Please help! Correct answer only, please! Consider the matrix shown below: Find the determinant of the matrix C. Group of answer choices A. -14 B. 14 C. -22 D. The determinant cannot be found for a matrix with these dimensions.

Answer:

(a) The confidence coefficient is 2.779.

(b) The desired sampling error for this study is the margin of error, ±3.

(c) The sample size required to obtain the desired estimate of the true mean is 55.

Step-by-step explanation:

The information provided is:

[tex]\bar x=53.4\ \text{grams}\\s=8.6\ \text{grams}\\n=27\\\text{Confidence Level} = 99\%[/tex]

(a)

As the sample size is n = 27 < 30 and the population standard deviation is not provided a t-interval would be used to estimate the true mean.

So, the confidence coefficient for the 99% confidence interval for true mean is:

[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (27-1)}=t_{0.005, 26}=2.779[/tex]

*Use the t-table.

Thus, the confidence coefficient is 2.779.

(b)

The error caused due to incorrect sample selection, i.e. a sample that is not a true representative of the population is known as the sampling error.

In this case the sampling error is known as the margin of error of the confidence interval.

The margin of error is:

MOE = 3 grams

Thus, the desired sampling error for this study is the margin of error, ±3.

(c)

Assume that the ample size required to obtain the desired estimate of the true mean is quite large, such that the sampling distribution of sample mean follows a normal distribution according to the central limit theorem.

Then the margin of error for a z-interval for mean is:

[tex]\text{MOE}=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]

MOE = 3 grams

The critical value of z for 99% confidence level is:

[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58[/tex]

*Use a z-table.

Compute the sample size as follows:

[tex]\text{MOE}=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]

      [tex]n=[\frac{z_{\alpha/2}\times s}{MOE}]^{2}[/tex]

         [tex]=[\frac{2.58\times 8.6}{3}]^{2}\\\\=54.700816\\\\\approx 55[/tex]

Thus, the sample size required to obtain the desired estimate of the true mean is 55.

Answer: A

Step-by-step explanation: 5y=x+5 simplified to slope-intercept form is y = 1/5x + 1

The equation 5y = x + 5 is a linear equaton, and the graph A represents the equation 5y = x + 5

The equation is given as:

5y = x + 5

Make x = 0.

So, we have:

5y = 0 + 5

This gives

5y = 5

So, we have:

y = 1

The point is represented as (0,1)

Make y = 0.

So, we have:

5 * 0 = x + 5

This gives

0 = x + 5

So, we have:

x = -5

The point is represented as (-5,0)

This means that the graph of 5y = x + 5 passes through the points (0,1) and (-5,0)

From the list of options, only graph (A) passes through the points (0,1) and (-5,0)

Hence, the graph A represents the equation 5y = x + 5

Read more about linear equations at:

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Answer:

(a)20

(b)Elastic

(c)8

(d) Elastic

Step-by-step explanation:

Elasticity of demand(E) indicates the impact of a price change on a product's sales.

The general formula for an exponential demand curve is given as:

[tex]y=ae^{-bp}[/tex]

Given the demand curve formula

[tex]q=f\left(p\right)=200e^{-0.4p}[/tex]

The formula for Elasticity of demand, E

[tex]E = -\dfrac{p}{q}\dfrac{\text{d}q}{\text{d}p}[/tex]

(a)When Price,  p = $50

p=50

[tex]q=200e^{-0.4*50}=200e^{-20}[/tex]

[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]

Therefore:

[tex]E = -\dfrac{50}{200e^{-20}}*-80e^{-0.4*50}\\=\dfrac{1}{4e^{-20}}*80e^{-20}\\\\E=20[/tex]

(b)At p = $50, Since elasticity is greater than 1, the demand is elastic.

An elasticity value of 20 means that a 1% increase in price causes a 20% decrease in demand.

(c)At p=$20

p=20

[tex]q=200e^{-0.4*20}=200e^{-8}[/tex]

[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]

Therefore:

[tex]E = -\dfrac{20}{200e^{-8}}*-80e^{-0.4*20}\\=\dfrac{1}{10e^{-20}}*80e^{-20}\\\\E=8[/tex]

(d)At p = $20, the demand is elastic.

An elasticity value of 8 means that a 1% increase in price causes a 8% decrease in demand.

Answer:

There were 89 retail stores in 1997.

