Trucks in a delivery fleet travel a mean of 120 miles per day with a standard deviation of 22 miles per day. The mileage per day is distributed normally. Find the probability that a truck drives between 100 and 157 miles in a day. Round your answer to four decimal places.

Answer: Down payment amount = $330

Step-by-step explanation:

Given in the question that Shawn is interested in purchasing a new computer system and he wants to to give a 20%  down payment.

Cost of Computer system = $1650

He would like to made a 20% down payment

So, the down payment amount is as follows:

20% of $1650 = [tex]\frac{20}{100}[/tex] × 1650

                        = $ 330 ⇒ Down payment amount

Answer:

The answers is 228 pennies.

Step-by-step explanation:

The values of the subunits of a dollar are:

1 penny = $0.01.

1 nickel = $0.05.

1 quarter = $0.25.

If all the pennies and nickels that you have are equal to 32 quarters:

Now, we have twice pennies as many nickels, and x is the total amount of coins:

Multiply and add the terms of the equation:

Divide by $0.07 in both terms of the equation:

When x = 114 means that you have 114 nickels and 228 pennies.

Replacing x=114

Answer:

[tex]y^2=2\ln (1+t^2)+4[/tex]

Step-by-step explanation:

Given that

[tex]\dfrac{dy}{dt}=\dfrac{2t}{y+yt^2}[/tex]

This is a differential equation.

Now by separating variables

[tex]y dy= \dfrac{2t}{1+t^2}dt[/tex]

Now by integrating both side

[tex]\int y dy=\int \frac{2t}{1+t^2}dt[/tex]

Now by soling above integration

We know that  integration of dx/x is lnx.

[tex]\dfrac{y^2}{2}=\ln (1+t^2)+C[/tex]

Where C is the constant.

[tex]y^2=2\ln (1+t^2)+C[/tex]

Given that when t=0 then y= -2

So by putting the above values of t and y we will find C

4=2 ln(1)+C     (we know that ln(1)=0)

So C=4

⇒[tex]y^2=2\ln (1+t^2)+4[/tex]

So solution of above equation is  [tex]y^2=2\ln (1+t^2)+4[/tex]

Answer:

a. 19x-13   b. [tex]2(32x^{15}y^{3}-1).[/tex]

Step-by-step explanation:

a. 3(-3+5x)-1(4-4x) = -9+15x-4+4x = 15x+4x-9-4 = 19x-13.

b. [tex]64x^{15}y^{3}-2(-15e^{5}\frac{t}{30}e^{-2}t^{-3})^{0}[/tex]

= [tex]64x^{15}y^{3}-2 = 2(32x^{15}y^{3}-1).[/tex]

a.  The power series representation for [tex]\( f(x) = \frac{1}{(10 + x)^2} \)[/tex] is:

[tex]f(x) = \sum_{n=0}^{\infty} (n + 1) \left(-\frac{1}{10}\right)^{n+1} x^n[/tex]

with a radius of convergence of 10.

b. The power series representation for [tex]\( f(x) = \frac{1}{(10 + x)^3} \)[/tex] is:

[tex]f(x) = \sum_{m=0}^{\infty} (m + 2)(m + 1) \left(-\frac{1}{10}\right)^{m+2} x^m[/tex]

with a radius of convergence of 10.

Question a:

To find a power series representation for the function [tex]\( f(x) = \frac{1}{(10 + x)^2} \)[/tex].

The sum of an infinite geometric series is given by:

[tex]\frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n[/tex]

where [tex]\( |r| < 1 \)[/tex] for convergence.

First, let's consider the function [tex]\( g(x) = \frac{1}{10 + x} \)[/tex]. Its power series can be found by rewriting it in a form similar to the geometric series:

The geometric series with [tex]\( r = -\frac{x}{10} \)[/tex]. Thus, its power series is:

[tex]g(x) = \frac{1}{10} \sum_{n=0}^{\infty} \left(-\frac{x}{10}\right)^n[/tex]

To find the power series for [tex]\( f(x) = \frac{1}{(10 + x)^2} \)[/tex], we can differentiate [tex]\( g(x) \)[/tex] term by term, as the derivative of [tex]\( g(x) \) is \( f(x) \)[/tex]. The derivative of [tex]\( g(x) \)[/tex] is:

[tex]g'(x) = \frac{1}{10} \sum_{n=0}^{\infty} n \left(-\frac{1}{10}\right)^n x^{n-1}[/tex]

Since [tex]\( g'(x) = f(x) \)[/tex], we have:

[tex]f(x) = \frac{1}{10} \sum_{n=0}^{\infty} n \left(-\frac{1}{10}\right)^n x^{n-1}[/tex]

Adjust the index and powers to start the series from [tex]\( n = 0 \)[/tex]. Let's change the index by setting [tex]\( m = n - 1 \)[/tex], so [tex]\( n = m + 1 \)[/tex].

Since the series actually starts from [tex]\( m = 0 \) (equivalent to \( n = 1 \))[/tex], we can rewrite it as:

[tex]$$f(x) = \sum_{m=0}^{\infty} (m + 1) \left(-\frac{1}{10}\right)^{m+1} x^m$$[/tex]

For the radius of convergence, [tex]\( R \)[/tex], we can use the ratio test. The ratio test states that for a series [tex]\( \sum a_n \)[/tex], if [tex]\( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \)[/tex], then the series converges if [tex]\( L < 1 \)[/tex]. The terms of our series are [tex]\( a_n = (n + 1) \left(-\frac{1}{10}\right)^{n+1} x^n \)[/tex].

The terms of our series are [tex]\( a_n = (n + 1) \left(-\frac{1}{10}\right)^{n+1} x^n \)[/tex]. Applying the ratio test:

[tex]\lim_{n \to \infty} \left| \frac{(n + 2)}{(n + 1)} \cdot \left(-\frac{1}{10}\right) \cdot x \right|[/tex]

As [tex]\( n \)[/tex] approaches infinity, the term [tex]\( \frac{(n + 2)}{(n + 1)} \)[/tex] approaches 1, so the limit simplifies to:

[tex]\lim_{n \to \infty} \left| -\frac{x}{10} \right| = \frac{|x|}{10}[/tex]

For the series to converge, this limit must be less than 1:

[tex]\frac{|x|}{10} < 1[/tex]

[tex]|x| < 10[/tex]

Thus, the radius of convergence, [tex]\( R \)[/tex], is 10.

Power series representation for [tex]\( f(x) = \frac{1}{(10 + x)^2} \)[/tex] is:

[tex]f(x) = \sum_{n=0}^{\infty} (n + 1) \left(-\frac{1}{10}\right)^{n+1} x^n[/tex]

Question b:

To find a power series representation for [tex]\( f(x) = \frac{1}{(10 + x)^3} \)[/tex], we can use the result from part (a), where we found a power series for [tex]\( \frac{1}{(10 + x)^2} \)[/tex], and differentiate it once more.

From part (a), we have:

[tex]\frac{1}{(10 + x)^2} = \sum_{n=0}^{\infty} (n + 1) \left(-\frac{1}{10}\right)^{n+1} x^n[/tex]

To find the power series for [tex]\( f(x) = \frac{1}{(10 + x)^3} \)[/tex], we differentiate the series for [tex]\( \frac{1}{(10 + x)^2} \)[/tex] term by term.

The derivative of [tex]\( (n + 1) \left(-\frac{1}{10}\right)^{n+1} x^n \)[/tex] with respect to [tex]\( x \)[/tex] is:

[tex](n + 1) n \left(-\frac{1}{10}\right)^{n+1} x^{n-1}[/tex]

power series for [tex]\( f(x) \)[/tex] is:

[tex]f(x) = \sum_{n=0}^{\infty} (n + 1) n \left(-\frac{1}{10}\right)^{n+1} x^{n-1}[/tex]

Change the index by setting [tex]\( m = n - 1 \)[/tex], so [tex]\( n = m + 1 \)[/tex]. Then, our series becomes:

[tex]f(x) = \sum_{m=-1}^{\infty} (m + 2)(m + 1) \left(-\frac{1}{10}\right)^{m+2} x^m[/tex]

Since the series actually starts from [tex]\( m = 0 \)[/tex] (equivalent to [tex]\( n = 1 \))[/tex], we can rewrite it as:

[tex]f(x) = \sum_{m=0}^{\infty} (m + 2)(m + 1) \left(-\frac{1}{10}\right)^{m+2} x^m[/tex]

For the radius of convergence, [tex]\( R \)[/tex], we can use the same approach as in part (a).

