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Leokris [45]
3 years ago
10

A new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries wil

l be independently selected and tested until two acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages.What is the probability that you test exactly 4 batteries
Mathematics
1 answer:
klio [65]3 years ago
3 0

Answer:

The probability that you test exactly 4 batteries is 0.0243.

Step-by-step explanation:

We are given that a new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found.

Suppose that 90% of all batteries have acceptable voltages.

Let the probability that batteries have acceptable voltages = P(A) = 0.90

So, the probability that batteries have unacceptable voltages = P(U) = 1 - P(A) = 1 - 0.90 = 0.10

Now, the probability that you test exactly 4 batteries is given by the three cases. Firstly, note that batteries will be tested until two acceptable ones have been found.

So, the cases are = P(AUUA) + P(UAUA) + P(UUAA)

This means that we have tested 4 batteries until we get two acceptable batteries.

So, required probability = (0.90 \times 0.10 \times 0.10 \times 0.90) + (0.10 \times 0.90 \times 0.10 \times 0.90) + (0.10 \times 0.10 \times 0.90 \times 0.90)

     =  0.0081 + 0.0081 + 0.0081 = <u>0.0243</u>

<u></u>

Hence, the probability that you test exactly 4 batteries is 0.0243.

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yanalaym [24]

Answer:

look at explanation

Step-by-step explanation:

To solve all of these questions you just need to scale. While scaling you can divide or multiply.

1. \frac{180 km}{4 hrs} (divide by 4 to get unit rate)

     unit rate = \frac{45km}{1hr}

2. If it takes one hour to travel 40 miles, then we don't have to scale or anything. We know \frac{40 miles}{1 hr}, the answer is...

"It'll only take one hour to travel 40 miles. This is correct because the unit rate is \frac{40 miles}{1 hr}. "

3. Knowing that the constant speed of the plane is \frac{800 km}{1 hour\\}, we can scale then add half of the unit rate to get our answer.

\frac{800 km}{1 hour\\} x 3 = \frac{2400 km}{3 hrs}   800 x 3 = 2400

then add half of unit rate which is \frac{400 km}{30 mins}  \sqrt[2]{800} = 400

so, our answer is \frac{2800 km}{3.5 hours}

4.  \frac{3km}{30 mins} <--- ( divided by 2) \frac{6 km}{1 hr} ( multiplied x 2.5) ---> \frac{15 km}{2.5 hrs}

To get an easier way to find the answer, if possible you can scale back farther than the 1 hour mark.

Elmer would take 2 and a half hours to ride 15 km.

5. The boys speed is \frac{4 km}{1 hr}. All we need to do is divide by 2. \sqrt[2]{8} = 4

3 0
3 years ago
The x-axis contains the base of an equilateral triangle RST. The origin is at S. Vertex T has coordinates (2h, 0) and the y-coor
frosja888 [35]
For the first part remember that an equilateral triangle is a triangle in which all three sides are equal & all three internal angles are each 60°. <span>So x-coordinate of R is in the middle of ST = (1/2)(2h-0) = h
And for the second </span><span> since this is an equilateral triangle the x coordinate of point R is equal to the coordinate of the midpoint of ST, which you figured out in the previous answer. Hope this works for you</span> 
7 0
4 years ago
Read 2 more answers
Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th
zalisa [80]
We have the following equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

y \leq -0.5x+5 \\ y \leq -1.25x+8

Region B.

Let's take a point that is in this region, that is:

P(0,6)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5  \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\  \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\  (2) \ y  \leq  -1.25x+8

Region C.

A point in this region is:

P(0,10)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \  \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Region D.

A point in this region is:

P(8,0)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \  \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2

So the inequalities are:

(1) \ y  \leq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\  \\ Solving \ for \ y: \\ y=-0.5(4)+5=3

Accordingly, the point is:

\boxed{p(4,3)}

Point q:

This point is the x-intercept of the line:

y=-0.5x+5

So let:

y=0

Then

x=\frac{5}{0.5}=10

Therefore, the point is:

\boxed{q(10,0)}

Part b) 

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

O(0,0)

Substituting in the inequalities:

y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}

It is true.

Region B.

Let's take a point that is in this region, that is:

P(0,7)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y  \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}

It is true

Region C.

Let's take a point that is in this region, that is:

P(0,11)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}

It is true

Region D.

Let's take a point that is in this region, that is:

P(9,0)

Substituting in the inequalities:

y  \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}

It is true

7 0
4 years ago
What is 0+-4????????????
Mumz [18]
0 + -4 = -4 

the 0 has no value and cannot add anything on top of the -4

So -4 stays the same:)

Answer = -4 

Good Luck! :)
4 0
3 years ago
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Alenkinab [10]

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