If all the beads in a jar are either orange or purple and the ratio of orange to purple beads is 6 to 5, what is the ratio of the following:
a. purple to orange
b. purple to all beads
c. all beads to orange
Please help ASAP on all 3 parts!!! :(

Answer:

150.51 dB

Step-by-step explanation:

Data provided in the question:

decibel level of sound at 161 km distance = 180 dB

d₁ = 161 km

d₂ = 4800 km

I₁ = 180 db

The formula for intensity of sound is given as:

I = [tex]10\log(\frac{I_2}{I_1})[/tex]

and the relation between intensity and distance is given as:

I ∝ [tex]\frac{1}{d^2}[/tex]

or

Id² = constant

thus,

I₁d₁² = I₂d₂²

or

[tex]\frac{I_2}{I_1}=\frac{d_1}{d_2}[/tex]

therefore,

I = [tex]10\log(\frac{d_1}{d_2})^2[/tex]

or

I = [tex]10\times2\times\log(\frac{161}{4,800})[/tex]

or

I = 20 × (-1.474)

or

I = -29.49

Therefore,

the decibel level on Rodriguez Island, 4,800 km away

= 180 - 29.49

= 150.51 dB

To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, you can use the inverse square law and the formula dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, you can calculate the decibel level on Rodriguez Island to be approximately 201.94 dB.

To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, we can use the inverse square law. According to the problem, the decibel level at a distance of 161 km is 180 dB. To find the decibel level at 4,800 km, we can use the formula: dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, we get: dB2 = dB1 - 20log10(d1/d2). Substituting dB1 = 180 dB, d1 = 161 km, and d2 = 4800 km, we can calculate the decibel level on Rodriguez Island.

dB2 = 180 - 20log10(161/4800) ≈ 180 - 20(-1.097) ≈ 180 + 21.94 ≈ 201.94 dB

Therefore, the decibel level on Rodriguez Island, 4,800 km away, is approximately 201.94 dB.

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Answer:

The lengths of the sides of a right triangle are

Longer leg = 0.4 units.

Shorter leg = 0.3 units.

Step-by-step explanation:

Given:

Hypotenuse = 0.5 units

Let the length of shorter leg of right triangle be x units then

According to the given condition,

length of longer leg will be (0.1 + x) units

Now,we know for a right triangle,by Pythagoras theorem we have

[tex](\textrm{Hypotenuse})^{2} = (\textrm{Longer leg})^{2}+(\textrm{Shorter leg})^{2}[/tex]

substituting the values we get

[tex]0.5^{2}= (x+0.1)^{2}+ x^{2}[/tex]

Applying [tex](a+b)^{2}= a^{2}+2ab+b^{2}[/tex]  we get

[tex]0.25= x^{2} +2\times 0.1\times x+ 0.1^{2} + x^{2} \\2x^{2} +0.2x+0.01-0.25=0\\2x^{2} +0.2x-0.24=0\\[/tex]

which is a quadratic equation

dividing the equation throughout by two we get

[tex]x^{2} +0.1x-0.12=0\\\textrm{on factorizing we get}\\x^{2} +0.4x-0.3x-0.12=0\\(x+0.4)(x-0.3)=0[/tex]

[tex]\therefore (x-0.3)= 0\\\therefore x=0.3[/tex]

Since x cannot be negative we  take

x = 0.3 units

∴ Longer leg = x + 0.1

                     = 0.3+0.1

                     =0.4 units

So, the lengths of the sides of a right triangle are

Longer leg = 0.4 units.

Shorter leg = 0.3 units.

Final answer:

To find the lengths of the sides of the right triangle with a hypotenuse of 0.5 units and one leg being 0.1 units longer than the other, you can use the Pythagorean theorem to set up an equation and solve for the lengths of the legs.

Explanation:

To find the lengths of the sides of the triangle, let's assume the length of the shorter leg is x units. Then, the length of the longer leg would be x + 0.1 units. Applying the Pythagorean theorem, where a and b are the legs and c is the hypotenuse:

a² + b² = c²

(x)² + (x + 0.1)² = (0.5)²

Simplifying the equation and solving for x, we get x ≈ 0.226 units for the shorter leg, and the longer leg would be x + 0.1 ≈ 0.326 units.

Answer:

Step-by-step explanation:

Hello!

The population variance is symbolized σ² and the standard deviation σ (remember that for estimations and statistics test, the parameter of study is always the variance)

The sentence "The standard deviation of pulse rates of adult males is less than 12 bpm." is symbolized σ <12

The sample variance is symbolized S² and the standard deviation is S.