Step-by-step explanation:

You know:

y =  11*x + 12

where y is the total number of stores and x is the number of years after 1990.

Being this a linear function (Linear functions are those functions that have the form y = mx + b as it happens in this case), to know the total number of stores and, then you must know the value of x, that is, the number of years after 1990, and replace that value in the given function.

To find out how many retail stores there were in 1997, then since 1990 7 years have passed, calculated by subtracting: 1997 - 1990 = 7

So, 7 is the number of years after 1990, then x=7

Replacing in the function y=11*x + 12 you get:

y=11*7+12

Solving:

y=77+12

y=89

There were 89 retail stores in 1997.

Answer:

Rotation = 1.

Reflection = 5 or 4.

Step-by-step explanation:

A regular octagon is the octagon that has eight, 8 sides and all these eight sides are equal.

So, if we are going to consider rotation and reflection distinct in order to color one diagonal red and another diagonal blue so that the two colored diagonals cross (in the interior) we can do that by following the steps Below;

=> Make sure that the first diagonal with the red color is drawn first.

=> to another make fixing to another end.

=> Determine the possibilities.

The numbers of times can be gotten by using Burnside Lemma which gives Rotation = 1 And Reflection = 5 or 4.

Answer:

2.5

Step-by-step explanation:

Answer:

2.75

Step-by-step explanation:

2 x 4=8 + 3=11/4=2.75

Answer: is that the full question?

Step-by-step explanation:

Answer:

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

P(x⁻>96) =0.0359

Step-by-step explanation:

Explanation:-

Given sample size 'n' =10

mean of the Population = 90 words per minute

standard deviation of the Population =10 wpm

we will use formula

                           [tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]

Let X⁻  = 96

                           [tex]Z = \frac{96-90 }{\frac{10}{\sqrt{10} } }[/tex]

                          Z =  1.898

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

[tex]P(X^{-}>x^{-} ) = P(Z > z^{-} )[/tex]

                   = 1- P( Z ≤z⁻)

                   = 1- P(Z<1.898)

                   = 1-(0.5 +A(1.898)

                   = 0.5 - A(1.898)

                   = 0.5 -0.4641 (From Normal table)

                  = 0.0359

Final answer:-

The probability that a random sample of 10 second grade students from  

                = 0.0359

Answer:

e=1o ajjajajajajkananwnwnwhhshssh

Answer:

Step-by-step explanation:

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Answer:

explanation is below happy to help ya!

Step-by-step explanation:

Answer:

X=1 x=-9

Step-by-step explanation:

(x^2)+8x=9

(x^2)+8x-9=0

(x-1)(x+9)=0

x=1 x=-9

Answer:

0.126 = 12.6% probability of selling less than 5 properties in one week.

Step-by-step explanation:

For each property, there are only two possible outcomes. The chance of selling any one property is independent of selling another property. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A real estate agent has 17 properties that she shows.

This means that [tex]n = 17[/tex]

She feels that there is a 40% chance of selling any one property during a week.

This means that [tex]p = 0.4[/tex]

Compute the probability of selling less than 5 properties in one week.

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{17,0}.(0.4)^{0}.(0.6)^{17} = 0.0002[/tex]

[tex]P(X = 1) = C_{17,1}.(0.4)^{1}.(0.6)^{16} = 0.0019[/tex]

[tex]P(X = 2) = C_{17,2}.(0.4)^{2}.(0.6)^{15} = 0.0102[/tex]

[tex]P(X = 3) = C_{17,3}.(0.4)^{3}.(0.6)^{14} = 0.0341[/tex]

[tex]P(X = 4) = C_{17,4}.(0.4)^{4}.(0.6)^{13} = 0.0796[/tex]

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0019 + 0.0102 + 0.0341 + 0.0796 = 0.126[/tex]

0.126 = 12.6% probability of selling less than 5 properties in one week.

Answer:

88 mins

Step-by-step explanation:

Divide 40 by 5 to get 8. That is your mile per minute ratio. Then multiply 11 by 8 to get the time of 11 miles!

