Colgate claims that 90% of dentists recommend Colgate toothpaste. Suppose you randomly choose 10 dentists and ask whether they recommend Colgate toothpaste. What is the probability that exactly 8 dentists in your sample recommend Colgate toothpaste

Answer:

62.93%

Step-by-step explanation:

We have to solve it by a Poisson distribution, where:

p (x = n) = e ^ (- l) * l ^ (x) / x!

Where he would come being the number of birds that there would be in 40 minutes, we know that in 2 hours, that is 120 minutes there are 13, therefore in 40 there would be:

l = 13 * 40/120

l = 4,333

Now, we have p (x> 3) and that is equal to:

p (x> 3) = 1 - p (x <= 3)

So, we calculate the probability from 0 to 3:

 p (x = 0) = 2.72 ^ (- 4.33) * 4.33 ^ (0) / 0! = 0.01313

p (x = 1) = 2.72 ^ (- 4.33) * 4.33 ^ (1) / 1! = 0.0568

p (x = 2) = 2.72 ^ (- 4.33) * 4.33 ^ (2) / 2! = 0.12310

p (x = 3) = 2.72 ^ (- 4.33) * 4.33 ^ (3) / 3! = 0.17767

If we add each one:

0.01313 + 0.0568 + 0.12310 + 0.17767 = 0.3707

replacing:

p (x> 3) = 1 - 0.3707

p (x> 3) = 0.6293

Which means that the probability is 62.93%

Answer:

1/25

Step-by-step explanation:

5^-2

We know that a^ -b = 1/ a^b

5^-2 = 1/ 5^2

       = 1/25

Answer:

meh

Step-by-step explanation:

Find the given attachments

Answer:

G

Step-by-step explanation:

Point G is coplanar with points B, C, H.

Answer:

B.

Step-by-step explanation:

A cone with a radius of 1-unit will be obtained by rotating the 3-D figure.

Answer:

A. 1.8 ×[tex]10^{30}[/tex] Kg

B i. 3.0 × [tex]10^{17}[/tex] seconds

  ii. 9.6 × [tex]10^{9}[/tex] years

C. After 9.2 × [tex]10^{9}[/tex] (9.2 billion) years

Step-by-step explanation:

Given that the mass of the Sun = 2× [tex]10^{30}[/tex] Kg.

Mass of hydrogen when Sun was formed = 76% of 2× [tex]10^{30}[/tex] Kg

                            = [tex]\frac{76}{100}[/tex]  ×2× [tex]10^{30}[/tex] Kg

                           = 1.52 × [tex]10^{30}[/tex] Kg

Mass of hydrogen available for fusion = 12% of 1.52 × [tex]10^{30}[/tex] Kg

                           = [tex]\frac{12}{100}[/tex] × 1.52 × [tex]10^{30}[/tex] Kg

                           = 1.824 ×[tex]10^{30}[/tex] Kg

A. Total mass of hydrogen available for fusion over the lifetime of the sun is 1.8 ×[tex]10^{30}[/tex] Kg.

B. Given that the Sun fuses 6 × [tex]10^{11}[/tex] Kg of hydrogen each second.

i. The Sun's initial hydrogen would last;

                                     [tex]\frac{1.8*10^{30} }{6*10^{11} }[/tex]

                                 = 3.04 × [tex]10^{17}[/tex] seconds

The Sun's hydrogen would last 3.0 × [tex]10^{17}[/tex] seconds

ii. Since there are 31536000 seconds in a year, then;

The Sun's initial hydrogen would last;

                                     [tex]\frac{3.04*10^{17} }{31536000}[/tex]

                                 = 9.640 × [tex]10^{9}[/tex] years

The Sun's hydrogen would last 9.6 × [tex]10^{9}[/tex] years.

C. Given that our solar system is now about 4.6 × [tex]10^{9}[/tex] years, then;

                               [tex]\frac{9.6*10^{9} }{4.6*10^{9} }[/tex]

                             = 2.09

So that;   2 × 4.6 × [tex]10^{9}[/tex] = 9.2 × [tex]10^{9}[/tex] years

Therefore, we need to worry about the Sun running out of hydrogen for fusion after 9.2 × [tex]10^{9}[/tex] years.

Part(A): The total mass of hydrogen available 9.6 billion years.

Part(B): The total time is 5.10 billion years.

Part(D): The hydrogen will last [tex]5.04\times 10^9 \ years[/tex]

Our sun is the largest object in our solar system. The mass of the sun is approximately [tex]1.988\times 1030[/tex] kilograms

Part(A):

Given that,

The total mass of the Sun =[tex]2\times10^{30} kg[/tex]

Mass of hydrogen in Sun =  [tex]2\times10^{30} \times0.76\ kg[/tex]

The mass of hydrogen ever available for fusion is,

[tex]2\times10^{30} \times 0.76 \times 0.12 kg = 1.824\times 10^{29}[/tex]

Mass of hydrogen fuses each second = 600 billion kg/second.