The sample standard deviation is S= 11.5.

I hope it helps!

The symbol σ is used to represent the standard deviation in statistics. So, the standard deviation of pulse rates of adult males being less than 12 bpm in symbolic form is σ < 12. This value tells us that the dispersion of pulse rates among adult males is quite small.

The original claim is that the standard deviation of pulse rates of adult males is less than 12 bpm. This can be expressed in symbolic form as follows:

σ < 12

where σ represents the standard deviation.

In the context of this question, the standard deviation is a measure of the dispersion or spread in the pulse rates of adult males. The sample that was taken had a standard deviation of 11.5 bpm, which is less than 12 bpm, thus supporting the original claim.

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Answer:

a) solved in attachment

b)  If disk is running properly time between failure will exceed then Upper 90 % would be good indicator.

Step-by-step explanation:

Answer:

3,359

Step-by-step explanation:

Let A, B, C and D the four groups of people.

Let us denote with |A| the number of elements in a set A.

Then the number of elements of A∪B∪C∪D is the sum of the elements in each group subtracting the elements that have been counted twice.

That is,

|A∪B∪C∪D |=|A|+|B|+|C|+|D| - |A∩B| - |A∩C| - |A∩D| - |B∩C| -  |B∩D|- |C∩D| - |A∩B∩C| - |A∩B∩D| - |A∩C∩D| - |B∩C∩ D| - |A∩ B∩C∩ D| =

1000+1000+1000+1000 - 100 - 100 - 100 - 100 - 100 - 100 - 10- 10- 10- 10 - 1 = 3359

r = (3V/2π)^(1/3) and h = V/(πr^2), respectively.

To determine the dimensions of the silo that requires the least amount of material to build, we need to find the values of the radius and height that minimize the surface area. Let's start by expressing the volume of the silo as a function of one variable. The volume is given by V = πr^2h + (2/3)πr^3, where r is the radius and h is the height. Since we're looking for the minimum surface area, we'll differentiate the surface area expression, equate it to zero, and solve the resulting equation to find the values of r and h that minimize the surface area.

Taking the derivative of the surface area function A(r, h) = π(3r^2 + 2rh) with respect to r and h, we get:

dA/dr = 6πr + 2πh = 0

dA/dh = 2πr = 0

Solving these equations simultaneously, we find that r = 0 and h = 0, which are not meaningful values in this context. Therefore, there are no critical points inside the domain. Instead, we need to examine the endpoints of the domain. Since r and h must be positive, we find that A(r, h) goes to infinity as r or h approaches zero. Therefore, the surface area does not have a minimum in the interior of the domain and we need to consider the endpoints.

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To determine the dimensions of the silo that requires the least material, set up an equation involving the volume and the surface area. Solve for the values of r and h that minimize the surface area. Use differentiation and the second derivative test to find the minimum point.

To determine the radius and height of the silo that requires the least amount of material to build, we need to find the dimensions that minimize the surface area of the silo. The surface area of the silo is given by A = π(3r^2 + 2rh), and the volume of the silo is given by V = πr^2h + (2/3)πr^3.

We are given that the volume of the silo is 505π ft^3. We can use this information to set up an equation involving the volume and the dimensions of the silo. Solving this equation will give us the values of r and h that minimize the surface area.

By differentiating the surface area equation with respect to r and h, and setting the derivatives equal to zero, we can find the critical points. By analyzing the second derivative test, we can determine which point is the minimum.

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By Green's theorem,

[tex]\displaystyle\int_C(7(x^2-y)\,\vec\imath+3(y^2+x)\,\vec\jmath)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial3(y^2+x)}{\partial x}-\frac{\partial7(x^2-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]

[tex]\displaystyle=10\iint_D\mathrm dx\,\mathrm dy[/tex]

where [tex]D[/tex] is the region bounded by the closed curve [tex]C[/tex]. The remaining integral is 10 times the area of [tex]D[/tex].

Since [tex]D[/tex] is a circle in both cases, and we're given the equations for them right away, it's just a matter of determining the radius of each one and plugging it into the well-known formula for the area of a circle with radius [tex]r[/tex], [tex]\pi r^2[/tex].

(a) [tex]C[/tex] is a circle with radius 3, so the line integral is [tex]10\pi(3^2)=\boxed{90\pi}[/tex].

(b) [tex]C[/tex] is a circle with radius [tex]R[/tex], so the line integral is [tex]\boxed{10\pi R^2}[/tex].