Applying the ratio test:

[tex]\lim_{m \to \infty} \left| \frac{(m + 3)}{(m + 1)} \cdot \left(-\frac{1}{10}\right) \cdot x \right|[/tex]

As [tex]\( m \)[/tex] approaches infinity, the term [tex]\( \frac{(m + 3)}{(m + 1)} \)[/tex] approaches 1, so the limit simplifies to:

[tex]\lim_{m \to \infty} \left| -\frac{x}{10} \right| = \frac{|x|}{10}[/tex]

For the series to converge, this limit must be less than 1:

[tex]\frac{|x|}{10} < 1[/tex]

[tex]|x| < 10[/tex]

Power series representation for [tex]\( f(x) = \frac{1}{(10 + x)^3} \)[/tex] is:

[tex]f(x) = \sum_{m=0}^{\infty} (m + 2)(m + 1) \left(-\frac{1}{10}\right)^{m+2} x^m[/tex]

Answer:

6.68 %.

Step-by-step explanation:

The standardised z-score = ( 10.95 - 10.5) /  0.3

= 1.5.

Looking up the Normal Distribution tables ( area to the left) 1.5 corresponds to  0.93319 so for   a  weight above 10.95 the proportion is

1 - 0.93319 = 0.06681

= 6.68%.

We can use the z-score to find the proportion of jars that are above a certain weight in a normal distribution. The z-score for 10.95 ounces is 1.5. Using a standard normal distribution table, we find that about 6.68% of the jars weigh more than 10.95 ounces.

In this problem, we are using the concept of normal distribution, specifically to find the proportion of jars that are above a certain weight. Given that the mean (average) weight of the jars is 10.5 ounces and the standard deviation (which measures the dispersion of the weights) is 0.3 ounce, we can calculate the z-score for 10.95 ounces.

The z-score is defined as the number of standard deviations a data point is from the mean. Compute it using the formula: Z = (X - μ) / σ where X is the value, μ is the mean and σ is the standard deviation.

Plugging into the formula we get: Z = (10.95 - 10.5) / 0.3 = 1.5

You can then look up this z-score in a standard normal distribution table (or use a calculator or computer software that calculates it), to find the proportion below this z-score. But we need the proportion above, so we subtract this from 1. Let's say the value from a z-table for 1.5 is 0.9332, the proportion of values above this would be 1 - 0.9332 = 0.0668 or about 6.68% of the jars weigh more than 10.95 ounces.

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Answer:

The no. of possible handshakes takes place are 45.

Step-by-step explanation:

Given : There are 10 people in the party .

To Find: Assuming all 10 people at the party each shake hands with every other person (but not themselves, obviously) exactly once, how many handshakes take place?

Solution:

We are given that there are 10 people in the party

No. of people involved in one handshake = 2

To find the no. of possible handshakes between 10 people we will use combination over here

Formula : [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

n = 10

r= 2

Substitute the values in the formula

[tex]^{10}C_{2}=\frac{10!}{2!(10-2)!}[/tex]

[tex]^{10}C_{2}=\frac{10!}{2!(8)!}[/tex]

[tex]^{10}C_{2}=\frac{10 \times 9 \times 8!}{2!(8)!}[/tex]

[tex]^{10}C_{2}=\frac{10 \times 9 }{2 \times 1}[/tex]

[tex]^{10}C_{2}=45[/tex]

No. of possible handshakes are 45

Hence The no. of possible handshakes takes place are 45.

Final answer:

The EPQ model differs from the EOQ model in terms of annual cost, maximum inventory level, inventory depletion rate, and formulas for computing holding costs.

Explanation:

In the EPQ (Economic Production Quantity) model, the costs and demands are assumed to be the same. Compared to the EOQ (Economic Order Quantity) model, there are a few differences:

Answer:

The first option is the correct answer.

Step-by-step explanation:

A PTIN is required to prepare or sign most tax returns.

The PTIN or full form Preparer Tax Identification Number (PTIN) is an identification number, used by the pre parers to claim for refund or compensation during tax return filing.