Time hydrogen will last in seconds=[tex]1.824\times 10^{29}seconds.[/tex]

[tex]= 0.00304\times 10^{29 }= 3.04\times 10^{17}.[/tex]

Time hydrogen will last in seconds =[tex]1 year = 31,536,000 seconds.[/tex]

[tex]31,536,000 x = 3.04\times 10^{17} = 9.6 \ billion \ years.[/tex]

(B) Present age of sun = [tex]9.6-4.5 \ billion \ years[/tex]

The time when we need to worry about Sun running out of hydrogen for fusion = 5.10 billion years.

(D) The solar system is 4.6 billion years old that is [tex]4.6\times 10^9\ years[/tex]

And in part (B) we have calculated that hydrogen will last [tex]9.64\times 10^9[/tex] then,

[tex]9.64\times 10^9-4.6\times 10^9=5.04\times 10^9[/tex]

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Answer:

Step-by-step explanation:Srry it's bit rough...

1 × 18/100 = 12/100(g+11), is the equation. The answer is 12/100 gallons

Answer:

2.54 seconds

Step-by-step explanation:

We can use the following equation to model the vertical position of the ball:

S = So + Vo*t + a*t^2/2

Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.

Then, using S = 2.5, So = 0.4, Vo = 14 and a = -9.8 m/s2, we have that:

2.5 = 0.4 + 14*t - 4.9t^2

4.9t^2 - 14t + 2.1 = 0

Solving this quadratic equation, we have that t1 = 2.6983  s and t2 = 0.1588 s.

Between these times, the ball will be higher than 2.5 m, so the amount of time the ball will be higher than 2.5 m is:

t1 - t2 = 2.6983  - 0.1588 = 2.54 seconds

Answer:

[tex]P(x>20)=9.67*10^{-10}[/tex]

Step-by-step explanation:

If we call x the number of correct answers, we can said that P(x) follows a Binomial distribution, because we have 25 questions that are identical and independent events with a probability of 1/4 to success and a probability of 3/4 to fail.

So, the probability can be calculated as:

[tex]P(x)=nCx*p^{x}*q^{n-x}=25Cx*0.25^{x}*0.75^{25-x}[/tex]

Where n is 25 questions, p is the probability to success or 0.25 and q is the probability to fail or 0.75.

Additionally, [tex]25Cx=\frac{25!}{x!(25-x)!}[/tex]

So, the probability that the student answers more than 20 questions correctly is equal to:

[tex]P(x>20)=P(21)+P(22)+P(23)+P(24)+P(25)[/tex]

Where, for example, P(21) is equal to:

[tex]P(21)=25C21*0.25^{21}*0.75^{25-21}=9.1*10^{-10}[/tex]

Finally, P(x>20) is equal to:

[tex]P(x>20)=9.67*10^{-10}[/tex]

Answer:

The system has one solution.

Step-by-step explanation:

We have two equations:

y = -2x - 4

y = 3x + 3

Equalling them:

y = y

-2x - 4 = 3x + 3

5x = -7

[tex]x = -\frac{7}{5}[/tex]

And

[tex]y = 3x + 3 = 3(-\frac{7}{5}) + 3 = \frac{-21}{5} + 3 = \frac{-21}{5} + \frac{15}{5} = -\frac{6}{5}[/tex]

Replacing in the other equation we should get the same result.

[tex]y = -2x - 4 = -2(-\frac{7}{5}) - 4 = \frac{14}[5} - 4 = \frac{14}{4} - \frac{20}{5} = -\frac{6}{5}[/tex]

So the system has one solution.

Answer:

The nearest whole is 122.99 repeated

Step-by-step explanation:

Answer 1750000

Step-by-step explanation:

The number "one and three-quarter million" can be written in figures as 1,750,000.

Given the number one and three quarter million.

"One" represents the digit 1.

"Three-quarter" means three-fourths or 3/4.

In decimal form, this fraction is 0.75.

"Million" signifies a value of 1,000,000.

To represent this number in figures, we combine these components:

We start with the digit 1, representing "one."

Then we append the value 750,000, which corresponds to "three-quarter million."

Combining these two parts, we have 1,750,000.

Therefore, "one and three-quarter million" can be written in figures as 1,750,000.

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Answer:

53

Step-by-step explanation:

8•7 is 56

56 - 3 is 53

Answer:

53

Step-by-step explanation:

z = 7

8z is the same as saying 8×z

8×7-3 (do multiplication first)

56-3 = 53

Answer:

Let's call the amount of 90% solution x.

We can write:

45% * 525 + 90%x = 55%(x + 525)

0.45 * 525 + 0.9x = 0.55(x + 525)

Solving for x we get x = 150 so the first blank is 150 and the second blank is 525 + 150 = 675.