Answer:

a) The third quartile Q₃ = 56.45

b) The variance = 2633.31

Step-by-step explanation:

a) The coefficient of skewness formula is given as follows;

[tex]SK = \dfrac{Q_{3}+Q_{1}-2Q_{_{2}}}{Q_{3}-Q_{1}}[/tex]

Plugging in the values, we have;

[tex]-0.38 = \dfrac{Q_{3}+30.8-2 \times 48.5_{_{}}}{Q_{3}-30.8}[/tex]

Solving gives Q₃ = 56.45

b) To determine the variance, we use the skewness formula as follows;

[tex]SK_{p} = \dfrac{Mean-\left (3\times Median - 2\times Mean \right )}{\sigma } = \dfrac{3\times\left ( Mean - Median \right )}{\sigma }[/tex]

Plugging in the values, we get;

[tex]-0.38= \dfrac{42-\left (3\times 48.5- 2\times 42\right )}{\sigma } = \dfrac{-19.5}{\sigma}[/tex]

[tex]\therefore \sigma =\dfrac{-19.5}{-0.38} = 51.32[/tex]

The variance = σ² = 51.32² = 2633.31.

Answer:

The area is 120 cm²

the perimeter is 52 cm

Step-by-step explanation:

The rhombus area is given by:

A = D * d / 2

that is, the larger diagonal D by the smaller diagonal d, between two, we know that D = 24 cm and d = 10 cm, replacing:

A = 24 * 10/2

A = 120

The area is 120 cm²

To calculate the perimeter use the Pythagorean theorem

h² = a² + b²

Since if you look at the rhombus it is formed by four right triangle we will take 1 of them with the following measures 5 cm (10/2) and height 12 cm (24/2) and replace:

h² = 5² + 12²

h² = 169

h = 13

now, the perimeter would be the sum of all its sides, which in this case are equal and measure 13 cm, therefore:

p = 4 * 13

p = 52

which means that the perimeter is 52 cm

Answer:

x^2+2x+3 I sent the answer

Answer:

[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]

Step-by-step explanation:

Given:  3x1 - 6x2 + 9x3 = 0

x2 and x3 are free variables

We have:

3x1 = 6x2 - 9x3

divide all sides by 3, we have:

x1 = 2x2 - 3x3

Finding the general solution, we have:

[tex] \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{ccc}2x_2 - 3x_3\\x_2\\x_3\end{array}\right] [/tex]

[tex] = \left[\begin{array}{ccc}2x_2\\x_2\\0\end{array}\right] + \left[\begin{array}{ccc}-3x_3\\0\\x_3\end{array}\right][/tex]

[tex]= x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]

The general solution is

[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]

Answer:

18

Step-by-step explanation:

pls mark brainliest

idk if this is a joke

Answer:

h = 400 ft

The bird is 400 ft above the submarine

Step-by-step explanation:

Given;

The submarine is 300 feet below sea level

Vertical distance between submarine and sea level h1 = 300 feet

The bird is flying at 100 feet above sea level

Vertical distance between the sea level and the bird h2 = 100 feet

The total vertical distance between the bird and the submarine is the sum of the vertical distance between submarine and sea level and the vertical distance between the sea level and the bird.

h = h1 + h2

h = 300 ft + 100 ft

h = 400 ft

The bird is 400 ft above the submarine

Answer:

The bird is 400 ft above the submarine

Step-by-step explanation:

Answer:

(o+y) + 3y = 61

to solve divide by 5 because there is 4 younger sister age + 1 for the older sister. 61-1=60. we subtract because the sister is older by 1 year

60/5=12 so the younger sister is 12

(13+12) + 36 = 61

Consecutive means in order like 3,4,5 or 11,12,13

Hope this helps

Step-by-step explanation:

Answer:

There is 55% probability that I order both the sandwich and soup

Step-by-step explanation:

P(sandwich) = 0.8

P(Soup) = 0.7

P(neither sandwich nor soup) = 0.05

P(sandwich or soup) = 1 - P(neither sandwich nor soup)

P(sandwich or soup) = 1 - 0.05 = 0.95

P(Sandwich & Soup) = x

P(Sandwich only) = 0.8 - x

P(Soup only) = 0.7 - x

P(sandwich or soup) = P(Sandwich only) + P(Soup only) + P(Sandwich & Soup)

Note that P(neither sandwich nor soup) has already been used to get the P(sandwich or soup) and should not be included in the above formula. Don't make that mistake!

0.95 = 0.8 - x + 0.7 - x + x

0.95 = 1.50 - x

x = 1.50 - 0.95

x = 0.55

There is 55% probability that I order both the sandwich and soup

Answer:

The probability of tossing a tail and then rolling a number greater than 5 is 0.188

Step-by-step explanation:

Independent events:

If two events, A and B, are independent, we have that:

[tex]P(A \cap B) = P(A)*P(B)[/tex]

In this question:

The coin and the die are independent. So

Event A: Tossing a tail.

