz helppppppp

(14.14) larry reads that 1 out of 6 eggs contains salmonella bacteria. so he never uses more than 5 eggs in cooking. if eggs do

melomori [17]

we know that

The probability that "at least one" is the probability of exactly one, exactly 2, exactly 3, 4 and 5 contain salmonella.

The easiest way to solve this is to recognise that "at least one" is ALL 100% of the possibilities EXCEPT that none have salmonella.

If the probability that any one egg has 1/6 chance of salmonella

then

the probability that any one egg will not have salmonella = 5/6.

Therefore

for all 5 to not have salmonella

= (5/6)^5 = 3125 / 7776

= 0.401877 = 0.40 to 2 decimal places

REMEMBER this is the probability that NONE have salmonella

Therefore

the probability that at least one does = 1 - 0.40

= 0.60

the answer is

0.60 or 60%

6 0

3 years ago

Calculate the sum of the multiples of 4 from 0 to 1000

allochka39001 [22]

Answer:

sum is 125,500

sum in summation notation is $\sum\limits_{n=0}^n a+nd$ = (2a+(n-1)d)n/2

Step-by-step explanation:

This problem can be solved using concept of arithmetic progression.

The sum of n term terms in arithmetic progression is given by

sum = (2a+(n-1)d)n/2

where

a is the first term

d is the common difference of arithmetic progression

_____________________________________________________

in the problem

series is multiple of 4 starting from 4 ending at 1000

so series will look like

series: 0,4,8,12,16..................1000

a is first term so

here a is 0

lets find d the common difference

common difference is given by nth term - (n-1)th term

lets take nth term as 8

so (n-1)th term = 4

Thus,

d = 8-4 = 4

d can also be seen 4 intuitively as series is multiple of four.

_____________________________________________

let calculate value of n

we have last term as 1000

Nth term can be described

Nth term = 0+(n-1)d

1000 = (n-1)4

=> 1000 = 4n -4

=> 1000 + 4= 4n

=> n = 1004/4 = 251

_____________________________________

now we have

n = 1000

a = 0

d = 4

so we can calculate sum of the series by using formula given above

sum = (2a+(n-1)d)n/2

= (2*0 + (251-1)4)251/2

= (250*4)251/2

= 1000*251/2 = 500*251 = 125,500

Thus, sum is 125,500

sum in summation notation is $\sum\limits_{n=0}^n a+nd$ = (2a+(n-1)d)n/2

3 0

3 years ago

Please help me with this question, image attached.

Brilliant_brown [7]

Answer:

AEG - BEA = BEG => 81-27=54

Step-by-step explanation:

3 0

3 years ago

Erik and Nita are playing a game with numbers. In the game, they each think of a random number from zero to 20. If the differenc

Mademuasel [1]

Let X is the random number Erik thinks of, and Y is the random number Nita thinks of.
Both X and Y are in the range from 0 to 20.
X<=20
Y<=20
If the difference between their two numbers is less than 10, then Erik wins.
The difference between the two numbers can be written X-Y, or Y-X depending on which number (X or Y) is greater. But we do not know that. In order not to get negative value, we calculate absolute value of X-Y, written |X-Y| which will give positive value whether X is greater than Y or not.
If |X-Y|<10 Erik wins.
If the difference between their two numbers is greater than 10, then Nita wins.
If |X-Y|>10 Nita Wins

6 0

3 years ago

(10 pts) (a) (2 pts) What is the difference between an ordinary differential equation and an initial value problem? (b) (2 pts)

laiz [17]

Answer:

Step-by-step explanation:

(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.

(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.

(C) Example of a second order linear ODE:

M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)

The equation will be homogeneous if K(t)=0 and heterogeneous if $K(t)\neq 0$

Example of a second order nonlinear ODE:

$Y=-3K(Y){2}$

(D) Example of a nonlinear fourth order ODE:

$K^4(x) - \beta f [x, k(x)] = 0$

4 0

4 years ago