A 5 1/2 quart pot is filled 2/3 of the way with water. How many more quarts of water can the pot hold?

a) P(identified as explosive) equals P(actual explosive & identified as explosive) + P(not explosive & identified as explosive) = (10/(4*10^6))*0.95+(1-10/(4*10^6))*0.005 = 0.005002363. Thus, the probability that it actually contains explosives given that it's identified as containing explosives is (10/(4*10^6))*0.95/0.005002363 = 0.000475. b) Let the probability of correctly identifying a bag without explosives be a. Therefore, a = 0.99999763, approximately 99.999763%. c) No, even if this becomes 1, the true proportion of explosives will always be below half of the total detected.

Answer: the likelihood of a randomly selected tire lasting exactly 47,500 miles is 0.067

Step-by-step explanation:

Since the expected lifespan of this tire brand follows a normal distribution, we will use the normal distribution formula:

z = (x - µ)/σ

Where

x = lifespan of the tire in miles.

µ = mean

σ = standard deviation

The given figures include,

µ = 40000 miles

σ = 5000 miles

The probability that a tire will last precisely 47,500 miles

P(x = 47500)

For x = 47500,

z = (40000 - 47500) / 5000 = -1.5

According to the standard normal distribution table, the probability associated with this z score is 0.067

Correct question:

An urn holds 3 red and 7 black balls. Players A and B take turns withdrawing balls until a red one is chosen. Calculate the probability that A picks the red ball. (A goes first, followed by B, with no replacement of drawn balls).

Answer:

The likelihood that A picks the red ball is 58.33 %

Step-by-step explanation:

A will select the red ball if it is drawn 1st, 3rd, 5th, or 7th.

1st draw: 9C2

3rd draw: 7C2

5th draw: 5C2

7th draw: 3C2

Calculating for all possible scenarios gives us:

9C2 = (9!) / (7!2!) = 36

7C2 = (7!) / (5!2!) = 21

5C2 = (5!) / (3!2!) = 10

3C2 = (3!) / (2!) = 3

Adding these possibilities results in 36 + 21 + 10 + 3 = 70.

The total outcomes for selecting a red ball = 10C3

10C3 = (10!) / (7!3!)

= 120.

The probability that A selects the red ball is determined by dividing the sum of possible events by the overall outcomes.

P( A selects the red ball) = 70 / 120

= 0.5833

= 58.33 %

Answer:

Alright, we can express the baseball's motion with an equation like:

h(x) = a*x^2 + b*x + c

Here, x denotes time, while h(x) indicates height.

Let’s construct this:

The acceleration is:

a(t) = a

For velocity, integrating over time results in:

v(x) = a*x + v0

Where v0 signifies the initial vertical velocity.

Subsequently, we can determine position or height by integrating once more:

h(x) = a*x^2 + v0*x + h0

Here, h0 is the initial height.

<pthus our="" equation="" is:="">

h(x) = a*x^2 + v0*x + h0.

<pexamining the="" table:="">

When x = 0s, h(0s) = 6ft

<pthus:>

h(0s) = a*0s^2 + v0*0s + h0 = 6ft

            h0 = 6ft.

It’s also noted that:

h(2s) = h(4s)

<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.&lt;/p"><pin a="" standard="" quadratic="" function:="">

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

<pin this="" instance:="">

b = v0

a = a

<ptherefore we="" derive:="">

3s  = -v0/(2*a)

v0 = -3s*(2a)

<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">

h(x) = a*x^2 - 6s*a*x + 6ft

<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

<pthus our="" resulting="" equation="" reads:="">

h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft

b) The height after 5 seconds is expressed as:

h(5s) =  (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft

</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>