Event B: Rolling a number greater than 5.

Probability of tossing a tail:

Coin can be heads or tails(2 outcomes), so the probability of a tail is [tex]P(A) = \frac{1}{2} = 0.5[/tex]

Probability of rolling a number greater than 5:

8 numbers(1 through 8), 3 of which(6,7,8) are greater than 5. So the probability of rolling a number greater than 5 is [tex]P(B) = \frac{3}{8} = 0.375[/tex]

Probability of A and B:

[tex]P(A \cap B) = P(A)*P(B) = 0.5*0.375 = 0.188[/tex]

The probability of tossing a tail and then rolling a number greater than 5 is 0.188

Answer:

[tex]d = \dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]

Step-by-step explanation:

Given the equation of velocity w.r.to time 't':

[tex]v=9t-\dfrac{2}{9}+t^2 ...... (1)[/tex]

Formula for Displacement:

[tex]Displacement = \text{velocity} \times \text{time}[/tex]

So, if we find integral of velocity w.r.to time, we will get displacement.

[tex]\Rightarrow \text{Displacement}=\int {v} \, dt[/tex]

[tex]\Rightarrow \int {v} \, dt = \int ({9t-\dfrac{2}{9}+t^2}) \, dt \\\Rightarrow \int{9t} \, dt - \int{\dfrac{2}{9}} \, dt + \int{t^2} \, dt\\\Rightarrow s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3} + C ....... (1)[/tex]

Here, C is constant (because it is indefinite integral)

Formula for integration used:

[tex]1.\ \int({A+B}) \, dx = \int {A} \, dx + \int{B} \, dx \\2.\ \int({A-B}) \, dx = \int {A} \, dx - \int{B} \, dx \\3.\ \int{x^{n} } \, dx = \dfrac{x^{n+1}}{n+1}\\4.\ \int{C } \, dx = Cx\ \{\text{C is a constant}\}[/tex]

Now, it is given that s = 0, when t = 0.

Putting the values in equation (1):

[tex]0=\dfrac{9\times 0^{2} }{2} - \dfrac{2}{9}\times 0 + \dfrac{0^3}{3} + C\\\Rightarrow C = 0[/tex]

So, the equation for displacement becomes:

[tex]s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]

Answer:

The standard deviation of the load distribution is of 5102.041 pounds.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 20000[/tex]

Also, the probability that the load is between 10,000 and 30,000 pounds is 0.95.

10,000 pounds and 30,000 pounds are equidistant from the mean. Due to this, and the probability of 0.95 of having a value in this range, 10000 is the (100-95)/2 = 2.5th percentile and 30000 is the (100+95)/2 = 97.5th percentile. Applying one of them, we find the standard deviation.

30,000 is the 97.5th percentile:

This means that when X = 30000, Z has a pvalue of 0.975. So when X = 30000, Z = 1.96. Then

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.96 = \frac{30000 - 20000}{\sigma}[/tex]

[tex]1.96\sigma = 10000[/tex]

[tex]\sigma = \frac{10000}{1.96}[/tex]

[tex]\sigma = 5102.041[/tex]

The standard deviation of the load distribution is of 5102.041 pounds.

Answer:

2,000 salmon

Step-by-step explanation:

The point estimate of the fraction of salmon that corresponds to the tagged Salmon released by the rangers, based on the fishermen report, is:

[tex]p = \frac{12}{80}\\p=0.15[/tex]

This means that the 300 salmon released into the lake correspond to 15% of the total salmon population. The estimate for the salmon population is:

[tex]P= \frac{300}{0.15}\\P=2,000\ salmon[/tex]

The estimate is 2,000 salmon.

Answer:

The estimate of the salmon population will be 2,000

Step-by-step explanation:

Rangers tagged and released 300 salmon into a Maine lake.

The fishermen reported catching 80 salmon of which 12 had tags

Thus we have 12/80 as a fraction of the tagged and released 300

 12/80 = 0.15 which is about 15% of the population of tagged and released 300.

To then estimate the salmon population, we have 300/0.15 = 2,000

Answer:

The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.

