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Anestetic [448]

3 years ago

6

If 8% of all cell phones are defective, and you have a random sample group of 100 Samsung Galaxy Vills,

1 answer:

Sati [7]3 years ago

5 0

Answer:

Mean = 8

Variance = 7.36

Standard Deviation = 2.7129

Step-by-step explanation:

This is a binomial distribution with parameters, n and p.

Where

n is sample size (given as 100)

p is the probability of success, or probability of defective (given as 8% or 0.08)

The mean, variance, standard deviation formula for binomial distribution is shown below:

Mean = $np$

Variance = $npq$

Standard Deviation = $\sqrt{npq}$

Where q would be probability of failure, or "1 - p"

Thus,

n = 100

p = 0.08

q = 1 - 0.08 = 0.92

SO, we have:

Mean = $np=100*0.08=8$

Variance = $npq=100*0.08*0.92 = 7.36$

Standard Deviation = $\sqrt{npq}=\sqrt{7.36}=2.7129$

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22. If 1 m,m/19 - 118°, and m/16 - 48°, calculate m/12.

Ulleksa [173]

Answer: Option 4

Step-by-step explanation:

$m\angle 20=62^{\circ}$ (linear pair)

$m\angle 11=70^{\circ}$ (angle sum in a triangle)

$m\angle 12=110^{\circ}$ (linear pair)

4 0

1 year ago

What is a algebraic expression for five less than three times the length,L

mina [271]

The expression is 5-3L or five minus three L

3 0

3 years ago

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the ref

hichkok12 [17]

Answer:

P is exactly 3km east from the oil refinery.

Step-by-step explanation:

Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:

$\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}$

So the cost of laying pipe under water with this distance is

$800000\sqrt{(3 - d)^2 + 4}$

And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:

$800000\sqrt{(3 - d)^2 + 4} + 400000d$

We can find the minimum value of this by taking the 1st derivative and set it to 0

$800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0$

We can move the first term over to the right hand side and divide both sides by 400000

$1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}$

$\sqrt{(3 - d)^2 + 4} = 6 - 2d$

From here we can square up both sides

$(3 - d)^2 + 4 = (6 - 2d)^2$

$9 - 6d + d^2 + 4 = 36 - 24d + 4d^2$

$3d^2-18d+27 = 0$

$d^2 - 6d + 9 = 0$

$(d - 3)^2 = 0$

$d -3 = 0$

d = 3

So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.

3 0

2 years ago

Good luck<br><br> 7•?=x <br><br><br> x+20=p <br><br><br> p+x=180

DiKsa [7]

Answer:

80/7 or 11.429

3 0

2 years ago

Read 2 more answers

2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.

Sergeu [11.5K]

Answer:

a) The maximum height is approx 15.5 unit.

b) The time it will take for Joey to reach the water is 1.45 hour.

Step-by-step explanation:

Given : When Joey dives off a diving board, the equation of his pathway can be modeled by $h(t)= -16t^2+15t + 12$

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled $h(t)= -16t^2+15t + 12$ ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

$h'= -32t+15$

For critical point put h'=0,

$-32t+15=0$

$t=\frac{15}{32}$

$t=0.46875$

Derivate again w.r.t. t,

$h''= -32$

It is maximum at t=0.46875.

Substitute t in equation (1),

$h(0.46875)= -16(0.46875)^2+15(0.46875)+12$

$h(0.46875)= -3.515625+7.03125+12$

$h(0.46875)= 15.515625$

The maximum height is approx 15.5 unit.

b) To find the time it will take for Joey to reach the water.

Put h=0 in equation (1),

$-16t^2+15t + 12=0$

Apply quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=-16 , b=15, c=12

$t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}$

$t=\frac{-15\pm\sqrt{225+768}}{-32}$

$t=\frac{-15\pm\sqrt{993}}{-32}$

$t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}$

$t=−0.515,1.453$

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.

5 0

3 years ago