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Ostrovityanka [42]
2 years ago
9

The probability that an egg on a production line is cracked is 0.01. Two eggs are selected at random from the production line. F

ind the probability that at least one egg is cracked. Write the entire decimal answer.
Mathematics
1 answer:
musickatia [10]2 years ago
5 0

Answer:

P(X \geq 1)=1-P(X

P(X=0)=(2C0)(0.01)^0 (1-0.01)^{2-0}=0.9801

And replacing we got:

P(X \geq 1) = 1-0.9801 = 0.0199

Step-by-step explanation:

Let X the random variable of interest "number of craked eggs", on this case we now that:

X \sim Binom(n=2, p=0.01)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(x \geq 1)=1-P(X

And we can find the probability:

P(X=0)=(2C0)(0.01)^0 (1-0.01)^{2-0}=0.9801

And replacing we got:

P(X \geq 1) = 1-0.9801 = 0.0199

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Let y in the form of a + ib, where a and b are real numbers, be the cubic roots of a complex number z 20, where z =2 4+i3. Find
GalinKa [24]

The possible values of a + b are 0.89, 3.01 and -3.9

<h3>How to determine the value of a + b?</h3>

The given parameters are:

y = a + ib

z = 24 + i3

Where:

y = ∛z

Take the cube of both sides

y³ = z

Substitute the values for y and z

(a + ib)³ = 24 + i3

Expand

a³ + 3a²(ib) + 3a(ib)² + (ib)³ = 24 + i3

Further, expand

a³ + i3a²b + i²3ab² + i³b³ = 24 + i3

In complex numbers;

i² = -1 and i³= -i

So, we have:

a³ + i3a²b + (-1)3ab² -ib³ = 24 + i3

Further expand

a³ + i3a²b - 3ab² - ib³ = 24 + i3

By comparing both sides of the equation, we have:

a³ - 3ab² = 24

i3a²b - ib³ = i3

Divide through by i

3a²b - b³ = 3

So, we have:

a³ - 3ab² = 24

3a²b - b³ = 3

Using a graphing tool, we have:

(a,b) = (-1.55, 2.44), (2.89,0.12) and (-1.34,-2.56)

Add these values

a + b = 0.89, 3.01 and -3.9

Hence, the possible values of a + b are 0.89, 3.01 and -3.9

Read more about complex numbers at:

brainly.com/question/10662770

#SPJ1

6 0
2 years ago
Two runners one averaging 5 miles per hour and the other one averaging 4 miles per hour, start at the same place and run along t
Vikentia [17]

Answer:

The Distance cover by both the Runners is same = 10 miles  

Step-by-step explanation:

According to question ,

The Speed of first runners (S 1) = 5 miles per hour

The speed of second runners (S 2)  = 4 miles per hour

Let The Time taken by First runner (T 1 ) = T hour

But the second runner  arrives half hour after the first runner ,

I.e The Time taken by Second runner (T 2) = ( T + \frac{1}{2} )

Now from Distance = Speed × Time

Since both the runners start from same place and run along the same trail

SO ,Both the Distance cover by both are same , D 1 = D 2

i.e Speed 1 × Time 1 = Speed 2 × Time 2

    5 mph × T            =  4 mph   ×  ( T + \frac{1}{2} )

    5 T =  4 T + ( 4 × \frac{1}{2} )

Or,  5 T - 4 T = 2

∴       T    =  2 hour  ,

Time take by first = T1 = T = 2 hour

Time take by second = T2 = T + \frac{1}{2} = (2  +  \frac{1}{2} )hour = \frac{5}{2}

Now the Distance cover = Speed × Time

              Distance   (D1)          = 5 mph  × 2 hour = 10 mile

And        Distance    (D2)         =  4 mph  × \frac{5}{2} = 10 miles

Hence, As The Distance cover by both the Runners is same = 10 miles  Answer

5 0
3 years ago
HELP ME PLS PLS PLS PLS
Kipish [7]

Answer:

y=-6/4x-4

Step-by-step explanation:

8 0
2 years ago
Rollie was succesful in losing weight he had a goal weight in mind he went on a diet for three months each month he would lose o
Llana [10]
Let's say that in the beginning he weighted x and at the end he weighted x-y, y being the number of kg he wanted to loose.

first month he lost
y/3

then he lost:  
(y-y/3)/3
this is
(2/3y)/3=2/9y
explanation: ((y-y/3) is what he still needed to loose: y minus what he lost already

and then he lost
 (y-2/9y-1/3y)/3+3 (the +3 is his additional 3 pounts)
 (y-2/9y-1/3y)/3-3=(7/9y-3/9y)/3+3=4/27y+3

it's not just y/3 because each month he lost one third of what the needed to loose at the current time, not in totatl

and  the weight at the end of the 3 months was still x-y+3, 3 pounds over his goal weight!


so:  x -y/3-2/9y-4/27y-3=x-y+3

we can subtract x from both sides:
-y/3-2/9y-4/27y-3=-y+3
add everything up:

-19/27y=-y+6

which means
-19/27y=-y+6

y-6=19/27y

8/27y=6
4/27y=3
y=20.25

so... that's how much he wanted to loose, but he lost 3 less than that, so 23.25

ps. i hope I didn't make a mistake in counting, let me know if i did. In any case you know HOW to solve it now, try to do the calculations yourself to see if they're correct!










5 0
3 years ago
HELPPPPPPPPPPP
liubo4ka [24]
<span>(A) Find the approximate length of the plank. Round to the nearest tenth of a foot.
Given that the distance of the ground is 3ft.
In order to get the length of the plank,
we can use the this one.

cos 49 = ground / plank
cos 49 = 3 / plank
plank = cos 49 / 3
plank = 0.10 ft

</span><span>(b) Find the height above the ground where the plank touches the wall. Round to the nearest tenth of a foot.
</span><span>
The remaining angle is equal to
angle = 180 - (90+49)
angle = 41

Finding the height.
tan 41 = height / ground
tan 41 = height / 3
height = tan 41 / 3
height = 0.05 ft.

(A) 0.10 feet
(B) 0.05 feet</span>
3 0
3 years ago
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