Solve and show all work: (−3)^2=

32% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

Tems11 [23]

Answer:

a) $P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

b) $P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

c) $P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.32)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

Part b

$P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]$

$P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211$

$P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

Part c

$P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264$

$P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218$

$P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123$

$P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

6 0

3 years ago

Claire made $160 babysitting last summer. She put the money in a savings account that pays 3% interest per year. If Claire doesn

aleksklad [387]

We can make and use the equation: A = 160( 1.03 )^y: A = amount, 160 = amount put in, 1 + .03 = 1.03 = interest, and y as the number of years, as the money is compounded every year. (This is the compound interest equation, but I put in the numbers)

A) 160 * 1.03 = $164.80, 160 * 1.03² = $169.744 ≈ $169.74

B) 160 * 1.03⁵ = 185.48385 ≈ $185.48

The expression is the same as saying (((((160 * 1.03)1.03)1.03)1.03)1.03). This is like multiplying 160 * 1.03, which gives the amount for the first year. Then, that amount is multiplied by 1.03, which is the amount for the second year. This isi repeated until 5 years.

C) 160(1.03)³⁰

7 0

3 years ago

A regression analysis between weight (y in pounds) and height (x in inches) resulted in the following least squares line: ŷ = 13

Lunna [17]

Answer:

$y=6x +135$

And for this case the interpretation for the slope would be that for every unit that the height in inches increase then the weight in pounds increase 6 units.

For the intercept of 135 represent the amount initial amount of weight for the scale.

Step-by-step explanation:

We assume that they use least squares in order to create the regression equation

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

And the slope would be:

$m=6$

The means for x and y are given by:

$\bar x= \frac{\sum x_i}{n}$

$\bar y= \frac{\sum y_i}{n}$

And we can find the intercept using this:

$b=\bar y -m \bar x$

For this case we know that the line adjusted is:

$y=6x +135$

And for this case the interpretation for the slope would be that for every unit that the height in inches increase then the weight in pounds increase 6 units.

For the intercept of 135 represent the amount initial amount of weight for the scale.

3 0

4 years ago

A small family home in tucson arizona has a rooftop area of 1967 square feet, and it is possible to capture rain falling on abou

strojnjashka [21]

Answer:

Volume = 36.4 m^3 or 9616.6 gallon

weight = 18025.47 N or 80213.33 lbs

Step-by-step explanation:

The complete question is

A small family home in Tucson, Arizona, has a rooftop area of 1967 square feet, and it is possible to capture rain falling on about 56% of the roof. A typical annual rainfall is about 14 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water weigh when the tank was full (in N and in lbf)?

The area of the roof = 1967 ft^2

Area of water to be possibly captured = 56% of 1967 ft^2 = 0.56 x 1967 = 1101.52 ft^2

Rainfall depth = 14 inches = 14/12 ft = 1.167 ft

Volume of water = (area of water) x (depth of water) = 1101.52 ft^2 x 1.167 ft = 1285.47 ft^3

35.315 ft^3 = 1 m^3

1285.47 ft^3 = $x$ m^3

==> 1285.47/35.315 = 36.4 m^3

1 ft^3 = 7.481 gallon

1285.47 ft^3 = $x$ gallon

==> 1285.47 x 7.481 = 9616.6 gallon

Weight of water W = ρv

where ρ is the density of water = 62.4 lbs/ft^3

v is the volume of water = 1285.47 ft^3

substituting values,

W = 62.4 x 1285.47 = 80213.33 lbs

but,

1 lb = 4.45 N

therefore,

80213.33 lbs = 80213.33/4.45 = 18025.47 N

7 0

3 years ago

I need help with the question 6(x+4)=36

Savatey [412]

The answer:
6(x+4)=36
x=10

4 0

4 years ago

Read 2 more answers