Let ,

First calculate the value for given equation by using green's theorem,

Since, the formula for greens theorem is:

The given equation is,

[tex]\int\ C [7(x^2-y)i +3(y^2+x)j]dr[/tex]

Here,

[tex]F=[7(x^2-y)i +3(y^2+x)j][/tex]

Now to calculate the value of [tex]Curl F[/tex],

[tex]Curl F=\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial k} \\7(x^2-y)&(y^2+x)&0\end{array}\right] \\\\CurlF=[\frac{\partial}{ \partial y} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (y^2+x)-\frac{\partial}{\partial y} (7x^2-y)]\\\\Curl F=0+0+10k\\\\CurlF=10k[/tex]

Substitute in equation (1),

[tex]\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10k *kdA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10dA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=10\int\ \int\ dA[/tex]

The remaining integral is [tex]10[/tex] times the area of region .

The general equation is,

[tex]x^2+y^2=r^2[/tex]

The area of a circle is [tex]\pi r^2[/tex]  .

Hence, area of region of circle is [tex]10\pi r^2[/tex].

Now,

(a) The given equation is [tex](x-2)^2+(y-3)^2=3^2[/tex],

  C is a circle with radius 3, so the line integral is

         [tex]10\pi (3)^2=90\pi[/tex] .

(b) The given equation is [tex](x-a)^2+(y-b)^2=R^2[/tex],

    C is a circle with radius [tex]R[/tex] , so the line integral is,

         [tex]10\pi (R)^2=10\pi R^2[/tex].

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Answer:

[tex]P(\bar x>60)=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]

Is a very improbable event.

Step-by-step explanation:

We want to calculate the probability that the total weight exceeds the limit when the average weight x exceeds 6000/100=60.

If we analyze the situation we this:

If [tex]x_1,x_2,\dots,x_100[/tex] represent the 100 random beggage weights for the n=100 passengers . We assume that for each [tex]i=1,2,3,\dots,100[/tex] for each [tex]x_i[/tex] the distribution assumed is normal with the following parameters [tex]\mu=49, \sigma=18[/tex].

Another important assumption is that the each one of the random variables are independent.

1) First way to solve the problem

The random variable S who represent the sum of the 100 weight is given by:

[tex]S=x_1 +x_2 +\dots +x_100 =\sum_{i=1}^{100} x_i[/tex]

The mean for this random variable is given by:

[tex]E(S)=\sum_{i=1}^{100} E(x_i)=100\mu = 100*49=4900[/tex]

And the variance is given by:

[tex]Var(S)=\sum_{i=1}^{100} Var(x_i)=100(\sigma)^2 = 100*(18)^2[/tex]

And the deviation:

[tex]Sd(S)=\sqrt{100(\sigma)^2} = 10*(18)=180[/tex]

So we have this distribution for S

[tex]S \sim (4900,180)[/tex]

On this case we are working with the total so we can find the probability on this way:

[tex]P(S>6000)=P(z>\frac{6000-4900}{180})=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]

2) Second way to solve the problem

We know that the sample mean have the following distribution:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}}[/tex]

If we are interested on the probability that the population mean would be higher than 60 we can find this probability like this:

[tex]P(\bar x >60)=P(\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{60-49}{\frac{18}{\sqrt{100}}})[/tex]

[tex]P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]

And with both methods we got the same probability. So it's very improbable that the limit would be exceeded for this case.

Answer:

0.77,0.60

Step-by-step explanation:

Given that in a population of 1000 subjects, 770 possess a certain characteristic.

A sample of 40 subjects selected from this population has 24 subjects who possess the same characteristic

To find out sample proportion:

Sample size n = 40

Favourable x = 24

Sample proportion p = [tex]\frac{24}{40} =0.60[/tex]

To find out population proportion:

Total population N = 1000

Favourable X = 770

population proportion P = [tex]\frac{770}{1000} =0.77[/tex]

Answer:

a) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{1.36}=1.166[/tex]

b) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{149}=12.21[/tex]

Step-by-step explanation:

Part a

So then the random variable is given by this table

X     | 1      | 2    | 3    | 4     |

P(X) | 0.2 | 0.2 | 0.2 | 0.4 |

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]

And if we use the values obtained we got:

[tex]E(X)=1*0.2 +2*0.2 +3*0.2 +4*0.4=2.8[/tex]

In order to find the standard deviation we need to find first the second moment, given by :

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]

And using the formula we got:

[tex]E(X^2)=(1^2 *0.2)+(2^2 *0.2)+(3^2 *0.2)+(4^2 *0.4)=9.2[/tex]

Then we can find the variance with the following formula:

[tex]Var(X)=E(X^2)-[E(X)]^2 =9.2-(2.8)^2 =1.36[/tex]

And then the standard deviation would be given by:

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{1.36}=1.17[/tex]

Part b

So then the random variable is given by this table

X     | -20  | -10 | 0   | 10  |20  |

P(X) | 0.1 | 0.2 | 0.4 | 0.1 | 0.2  |

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]

And if we use the values obtained we got:

[tex]E(X)=(-20*0.1) +(-10*0.2) +(0*0.4) +(10*0.1)+(20*0.2)=1[/tex]

In order to find the standard deviation we need to find first the second moment, given by :

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]

And using the formula we got:

[tex]E(X^2)=((-20)^2 *0.1)+((-10)^2 *0.2)+(0^2 *0.4)+(10^2 *0.1)+(20^2 *0.2)=150[/tex]

Then we can find the variance with the following formula:

[tex]Var(X)=E(X^2)-[E(X)]^2 =150-(1)^2 =149[/tex]

And then the standard deviation would be given by:

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{149}=12.21[/tex]

The standard deviation of the first probability distribution is 1.10 and of the second distribution is 15.23.

The standard deviation σ of a probability distribution is calculated by first finding its mean μ, then using the formula:

σ = √[Σ(x-μ)^2 * P(X = x)]

For the first distribution, the mean μ = (1*0.2) + (2*0.2) + (3*0.2) + (4*0.4) = 3.2. The standard deviation σ = √[(1-3.2)^2 * 0.2 + (2-3.2)^2 * 0.2 + (3-3.2)^2 * 0.2 + (4-3.2)^2 * 0.4] = 1.10.

For the second distribution, the mean μ = (-20*0.1) + (-10*0.2) + (0*0.4) + (10*0.1) + (20*0) + (30*0.2) = -2. The standard deviation σ = √[(-20+2)^2 * 0.1 + (-10+2)^2 * 0.2 + (0+2)^2 * 0.4 + (10+2)^2 * 0.1 + (20+2)^2 * 0 + (30 + 2)^2 * 0.2] = 15.23.

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Answer:

a) 7,858,539,612

b) 2,080,201,662

c) 346,700,277

d) 7,511,839,335

e) 410,040

Step-by-step explanation:

a. How many ways are there to extend the 6 offers to 6 of the 136 candidates?

Combinations of 136 (candidates) taken 6 (offers) at a time without repetition:

[tex]\large \binom{136}{6}=\frac{136!}{6!(136-6)!}=\frac{136!}{6!130!}=7,858,539,612[/tex]

b. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer, but we do not know which?

There are 6 ways Computer Joe can get an offer. Now there are left 5 offers and 135 candidates. So there are  

6 times combinations of 135 taken 5 at a time without repetition:

[tex]\large 6*\binom{135}{5}=6*\frac{135!}{5!(135-5)!}=6*\frac{135!}{5!130!}=2,080,201,662[/tex]

c. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer for job number 2?

Now, we only have 5 offers and 135 candidates. So there are combinations of 135 taken 5 at a time without repetition:

[tex]\large \binom{135}{5}=\frac{135!}{5!(135-5)!}=\frac{135!}{5!130!}=346,700,277[/tex]

d. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is not getting any offers?

Here we have 6 offers and 135 candidates, given that  Computer Joe is out. So there are combinations of 135 taken 6 at a time without repetition:

[tex]\large \binom{135}{6}=\frac{135!}{6!(135-6)!}=\frac{135!}{6!129!}=7,511,839,335[/tex]

e. How many ways are there for 3 interviewers to select 3 resumes (one resume for each interviewer) from the pile of 136 resumes for the first interview round?

There are combinations of 136 taken 3 at a time without repetition:

[tex]\large \binom{136}{3}=\frac{136!}{3!(136-3)!}=\frac{136!}{3!133!}=410,040[/tex]

Answer: y = - 466.375

Step-by-step explanation:

y varies jointly as x and z.

This means that y varies directly as x and also varies directly as z.

In order to remove the proportionality symbol, we will introduce a constant of proportionality, k. Therefore,

y = kxz

The next step is to determine the value of k

if y = 205 when x = –5 and z = –8.

we will substitute these values into the equation to determine k.

205 = k × -5 × -8

205 = 40k

k = 205/40 = 5.125

Therefore, the equation becomes

y = 5.125xz

We want to determine y when x = - 13 and z = 7

y = 5.125 × - 13 × 7

y = 5.125 × - 91

y = - 466.375

Answer:

  8

Step-by-step explanation:

You can skip directly to the formula for the sum of an infinite sequence with first term a₁ and common ratio r:

  S = a₁/(1-r)

Your values of the variables in this formula are a₁ = 6 and r = 2/8. Putting these into the formula gives ...