So, a person who has to claim refund must have his or her own PTIN and each tax return pre parer may only obtain one PTIN.

Answer with explanation:

Equation of Plane having Direction cosines A, B and C passing through points, p, q and r is

⇒A (x-p)+B(y-q)+C(z-r)=0

The plane passes through the points A(-2, 3,-1), B(0, 5, 2) and C(-1, -2, 1).

→A(x+2)+B(y-3)+C(z+1)=0----------(1)

→A(0+2)+B(5-3)+C(2+1)=0

2 A +2 B+3 C=0

→A(-1+2)+B(-2-3)+C(1+1)=0

A -5 B+2 C=0

[tex]\Rightarrow \frac{A}{4+15}=\frac{B}{3-4}=\frac{C}{-10-2}\\\\\Rightarrow \frac{A}{19}=\frac{B}{-1}=\frac{C}{-12}=k\\\\A=19 K,B=-K, C=-12K[/tex]

Substituting the value of A , B and C in equation (1)

⇒19 K(x+2)-K(y-3)-12 K(z+1)=0

⇒19 x +38 -y +3-12 z-12=0

⇒19 x -y -12 z +29=0, is the required equation of Plane in Cartesian form.

⇒(19 i -j -12 k)(xi +y j+z k)+29=0 ,is the required  vector equation of the plane.

Answer:

[tex] \frac{1}{ {8}^{3} } [/tex]

Answer: There are 85 Helena's client own stocks, bonds and mutual funds.

Step-by-step explanation:

Since we have given that

Let A: who own stocks

B : who own bonds

C : who own mutual fund

So, According to question,

n(A) = 288

n(B) = 200

n(C) = 184

n(A∩B) = 123

n(B∩C) = 106

n( A∩C) = 102

n(A∪B∪C) = 426

As we know the formula :

[tex]n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\426=288+200+184-123-106-102+n(A\cap B\cap C)\\\\426-341=n(A\cap B\cap C)\\\\85=n(A\cap B\cap C)[/tex]

Hence, there are 85 Helena's client own stocks, bonds and mutual funds.

To determine the number of clients who own stocks, bonds, and mutual funds, we can use the principle of inclusion-exclusion. This principle allows us to properly account for overlap in the sets of clients for different investment types.
We were told the following:
- The total number of clients is 426.
- The number of clients who own stocks is 288.
- The number of clients who own bonds is 200.
- The number of clients who own mutual funds is 184.
- The number of clients who own both stocks and bonds is 123.
- The number of clients who own both stocks and mutual funds is 106.
- The number of clients who own both bonds and mutual funds is 102.
Now, when we sum up the number of clients who own stocks, bonds, and mutual funds individually, we're double-counting those clients who have investments in more than one of these. We need to subtract the clients who are counted twice.
So, let's add up all the individuals:
S + B + M = 288 + 200 + 184 = 672
Now, let's subtract the number of clients who were double-counted:
(S + B + M) - (SB + SM + BM) = 672 - (123 + 106 + 102) = 672 - 331 = 341
However, in this calculation, we've subtracted clients who own all three: stocks, bonds, and mutual funds, three times (once for each pair), and then added them back in only once, so we've subtracted them two times too many.
Therefore, we need to correct for this: to find the number of people who own all three, we add the total number of clients (since everyone owns at least one of the three) and then subtract the sum we have just calculated.
Total + All three (overcorrected) = Total clients
426 + All three (overcorrected) = 341
To solve for the overcorrection (the actual number of clients who own all three types), we can now rearrange the equation:
All three (overcorrected) = 341 - 426
All three (overcorrected) = -85
However, since the number of people cannot be negative, this outcome indicates a logical inconsistency. Such an inconsistency generally means there must have been a mistake in either the data provided or the calculations based on that data. Under normal circumstances, you would go back and verify the numbers. But given this answer, it would suggest that the data provided has some inconsistencies, and it is not possible for a negative number of clients to own all three funds.

Final answer:

The area under the standard normal distribution curve between z = -2 and z = 2 corresponds to approximately 84.4 %.

Explanation:

To find the indicated area under the curve of the standard normal distribution between z = -2 and z = 2, we refer to a z-table that provides us with the area under the curve to the left of a given z-score.