Step-by-step explanation:

Suppose we have a function y = f(x). The zeros, which are the values of x for which y = 0, are also called the x-intercepts of the function.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

[tex]ax^{2} + bx + c, a\neq0[/tex].

This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:

[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]

[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]

[tex]\bigtriangleup = b^{2} - 4ac[/tex]

In this question:

I will write the function as a function of x, just exchanging a for x.

[tex]f(x) = x^{2} + 5x + 6[/tex]

[tex]\bigtriangleup = 5^{2} - 4*1*6 = 1[/tex]

[tex]x_{1} = \frac{-5 + \sqrt{1}}{2*1} = -2[/tex]

[tex]x_{2} = \frac{-5 - \sqrt{1}}{2*1} = -3[/tex]

The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.

Answer:

Consider the following system of linear equations: 2 + 3y + 2z = 5 - 2x + y - z= -2 2x + 3z = 11 Instructions: • Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. • Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. • At the end of the procedure, clearly state the solution to the system outside of a matrix. • If the solution is unique, express the solution in real numbers. • If there are infinitely many solutions, express the solution in parameter(s). . If there is no solution, say so, and explain why.

All of the following are possible ranks of a 4x3 matrix EXCEPT O 1 2 3 4

How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-r

Step-by-step explanation:

x + 3y +2z = 5

-2x + y - z = -2

2x + 3z = 11

Here,

[tex]A = \left[\begin{array}{ccc}1&3&2\\-2&1&-1\\2&0&3\end{array}\right][/tex]

[tex]B =\left[\begin{array}{ccc}5\\-2\\11\end{array}\right][/tex]

[tex]X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex]

i.e AX=B

We can write as augmented matrix

[tex]\left[\begin{array}{ccc|c}1&3&2&5\\-2&1&-1&-2\\2&0&3&11\end{array}\right][/tex]

[tex]\frac{R_2\rightarrow R_2+2R_1}{R_3\rightarrow R_3-2R_1} \left[\begin{array}{ccc|c}1&3&2&5\\0&7&3&8\\0&-6&-1&1\end{array}\right][/tex]

[tex]\frac{R_3\rightarrow R_3+\frac{6R_2}{7} }{R_1\rightarrow R_1-\frac{3R_2}{7} } \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&7&3&8\\0&0&11/7&55/7\end{array}\right][/tex]

[tex]\frac{R_2\rightarrow\frac{R_2}{7}}{R_3\rightarrow\frac{7}{11}R_3} \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&1&3/7&8/7\\0&0&11/7&55/7\end{array}\right][/tex]

[tex]\frac{R_1\rightarrow R_1 -\frac{5}{7}R_3}{R_2\rightarrow R_2 -\frac{3}{7}R_3} \left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&-1\\0&0&1&5\end{array}\right][/tex]

Since Rank (A|B) = Rank (A) = 3 = number of variables

Answer:

8cm

Step-by-step explanation:

If the pieces are 8cm and 10cm they will still add up to 18cm but the longer piece is 2cm longer so both conditions are satisfied.

Complete Question:

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

Answer:

a) P(exactly 5) = 0.209

b) P(at least six) = 0.183

c) P(less than four) = 0.358

Step-by-step explanation:

Sample size, n = 10

Proportion of adults that have very little confidence in newspapers, p = 41% p = 0.41

q = 1 - 0.41 = 0.59

This is a binomial distribution question:

[tex]P(X=r) = nCr p^{r} q^{n-r}[/tex]

a) P(exactly 5)

[tex]P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 252 * 0.01159 * 0.072\\P(X=5) = 0.209[/tex]

b) P(at least six)

[tex]P(X \geq 6) = P(6) + P(7) + P(8) + P(9) + P(10)[/tex]

[tex]P(X\geq6) = (10C6 * 0.41^6*0.59^4) + (10C7*0.41^7*0.59^3) + (10C8*0.41^8*0.59^2) + (10C9 *0.41^9*0.59^1) + (10C10 *0.41^{10})\\P(X\geq6) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001\\P(X\geq6) = 0.183[/tex]

c) P(less than four)

[tex]P(X < 4) = 1 - [x \geq 4][/tex]

[tex]P(X<4)= 1 - [P(4) + P(5) + P(x \geq 6)][/tex]

[tex]P(X <4)= 1 - [(10C4*0.41^4*0.59^6) + 0.209 + 0.183]\\P(X <4)= 0.358[/tex]