  S = 6/(1 -2/8) = 6/(6/8) = 8

The sum of the infinite geometric sequence is 8.

_____

The above formula is the degenerate form of the formula for the sum of a finite sequence:

  S = a₁((rⁿ -1)/(r -1))

When the common ratio r has a magnitude less than 1, the term rⁿ tends to zero as n gets very large. When that term is zero, the sum of the infinite sequence is ...

  S = a₁(-1/(r-1)) = a₁/(1-r)

Answer:

34% of lightbulb replacement requests numbering between 39 and 50.

Step-by-step explanation:

The 68-95-99.7 rule states that, for a normally distributed random variable:

68% are within 1 standard deviation of the mean(34% between one standard deviation below and the mean, 34% between the mean and one standard deviation above the mean).

95% are within 2 standard deviations of the mean.

99.7% are within 3 standard deviations of the mean.

In this problem, we have that:

The distribution of the number of daily requests is bell-shaped and has a mean of 50 and a standard deviation of 11.

Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 50?

50 is the mean

39 is one standard deviation below the mean.

This means that 34% of lightbulb replacement requests numbering between 39 and 50.

Creating a 95% confidence interval involves capturing 95% of the probability deviation in a normal distribution. The formula for doing so is (x ± (standard deviation/√n)x critical value). By adjusting the sample size or confidence level accordingly, different intervals can be created.

The question inquires about the construction of a

confidence interval

for the true average stray-load loss for a certain type of induction motor. When constructing a 95% confidence interval, we are looking to capture the central 95 percent of the probability on a normal distribution, excluding 2.5 percent on each tail of the distribution.

For point one, with n = 25, x = 58.3, and o = 3.0, use the following formula to construct a 95% Confidence Interval: (x ± (standard deviation/√n)x critical value), resulting in (58.3 ± (3/√25)x 1.96). The same method applies to the rest of the points, only changing the sample size, n, or confidence level.

For the final query, the required sample size, n, can be found by rearranging the formula for confidence intervals and solving for n with the half-width of the interval set at 0.5.

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(1). 95% CI (n=25): Standard deviation of 3.0, sample mean of 58.3, yielding a 95% confidence interval of (57.124, 59.476).

(2). 95% CI (n=100): With the same standard deviation and mean, a larger sample size leads to a narrower 95% confidence interval of (57.712, 58.888).

(3). 99% CI (n=100): Expanding the confidence level to 99% increases the margin of error, resulting in a wider interval of (57.5272, 59.0728).

(4). Sample Size (99% CI, ME=0.5): To achieve a half-width of 0.5 for a 99% confidence interval, approximately 956 samples are required.

Given the problem requirements, we will calculate the confidence intervals based on the provided data:

Using these values, we’ll calculate the 95% and 99% confidence intervals (CIs) for the population mean, μ:

1. Constructing a 95% CI when n = 25

SE = σ/√n

= 3.0/√25

= 3.0/5

= 0.6

= 1.96 * 0.6

= 1.176

2. Constructing a 95% CI when n = 100

SE = σ/√n

= 3.0/√100

= 3.0/10

= 0.3

3. Constructing a 99% CI when n = 100

4. Determining the sample size n for a half-width of 0.5 in the 99% CI

Therefore, n ≈ 956 to achieve a half-width of 0.5 for a 99% confidence interval.

Answer: The upper confidence limit for the 90% confidence interval would be 0.415.

Step-by-step explanation:

Since we have given that

n = 400

x = 150

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{150}{400}=0.375[/tex]

At 90% confidence interval, z = 1.645

So, margin of error would be

[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.375\times 0.625}{400}}\\\\=0.0398[/tex]

So, the upper limit would be

[tex]\hat{p}+0.0398\\\\=0.375+0.0398\\\\=0.415[/tex]

Hence, the upper confidence limit for the 90% confidence interval would be 0.415.

Answer:

(a) 0.28347

(b) 0.36909

(c) 0.0039

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

[tex]P(x=r)=^nC_rp^rq^{n-r}[/tex]

(a) Exactly two will be drunken drivers.