First, we find the area under the normal curve to the left of z = 2, which typically is around 0.8672.

Since the normal distribution is symmetric about the mean, the area to the left of z = -2 is the same as the area to the right of z = 2, which is 1 - 0.8672 = 0.0228.

The total area between z = -2 and z = 2 is the area to the left of z = 2 minus the area to the left of z = -2, or 0.8672 - 0.0228.

The difference gives us approximately 0.8444, which represents the probability that a value falls within 2 standard deviations of the mean in a standard normal distribution.

Converting this to a percentage, we multiply by 100 to find that about 84.4 % of the area is within 2 standard deviations of the mean.

Answer:

A. Nonnegative

Step-by-step explanation:

By definition, the absolute value of any number must be positive (i.e non-negative).Hence A is the answer.

Answer:

Step-by-step explanation:

Angles are measured from the direction of motion, so the "force made good" is the force in the rope multiplied by the cosine of the angle. If the forces in the ropes (in Newtons) are represented by x and y, then we have ...

  x·cos(40°) +y·cos(55°) = 700

In order for the resultant to be in the direction of motion, the forces perpendicular to the direction of motion must cancel.

  x·sin(40°) - y·sin(55°) = 0

Here, we have assumed that the positive direction for measuring 40° is the negative direction for measuring 55°. That is, the angles are measured in opposite directions from the direction of motion.

Any of the usual methods for solving systems of linear equations can be used to solve this set. My preference is to use a graphing calculator. It gives the answers summarized above.

Answer:

D

Step-by-step explanation:

[tex]x+x+x+x=4 \cdot x \text{ or } 4x[/tex]

So [tex]2^{30}+2^{30}+2^{30}+2^{30}=4 \cdot 2^{30}[/tex].

Now 4 can be rewritten so that it is 2 to some power.

4 is actually 2 to the second.

That is, [tex]4=2^2[/tex].

So [tex]2^{30}+2^{30}+2^{30}+2^{30}=2^2 \cdot 2^{30}[/tex].

Now there is a law of exponents that says if the bases are the same and your multiplying add the exponents.

[tex]2^{30}+2^{30}+2^{30}+2^{30}=2^2 \cdot 2^{30}[/tex].

[tex]2^{30}+2^{30}+2^{30}+2^{30}=2^{32}[/tex].

Answer:

The answer is D! Hope this helped.

Step-by-step explanation:

Answer: 0.9890

Step-by-step explanation:

Given : Mean : [tex]\mu=50[/tex]

Standard deviation : [tex]\sigma =7[/tex]

We assume the random variable X is normally distributed

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x=34.

[tex]z=\dfrac{34-50}{7}=-2.2857142\approx-2.29[/tex]

The p-value =[tex]P(z>-2.29)=1-P(z<-2.29)[/tex]

[tex]=1-0.0110107=0.9889893\approx0.9890[/tex]

Hence, [tex]P(X>34)=0.9890[/tex]

Answer:

Explanation contains the proof.

Step-by-step explanation:

[tex]a \equiv 5 (mod 8) \text{ means there is integer } k \text{ such that } a-5=8k[/tex].

[tex]b \eqiv 3 (mod 8) \text{ means there is integer } m \text{ such that } b-3=8m[/tex].

We want to show that [tex]8 \text{ divides } ab+1[/tex].  So we are asked to show that there exist integer [tex]n \text{ such that } 8n=ab+1 \text{ or 8n-1=ab[/tex]

So what is [tex]ab[/tex]?

[tex]a-5=8k \text{ gives us } a=8k+5[/tex].

[tex]b-5=8m \text{ gives us } b=8m+5[/tex].

So back to [tex]ab[/tex]....

[tex]ab[/tex]

[tex]=(8k+5)(8m+5)[/tex]

[tex]=64km+40k+40m+25[/tex]  (I use foil to get this)

Factoring out 8 gives us:

[tex]=8(8km+5k+5m)+25[/tex]

Now I could have factored some 8's out of 25.  There are actually three 8's in 25 with a remainder of 1.

[tex]=8(8km+5k+5m+3)+1[/tex]

We have shown that there is integer [tex]n \text{ such that } ab=8n-1[/tex].