Answer:

D. Pilot error isPilot error is the most serious threat to aviation safety. Pilots could be better trained.

Step-by-step explanation:

We can construct the relative frequency distribution dividing the amount of incidents for each category by the total amount of incidents.

This amount is:

[tex]\sum x_i=627+64+113+382+481=1667[/tex]

Then, the relative frequency for each category is:

[tex]\text{Pilot error}=627/1667=0.38\\\\\text{Human error}=64/1667=0.04\\\\\text{Weather}=113/1667=0.07\\\\\text{Mechanical problems}=382/1667=0.23\\\\\text{Sabotage}=481/1667=0.29\\\\[/tex]

As the pilot error has the largest relative frequency, we can conclude that pilot error is the most serious threat to aviation safety.

Answer:

D. G(x) = |x| + 7

Step-by-step explanation:

→For the function G(x) to shift upwards, there needs to be a number being added to the whole function.

→The answer isn't "A," because the 1 is being subtracted, making it shift downwards 1 unit, not upwards.

→The answer isn't "B," because adding the 2 there would cause the function to shift to the left for 2 units, not upwards.

→The answer isn't "C," because 10 is being multiplied, which would cause the function to narrow, and not shift upwards.

This means the correct answer is "D," because the 7 is being added, making the function shift upwards 7 units.

Answer:

105.12 ft^2

Step-by-step explanation:

Area of a rectangle: bh

In this case 8*10.... so area of the rectangle is 80

Area of a circle: pir^2

Half it for a semicircle.

so 1/2 pi r^2

radius is 4 cuz its half of 8.

so 1/2(3.14)(4^2)=(0.5)(3.14)(16)=25.12

Now add up 80+25.12

Total is 105.12

Hope I helped :)

Answer:

Option (2).

Step-by-step explanation:

From the figure attached,

EFGH is a quadrilateral and FH is line which divides the quadrilateral into two right triangles, ΔFEH and ΔHGF.

In ΔFEH and ΔHGF,

Sides EH ≅ FG [Given]

FH ≅ FH [reflexive property]

ΔFEH ≅ ΔHGF [HL (Hypotenuse - length) postulate of congruence]

Option (2) will be the answer.

Answer:

The annual interest tax shield to Salinas Corp would be of $560,000

Step-by-step explanation:

In order to calculate the annual interest tax shield to Salinas Corp if it goes through with the debt issuance we would have to calculate the following formula:

Annual Interest tax shield = Interest * tax

Interest = debt *rate of interest

Interest=$20 million * 0.07

Interest= $ 1.40 million

tax= 40%

Therefore, Annual Interest tax shield =$1.40 million * 0.40

Annual Interest tax shield = $560,000

The annual interest tax shield to Salinas Corp would be of $560,000

Answer:

19.82 units

Step-by-step explanation:

The number of units of tile simply refers to the perimeter.

So, we need to find all the sides of the rectangle.

Now, we have AB = 4.24 units and BD = 7.07 units.

So, we can find AD using pythagoras theorem.

So,

(AD)² + 4.24² = 7.07²

(AD)² + 17.978 = 49.985

(AD)² = 49.985 - 17.978

AD = √32.007

AD = 5.66 units

AD = BC = 5.66 units

Likewise, AB = DC = 4.24 units

Thus,

perimeter = 2(5.66) + 2(4.24) = 19.8 units

Closest answer among the options is approximately 19.82

Answer:

19.82 units

Step-by-step explanation:

just took test and got it right

Answer:

85.56% probability that less than 6 of them have a high school diploma

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have a high school diploma, or they do not. The probability of an adult having a high school diploma is independent of other adults. 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.

50% of adult workers have a high school diploma.

This means that [tex]p = 0.5[/tex]

If a random sample of 8 adult workers is selected, what is the probability that less than 6 of them have a high school diploma

This is P(X < 6) when n = 8.

[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]

In which

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

[tex]P(X = 0) = C_{8,0}.(0.5)^{0}.(0.5)^{8} = 0.0039[/tex]

[tex]P(X = 1) = C_{8,1}.(0.5)^{1}.(0.5)^{7} = 0.0313[/tex]

[tex]P(X = 2) = C_{8,2}.(0.5)^{2}.(0.5)^{6} = 0.1094[/tex]

[tex]P(X = 3) = C_{8,3}.(0.5)^{3}.(0.5)^{5} = 0.2188[/tex]

[tex]P(X = 4) = C_{8,4}.(0.5)^{4}.(0.5)^{4} = 0.2734[/tex]

[tex]P(X = 5) = C_{8,5}.(0.5)^{5}.(0.5)^{3} = 0.2188[/tex]

[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0039 + 0.0313 + 0.1094 + 0.2188 + 0.2734 + 0.2188 = 0.8556[/tex]

85.56% probability that less than 6 of them have a high school diploma

Answer:

The probability that a study participant has a height that is less than 65 inches is 0.1103.

