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3 years ago

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Sixty nine percent of U.S households play video or computer games. Chose 4 heads of households at random. Find the probability t

hat none play video/computer games?

2 answers:

leva [86]3 years ago

6 0

Answer: 1*(69/100)^0*(31/100)^4 = 0.0092 or 0.0092*100 = 0.92%

statuscvo [17]3 years ago

3 0

Answer: 0.00923521

Step-by-step explanation:

Given : The probability U.S households play video or computer games=69%=0.69

here, the probability of each U.S household play video or computer games is fixed as 0.69

Then, the probability of each U.S household not play video or computer games= 1-0.69=0.31

For independent events the probability of their intersection is product of probability of each event.

Now, the probability that none play video/computer games will be :-

$(0.31)^4=0.00923521$

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malfutka [58]

Answer:

2x^3+7=56

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

$V = \frac{1}{3}\pi r^{2}h$

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

3 0

2 years ago

Raj paid $31.86 for a new basketball. The price he paid included sales tax of 6.25%.

Marianna [84]

<span>D.
p + 6.25p = 31.86</span>

3 0

3 years ago

The function f(x)=-(x-3)^2 +9 can be used to represent the area of a rectangle with the perimeter of 12 units, as a function of

Ivenika [448]

Maximum area of the rectangle is $9cm^{2}$

<u>Explanation:</u>

<u></u>

Considering the dimensions to be in cm

$f(x) = -(x-3)^{2} +9\\f(x) = -(x^{2} +9 - 6x)+9\\f(x) = -x^{2} +6x\\f'(x) = -2x+6\\-2x+6 = 0\\2x=6\\x=3cm\\\\$

Putting the value of x = 3

$Perimeter = 2(x+b)\\12 = 2(3+b)\\6 = 3+b\\b= 3cm$

$Area of rectangle = x X b\\ = 3 X 3\\ = 9cm^{2}$

Therefore, maximum area of the rectangle is $9cm^{2}$

7 0

2 years ago

TotCo is developing a new deluxe baby bassinet. If the length of a newborn baby is normally distributed with a mean of 50 cm and

slavikrds [6]

Answer:

91.63 cm is the interior length of the bassinet to ensure that 99 percent of newborn babies will fit, with a safety margin of 15 cm on each end of the bassinet.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50 cm

Standard Deviation, σ = 5 cm

We are given that the distribution of length of a newborn baby is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(X<x) = 0.99

We have to find the value of x such that the probability is 0.99

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 50}{5})=0.99$

Calculation the value from standard normal table, we have,

$P(z$

$\displaystyle\frac{x - 50}{5} = 2.326\\x = 61.63$

Thus, 99% of newborn babies will have a length of 61.63 cm or less.

There is a safety margin of 15 cm on each end of the bassinet

Length of bassinet =

$61+63 + 15 +15 = 91.63\text{ cm}$

6 0

2 years ago