The integer I found that is n is 8km+5k+5m+3.

Therefore [tex]8|(ab+1)[/tex].

//

Answer:

See below.

Step-by-step explanation:

If a = 5 mod 8  and b = 3 mod 8

then ab = 5*3 mod 8 = 15 mod 8 = 7 mod 8.

ab + 1 =  8 mod 8 =  0 mod 8  so it is divisible by 8.

[tex]\[ \int_{0}^{5} \frac{1}{5} x^5 \, dx \approx 520.8333 \][/tex]

Answer:

0.85 M + 22.55

Step-by-step explanation:

We know that the total cost is the standard cost plus the insurance cost

C(M) = S(M) + I(M)

        = 17.75 + .60M + 4.80+.25M

Combine like terms

        = 0.85 M + 22.55

For this case we have that the standard charge, in dollars, of a company that rents vehicles is given by:

[tex]S = 17.75 + 0.60M[/tex]

M: Number of miles traveled.

On the other hand, the insurance charge is given by:

[tex]I = 4.80 + 0.25M[/tex]

If we want to find the total cost of renting the vehicle, we must add both equations:

[tex]C = 17.75 + 0.60M + 4.80 + 0.25M[/tex]

We add similar terms:

[tex]C = 17.75 + 4.80 + 0.60M + 0.25M\\C = 22.55 + 0.85M[/tex]

Answer:

[tex]C = 22.55 + 0.85M[/tex]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : [tex]n=45[/tex]

[tex]\overline{x}=148.79\text{ lb}[/tex]

[tex]\sigma=31.37\text{ lb}[/tex]

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = [tex]\mu=148.79\text{ lb}[/tex]

b) The confidence interval for the population mean is given by :-

[tex]\mu\ \pm E[/tex], where E is the margin of error.

Formula for Margin of error :-

[tex]z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}[/tex]

Given : Significance level : [tex]\alpha=1-0.90=0.1[/tex]

Critical value : [tex]z_{\alpha/2}=z_{0.05}=\pm1.645[/tex]

Margin of error : [tex]E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69[/tex]

Now, the 90% confidence interval for the population mean will be :-

[tex]148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)[/tex]

Hence, the 90​% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

Answer: There is a probability of no correct answers is 0.4096.

Step-by-step explanation:

Since we have given that

Number of trials = 4

Probability of success i.e. getting correct answer = 0.20

We need to find the probability of no correct answers.

We would use "Binomial distribution".

Let X be the number of correct answers.

So, it becomes,

[tex]P(X=0)=(1-0.20)^4=(0.80)^4=0.4096[/tex]

Hence, there is a probability of no correct answers is 0.4096.

The probability of guessing all answers incorrectly in a multiple-choice test with 4 questions, each with 5 options, is approximately 0.41 or 41% when answers are randomly guessed, according to the binomial probability distribution.

The question you're asking pertains to the concept of binomial probability, which is a type of probability that applies when there are exactly two mutually exclusive outcomes of a trial, often referred to as 'success' and 'failure'. In this case, a 'success' refers to correctly guessing an answer, which has a probability of p = 0.20. Conversely, a 'failure' refers to incorrectly guessing an answer, and this has a probability of q = 1 - p = 0.80.

To find the probability of no correct answers from 4 trials, we employ the formula for binomial probability: P(x) = (n C x)×(p×x)*(q×(n-x)). Here, 'n' represents the number of trials (4), 'x' represents the number of successes (0 for our case), and 'n C x' denotes the number of combinations of n items taken x at a time.

By plugging in the relevant values, the binomial probability distribution gives us P(0)= (4 C 0)×(0.20×0)×(0.80×4) = 1 × 1 × 0.4096 = 0.4096. So, the probability of guessing all answers incorrectly is approximately 0.41 or 41% when answers are randomly guessed.

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Answer:

The product of the y-coordinates of the solutions is equal to zero

Step-by-step explanation:

we have

[tex]x^{2}+y^{2}=9[/tex] -----> equation A

[tex]x+y=3[/tex] ------> equation B

Solve by graphing

Remember that the solutions of the system of equations are the intersection point both graphs

using a graphing tool

The solutions are the points (0,3) and (3,0)

see the attached figure

The y-coordinates of the solutions are 3 and 0

therefore

The product of the y-coordinates of the solutions is equal to

(3)(0)=0

The steps to solve the system of equations involve isolating x in one equation and substituting into the other. Solving yields two solutions for y, y = 0 and y = 3. Their product is 0.