[tex]P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}[/tex]

[tex]P(x=2)=66(0.2)^{2}(0.8)^{10}[/tex]

[tex]P(x=2)=\approx 0.28347[/tex]

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

[tex]P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)[/tex]

[tex]P(x=3\text{ or }x=4)=P(x=3)+P(x=4)[/tex]

Using binomial we get

[tex]P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}[/tex]

[tex]P(x=3\text{ or }x=4)=0.236223+0.132876[/tex]

[tex]P(x=3\text{ or }x=4)\approx 0.369099[/tex]

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

[tex]P(x\geq 7)=1-P(x<7)[/tex]

[tex]P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)][/tex]

[tex]P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155][/tex]

[tex]P(x\leq 7)=1-[0.9961][/tex]

[tex]P(x\leq 7)=0.0039[/tex]

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

[tex]P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)[/tex]

[tex]P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315[/tex]

[tex]P(x\leq 5)=0.9806[/tex]

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

a) Exactly two will be drivers: 0.2835. b) Three or four will be drivers: 1.5622; c) At least 7 will be drivers: 32.5669 (rounded to four decimal places). d) At most 5 will be drivers: 0.8749

a) Exactly two will be drivers:

For this case, we'll use the binomial probability formula:

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)

Where:

n is the number of trials (in this case, the number of drivers in the sample), which is 12.

k is the number of successful trials (in this case, the number of drivers), which is 2.

p is the probability of success in a single trial (in this case, the probability of a driver being), which is 0.20.

(n choose k) is the number of combinations of n items taken k at a time.

Calculating:

P(X = 2) = (12 choose 2) * (0.20)^2 * (0.80)^10

= 66 * 0.04 * 0.1073741824

= 0.2834678413

b) Three or four will be drivers:

For this case, we'll find P(X = 3) and P(X = 4) and then add them together.

For P(X = 3):

P(X = 3) = (12 choose 3) * (0.20)^3 * (0.80)^9

= 220 * 0.008 * 0.134456

= 0.23757696

For P(X = 4):

P(X = 4) = (12 choose 4) * (0.20)^4 * (0.80)^8

= 495 * 0.016 * 0.16777216

= 1.3245924272

Adding them together:

P(X = 3 or 4) = 0.23757696 + 1.3245924272

= 1.5621693872

c) At least 7 will be drivers:

To find this probability, we need to calculate P(X = 7) + P(X = 8) + ... + P(X = 12).

For P(X = 7):

P(X = 7) = (12 choose 7) * (0.20)^7 * (0.80)^5

= 792 * 0.128 * 0.32768

= 32.5669376

Similarly, find P(X = 8), P(X = 9), P(X = 10), P(X = 11), and P(X = 12) using the same method.

Finally, add all these probabilities together.

d) At most 5 will be drivers:

To find this probability, we need to calculate P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5).

For P(X = 0):

P(X = 0) = (12 choose 0) * (0.20)^0 * (0.80)^12

= 1 * 1 * 0.0687194767

= 0.0687194767

Similarly, find P(X = 1), P(X = 2), P(X = 3), P(X = 4), and P(X = 5) using the same method.

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Answer:

$ 725

Step-by-step explanation:

Price of call option = 7.25

buyer have to pay for one call option contract. Assume each contract is for 100 = 100 * 7.25 =  $ 725

To define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's, we can assign each element of the subset to a bit position in the string.

In mathematics, a bijection is a one-to-one correspondence between two sets, such that each element in one set is paired with a unique element in the other, and vice versa. It implies both injective (no duplicates) and surjective (every element has a match) properties.

To define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's, we can assign each element of the subset to a bit position in the string.  For example, the subset {1, 3, 4, 5, 8} corresponds to the 8-bit string 11011001, where the 1st, 3rd, 4th, 5th, and 8th bits are set to 1, representing the elements in the subset.

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Answer:

1.  20

2. 23

3. 6

Step-by-step explanation:

We have that:

f(x) = 2x

g(x) = x² + 1

f(g(x)) is the composite function of f and g. So

f(g(x)) = f(x²-1) = 2(x²+1) = 2x² + 2

1. f(g(3))

f(g(x)) = 2x² - 2 = 2(3)² + 2 = 18 + 2 = 20

2. f(3)+g(4)

f(3) = 2(3) = 6

g(4) = 4² + 1 = 17

f(3) + g(4) = 6 + 17 = 23

3. f(5) - 2g(1)​

f(5) = 2(5) = 10

g(1) = (1)² + 1 = 2

f(5) - 2g(1) = 10 - 2*2 = 10 - 4 = 6

Answer:

  If lines have the same slope, then they are parallel.

Step-by-step explanation:

The first part of the given statement sets up the condition for the second part to be true. A conditional statement makes that explicit.

  (lines described by first part) are (lines described by second part) ⇒

  if (lines are described by first part), then (lines are described by second part)

Your conditional statement can be written ...

