Find the measures of other angles for the rhombus below

Suppose that 80% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou

Gwar [14]

Answer:

P=0.147

Step-by-step explanation:

As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2

We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.

We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.

So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7

P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02

P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007

P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12

P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147

7 0

2 years ago

Can someone help ASAP !!

Olin [163]

Answer:

I can't help.

Step-by-step explanation:

There is no question, so therefore, I can not help.

6 0

3 years ago

Read 2 more answers

Find derivative problem Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago

School is Tomorrow I need help

german

Answer:

is their any way I can get some more info on this problem

Step-by-step explanation:

6 0

3 years ago

indicate the equation of the line, in standard form, that is the perpendicular bisector of the segment with endpoints (-1,6) and

vladimir1956 [14]

The equation of the line, in standard form, that is the perpendicular bisector of the segment with endpoints (-1,6) and (5, 5) is 12x - 2y = 13.

In the question, we are asked to indicate the equation of the line, in standard form, that is the perpendicular bisector of the segment with endpoints (-1,6) and (5, 5).

The slope of the line with endpoints (-1,6) and (5,5), can be calculated as:

m = (6 - 5)/(-1 - 5) = 1/(-6) = -1/6.

Thus, the slope of the perpendicular bisector = -1/m = -1/(-1/6) = 6.

The perpendicular bisector passes through the midpoint of the line with endpoints (-1,6) and (5,5), which can be calculated as:

(x₁, y₁) = ( {(-1 + 5)/2},{(6 + 5)/2} ),

or, (x₁,y₁) = (2, 11/2).

Thus, the required equation can be shown as:

(y - 11/2) = 6(x - 2), which can be shown in the standard form as follows:

(2y - 11)/2 = 6x - 12,

or, 2y - 11 = 12x - 24,

or, 12x - 2y = 13.

Thus, the equation of the line, in standard form, that is the perpendicular bisector of the segment with endpoints (-1,6) and (5, 5) is 12x - 2y = 13.

Learn more about the equation of perpendicular bisector at

brainly.com/question/20608689

#SPJ4

4 0

1 year ago

Find the measures of other angles for the rhombus below​

Find the measures of other angles for the rhombus below