The system of equations given are [tex]x^2+y^2=9[/tex] and x+y=3. From the second equation, we can isolate x as x = 3 - y and substitute into the first equation, yielding: [tex](3 - y)^2 + y^2 = 9[/tex]. This simplifies to [tex]2y^2 - 6y + 9 = 9,[/tex]and then to [tex]2y^2 - 6y = 0[/tex]. If we factor y from this equation, we get y(2y - 6) = 0, giving two possible solutions for y: y = 0, and y = 3. As asked, the product of these y-coordinates is 0 * 3 = 0.

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Answer:

Step-by-step explanation:

The only possibilities where there is exactly 1 tail are:

those are 5 favorable outcomes.

where h represent heads and t represent tails. There are [tex]2^5 32[/tex] total number of outcomes after tossing the coin 5 times. Because every time you toss the coin, you have 2 possibilities, and as you do it 5 times, those are [tex]2^5[/tex] options. We can conclude from this that

The probability that exactly 1 is a tail is [tex]5/32[/tex].

We already know the total number of outcomes; 32.  Now we need to find the number of favorable outcomes. In order to do that, we can divide our search in three cases: 1.-there are no tails, 2.-exactly 1 is a tail, 3.- exactly 2 are tails.

The first case is 1 when every coin is a head. The second case we already solved it, and there are 5. The third case is the interesting one, we can count the outcomes as we did in the previous questions, but that's only because there are not too many outcomes.  Instead we are going to use combinations:

We need to have 2 tails, the other coins are going to be heads. We made 5 tosses, then the possible combinations are [tex]C_{5,2} = \frac{5!}{3!2!} = \frac{120}{6*2} = 10[/tex]

Finally, we conclude that there are 1 + 5 + 10 favorable outcomes, and this implies that

The probability that at most 2 are tails is [tex]\frac{16}{32} = \frac{1}{2}[/tex].

In a five-coin toss, the probability of getting exactly one tail is 5/32 and the probability of getting at most two tails is 0.5. These probabilities are calculated considering all possible outcomes and arranging the heads and tails in distinct manners.

The question you've asked involves calculating the probabilities in coin flipping, a common concept in mathematics and particularly in statistics. This falls under the topic of probability theory.

When a fair coin is tossed 5 times, there are 2^5 or 32 equally likely outcomes. If we want exactly 1 tail, there are 5 ways this can happen (one for each position the tails can be in). Thus, the probability for this occurrence is 5/32.

To find out the probability of getting at most 2 tails, we need to calculate the probability for getting exactly 0, 1, or 2 tails. As we already know that the probability for 1 tail is 5/32 and for 0 tails is 1/32 (only 1 way to get this outcome, getting heads every time). The probability for exactly 2 tails can be found in the same manner as for 1 tail, now we have 2 tails and it can be arranged in 5C2 ways which is 10 ways. Therefore, the probability of 2 tails is 10/32. Hence, the probability of getting at most 2 tails is the sum of probabilities of 0,1 or 2 tails, which is (1 + 5 + 10 )/32 = 16/32 = 0.5.

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Answer:

3) 28.28.

Step-by-step explanation:

In order to find the answer we need to establish the EOQ equation which is:

[tex]EOQ=\sqrt{2*s*d/h}[/tex] where:

s=the cost of the setup

d=demand rate

h=cost of holding

Because demand is 200/month so d=200,

the ordering cost is $20/month  so s=20, and

the carrying cost in $10/month so h=10.

Using the equation we have:

[tex]EOQ=\sqrt{2*20*200/10}[/tex]

[tex]EOQ=\sqrt{800}[/tex]

[tex]EOQ=28.28[/tex]

So, answer to 'the order quantity for product' is 3) 28.28.

Answer:  

The cardinality of set (a) is 0,

the cardinality of set (b) is 1

and

the cardinality of set (c) is 3.

Step-by-step explanation:  We are given to find the cardinality of each of the following sets :

(a) { }.

(b) { { } }.