  If lines have the same slope, then they are parallel.

Answer:

The correct option is C. either a t test or an analysis of variance can be performed.

Step-by-step explanation:

Consider the provided information.

Hence, Either a t test or an analysis of variance can be performed to determine whether the means of two population are equal.

Therefore, the correct option is C. either a t test or an analysis of variance can be performed.

To determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.

The t-test is a statistical test that is used to compare the means of two groups, to determine if there is any significant difference between the two groups or populations.

The ANOVA, like the t-test is also used to compare means, however, it is used when the groups or populations involved are more than two.

Therefore, to determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.

Learn more about t-test and ANOVA on:

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Answer: height of kite is 147.042 feets

Step-by-step explanation:

The diagram of the kite is shown in the attached photo

Triangle ABC is formed and it is a right angle triangle.

The kite string made an angle of 33 degrees with the ground. The string used was 90 yards We will convert the 90 yards to feets.

I yard = 3 feets

90 yards would become

90×3 = 270 feets

This 270 feets form the hypotenuse of the triangle.

To determine the height of the kite h, we will use trigonometric ratio

Sin# = opposite / hypotenuse

Where

# = 33 degrees

Hypotenuse = 270 feets

Opposite = h feets

Sin 33 = h/270

h = 270sin33

h = 270 × 0.5446 = 147.042 feets

Answer:

Step-by-step explanation:

The F statistic, calculated through one-way ANOVA, for this problem is 1.07, which aims to examine the differences in the averages of multiple groups. However, without details such as the significance level and degrees of freedom, this task cannot determine if there's a significant difference between the average gas mileages of the four car models.

The average gas mileage comparison across four car models represented by the car dealer is an example of a problem solved by the One-Way ANOVA statistical approach. This test aims to determine if there is a statistically significant difference between the means of multiple groups, in this case, the average mileage of four different car models.

To find the test statistic, we consider the between-group mean square (MS Between) and the within-group mean square (MS Within). In ANOVA, the F statistic is used which performs the test of two variances, and is calculated as the ratio of MS Between to MS Within. In this case, MS Between is represented by 'MS Rows' (47.5500) and MS Within by 'MS Error' (44.5167). So, the F statistic = MS Between / MS Within = 47.5500 / 44.5167 = 1.07 (rounded to two decimal places).

However, the value of the F statistic alone is not enough to conclude the test. The conclusion depends on the significance level, degrees of freedom, and the value from the F-distribution table. Without these details, we cannot conclude whether there's a significant difference in the average gas mileages of the four car models.

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Answer:

m= -0.1

b= -1.2

Step-by-step explanation:

The equation of the given line is

y=10x, slope of the given line is 10.

As the line is perpendicular to this line, the slope  is [tex]-\frac{1}{10}[/tex]

m= -0.1

( product of slope of two perpendicular lines is -1)

thus the equation is

[tex]y = (-\frac{1}{10})x + b[/tex]

[tex]y = (-0.1)x + b[/tex]

now to find the value of b, input the point coordinates (8, -2) in the above equation.

[tex]-2 = (-0.1)(8) + b[/tex]

[tex]b = -2 + (0.8)[/tex]

b = -1.2

Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Step-by-step explanation:

Since we have given that

n = 84

x = 11

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13[/tex]

p = 0.20

So, hypothesis:

[tex]H_0:p=\hat{p}\\\\H_a:\hat{p}<p[/tex]

so, test statistic value would be

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604[/tex]

At 1% level of significance, critical value would be

z= 2.58

Since 2.58>-1.604

So, We will accept the null hypothesis.

Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Answer:

a) Critical value = -1.285

b) We should reject null hypothesis that the mean equals 0.32

Step-by-step explanation:

Given that the  statistic of z equals negative 2.40 is obtained when testing the claim that less than 0.32

i.e. for hypotheses

[tex]H_0: \bar = 0.32\\H_a: \bar x <0.32\\[/tex]

(one tailed test at 10% significance level)

Z critical value for 90% one tailed = -1.285

Since our test statistic is less than -1.285 we reject null hypothesis

a) Critical value = -1.285

b) We should reject null hypothesis that the mean equals 0.32

Answer:

As the sample size increases, the variability decreases.

Step-by-step explanation:

Variability is the measure of actual entries from mean.  The less the deviations the less would be the variance.

For a sample  of size n, we have by central limit theorem the mean of sample follows a normal distribution for random samples of large size.