(c) {a, {a}, {a, {a}} }.

We know that

CARDINALITY of a set is the number of elements present in the set.

(a) The given set is A = { }.

The set A is an empty set, so it does not contain any element. Hence, the cardinality of the set A is 0.

(b) The given set is B = { { } }.

The set is B is singleton set, contains only one element (that is empty set). So, the cardinality of the set B is 1.

(c) The given set is C = {a, {a}, {a, {a}} }.

The set C has three elements, a, the set {a} and the set {a, {a}}. So,  the the cardinality of set C is 3.

Thus,

the cardinality of set (a) is 0,

the cardinality of set (b) is 1

and

the cardinality of set (c) is 3.

Answer:

The probability that BOTH of them have the secret decoder ring is [tex]\frac{55}{1431}[/tex].

Step-by-step explanation:

From the given information it is clear that the total number of boxes is 54.

Total number of boxes that have the secret decoder ring = 11

Total number of boxes that have a different gift inside = 43

Total number of ways to select 2 boxes from the boxes that have the secret decoder ring is

[tex]\text{Favorable outcomes}=^{11}C_2=\frac{11!}{2!(11-2)!}=\frac{11\times 10\times 9!}{2!9!}=55[/tex]

Total number of ways to select 2 boxes from the total number of boxes is

[tex]\text{Total outcomes}=^{54}C_2=\frac{52!}{2!(52-2)!}=\frac{52\times 51\times 50!}{2!50!}=1431[/tex]

The probability that BOTH of them have the secret decoder ring is

[tex]P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]

[tex]P=\frac{55}{1431}[/tex]

Therefore the probability that BOTH of them have the secret decoder ring is [tex]\frac{55}{1431}[/tex].

Probability = 5/131, which is the final answer.

To calculate the probability that both boxes contain the secret decoder ring, we need to consider that the boxes are selected one after the other without replacement. Since there are 11 boxes with secret decoder rings out of 54, the probability of picking a secret decoder ring on the first try is 11/54. If a secret decoder box is picked first, there will be 10 secret decoder rings left out of the remaining 53 boxes. So the probability of picking another secret decoder ring is 10/53.

The overall probability of both events happening is the product of the two probabilities because we need both events to occur. Therefore, we multiply the separate probabilities together:

Probability = (11/54)  imes (10/53)

When we calculate this, we get:

Probability = 110/2862

This fraction can be simplified to:

Probability = 5/131, which is the final answer.

Answer:

Step-by-step explanation:

For this problem you have to use combinations. from 10 choices you are choosing 3. This means you are doing 10 choose 3. If you don't know what choose is I can explain.

Any number x choose y is the same as (x factorial)/(y factorial)(x-y factorial).

In this case that is 10 factorial/3 factorial times 7 factorial. ten factorial is the same as 10*9*8*7 factorial. So in the original equation you can factor away the seven factorials to get 10*9*8/3*2*1 factoring again you get 10*3*4  which is 120.

There are 120 different ways the chef can pick 3 brands of hot sauce to mix into a gumbo.

To find the number of ways the chef can pick 3 brands of hot sauce out of 10, we can use the combination formula:

[tex]nCr = n! / (r! * (n-r)!)[/tex]

where n is the total number of items (brands of hot sauce), and r is the number of items to be chosen (3 in this case).

In this problem, n = 10 and r = 3:

10C3 = 10! / (3! * (10-3)!)

Calculating the factorials:

10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800

3! = 3 × 2 × 1 = 6

(10-3)! = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040

Now, substitute the values:

10C3 = 3,628,800 / (6 * 5,040)

10C3 = 3,628,800 / 30,240

10C3 = 120

So, there are 120 different ways the chef can pick 3 brands of hot sauce to mix into a gumbo.

To know more about combination formula here

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To calculate the z-score, subtract the mean from the score and divide by the standard deviation. The calculated z-scores for the given test scores are -1.64, -0.73, 0.18, 1.09, -1.64, -1.18, -0.73, 0.18, 1.91, 1.00, and -0.27.

The z-score is a standardized value that measures how many standard deviations a score is above or below the mean. To calculate the z-score, we use the formula z = (X - µ) / σ, where X is the score, µ is the mean, and σ is the standard deviation.