X bar will have std deviation as [tex]\frac{s}{\sqrt{n} }[/tex]

where s is the square root of variance of sample

Thus we find the variability denoted by std deviation is inversely proportion of square root of sample size.

Hence as sample size increases, std error decreases.

As the sample size increases, the variability decreases.

In statistics, as the sample size increases, the variability typically decreases because more data points allow a closer approximation of the true population mean. Therefore, larger sample sizes provide a narrower confidence interval, leading to less variability.

The variability of a distribution is a measure of the differences from the mean that occur in the data points. The sample size refers to the number of data points collected in your sample from a population. In statistics, as the sample size increases, the variability or scatter of your dataset normally decreases, because a larger number of data points give a more accurate representation of the population you are studying.

Variability is affected by sample size in the following way: Increasing the sample size leads to a decrease in the error bound and makes a narrower confidence interval. This is because more data points enable a closer estimation of the true population mean. Thus, as your sample size grows larger, the variability decreases, and your data forms a tighter grouping around the mean.

For example, if you are conducting a survey, and you take four different samples of 50 people each from the same population, you might see differing outcomes due to sample variability. However, if you were to increase your sample size to perhaps 500 people, the results are likely to have less sample variability.

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Answer:

Step-by-step explanation:

a) What is the size of each cache?

Direct mapped cache= 2^index * size of cache line= 2^10 * 1B lines = 1KB.

2-way set associative cache= 2^index * size of cache line * 2 ways=2^7 * 4 words *2ways= 128 4B lines * 2 ways = 1KB

Fully associative cache= number of cache lines* size of each line= 256 * 4B lines = 1KB

b) How much space does each cache need to store tags?

Direct mapped cache= 1024 * 6-bit tags = 6Kb

2-way set associative cache= 256 * 7-bit tags = 1792 bits

Fully associative cache= 256 * 14-bit tags = 3584 bits

c) Which   cache   design   has   the   most   conflict   misses?   Which   has   the   least?    

Direct mapped cache has likely the most conflict misses, because it is direct mapped. Fully associative cache has the least since it is fully associative so it can never have conflict misses.

d) The   following   information   is   given   to   you: hit   rate   for   the   3   caches   is   50%,   70%   and   90%  but   did   not   tell   you   which   hit   rate   corresponds   to   which   cache,   which   cache   would   you   guess  corresponded   to   which   hit   rate?   Why?    

Since the size of all three caches is same size and as we said in the previous answer that direct mapped cache has more conflict misses and fully associative has the least so direct mapped will have 50%, 2-way set associative 70%, and Fully associative will have 90% hit rate.

e) Assuming   the   miss   time   for   each   is   20   cycles,   what   is   the   average   service   time   for   each? (Service   Time   =   (hit   rate)*(hit   time)   +   (miss   rate)*(miss   time)

We are given hit rates and miss rates. Also miss time=2o cycles for each cache and hit time= 1, 2, 5 for direct mapped, 2-way set associative and fully associative cache respectively.

Direct mapped= 0.5*1 + 0.5*20 = 10.5 cycles

2-way set associative= 0.7*2 + 0.3*20 = 7.4 cycles

Fully associative cache= 0.9*5 + 0.1*20 = 6.5 cycles.

The size of each cache, the space needed to store tags, the cache design with the most and least conflict misses, guessing which cache corresponds to each hit rate, and calculating the average service time for each cache.

Each cache design has different levels of conflict misses. The direct-mapped cache (D1) has the most conflict misses because multiple memory locations map to the same cache line. The 2-way set associative cache (D2) has fewer conflict misses because each set can hold two cache lines, reducing the chance of multiple memory locations mapping to the same set. The fully associative cache (D3) has the least conflict misses because any memory location can be stored in any cache line, reducing conflicts.

Answer:

[tex]z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]  

[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]  

Step-by-step explanation:

1) Data given and notation n  

n=1004 represent the random sample taken

X=482 represent the Chevrolet owners who said they would buy another Chevrolet.

[tex]\hat p=\frac{482}{1004}=0.480[/tex] estimated proportion of Chevrolet owners who said they would buy another Chevrolet.

[tex]p_o=0.47[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level  

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of Chevrolet owners who said they would buy another Chevrolet is higher than 47%:  

Null hypothesis:[tex]p\leq 0.47[/tex]  

Alternative hypothesis:[tex]p > 0.47[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]  

4) Statistical decision  

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

If we use the significance level provided, [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a one side upper test the p value would be:  

[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]  

So based on the p value obtained and using the significance level assumed [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of Chevrolet owners who said they would buy another Chevroletis not significantly higher than 0.47 or 47% .