Step-by-step explanation:

We are given that the heights in the​ 20-29 age group were normally​ distributed, with a mean of 69.9 inches and a standard deviation of 4.0 inches.

A study participant is randomly selected.

Let X = heights in the​ 20-29 age group.

So, X ~ Normal([tex](\mu=69.9,\sigma^{2} =4.0^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                             Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = mean height = 69.9 inches

           [tex]\sigma[/tex] = standard deviation = 4.0 inches

Now, the probability that a study participant has a height that is less than 65 inches is given by = P(X < 65 inches)

      P(X < 65 inches) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{65-69.9}{4}[/tex] ) = P(Z < -1.225) = P(Z [tex]\leq[/tex] 1.225)

                                                                 =  1 - 0.8897 = 0.1103

The above probability is calculated by looking at the value of x = 1.225 in the z table which lies between x = 1.22 and x = 1.23 which has an area of 0.88877 and 0.89065 respectively.

Answer:

a = 8

Step-by-step explanation:

Explanation:-

Given P(x) = 2 x+4 a

          Q(x)=4 x - 2

          P( Q(4)) = 60

         P(4 (4) - 2) = 60

        P( 14 ) = 60

       2 (14) + 4 a = 60

        4 a + 28 = 60

Subtracting '28' on both sides , we get

       4 a +28 - 28 = 60 - 28

       4 a = 32

Dividing '4' on both sides , we get

        a = 8

Answer:

y=cosx

Step-by-step explanation:

cosx has a domain of all real numbers

Answer:

a. 50 turns

b. 0.0114 A

c. 0.25 A, 10 V

Step-by-step explanation:

Given:-

- The required current ( Is ) = 0.5 A

- The required voltage ( Vs ) = 5 V

- Transformer is 100% efficient ( ideal )

- The number of turns in the primary coil, ( Np ) = 2200

- The Voltage generated by power station, ( Vp ) = 220 V

Find:-

a. The number of turns in the secondary coil of the transformer

b. The current supplied by the power station

c. The effect on output current and voltage when the number of turns of secondary coil are doubled.

Solution:-

- For ideal transformers that consists of a ferromagnetic core with two ends wounded by a conductive wire i.e primary and secondary.

- The power generated at the stations is sent to home via power lines and step-down before the enter our homes.

- A household receives a voltage of 220 V at one of it outlets. We are to charge a phone that requires 0.5 A and 5V for the process.

- The outlet and any electronic device is in junction with a smaller transformer.

- All transformers have two transformation ratios for current ( I ) and voltage ( V ) that is related to the ratio of number of turns in the primary and secondary.

                     Voltage Transformation = [tex]\frac{N_p}{N_s} = \frac{V_p}{V_s}[/tex]

Where,

                 Ns : The number of turns in secondary winding

- Plug in the values and evaluate ( Ns ):

                           [tex]N_s = N_p*\frac{V_s}{V_p} \\\\N_s = 2200*\frac{5}{220} \\\\N_s = 50[/tex]

Answer a: The number of turns in the secondary coil should be Ns = 50 turns.

- Similarly, the current transformation is related to the inverse relation to the number of turns in the respective coil.

                  Current Transformation = [tex]\frac{N_p}{N_s} = \frac{I_s}{I_p}[/tex]

Where,

                 Ip : The current in primary coil

- Plug in the values and evaluate ( Ip ):

                        [tex]I_p = \frac{N_s}{N_p}*I_s\\\\I_p = \frac{50}{2200}*0.5\\\\I_p = 0.0114[/tex]  

Answer b: The current in the primary coil should be Ip = 0.0114 Amp.

- The number of turns in the secondary coil are doubled . From the transformation ratios we know that that voltage is proportional to the number of turns in the respective coils. So if the turns in the secondary are doubled then the output voltage is also doubled ( assuming all other design parameters remains the same ). Hence, the output voltage is = 2*5V = 10 V

- Similary, current transformation ratio suggests that the current is inversely proportional to the number of turns in the respective coils. So if the turns in the secondary are doubled then the output current is half of the required ( assuming all other design parameters remains the same ). Hence, the output current is = 0.5*0.5 A = 0.25 A

Answer:

(C) 10

Step-by-step explanation:

x+y−z=8

Subtract y from both sides

x-z= -y+8

Add z to both sides

x= -y+z+8

Subtract 8 from both sides

x-8= -y+z

x−y+z=12

Add y to both sides

x+z=12+y

Subtract z from both sides

x=y-z+12

Subtract 12 from both sides

x-12=y-z

Multiply both sides by -1

-x+12= -y+z

Combine equations:

x-8= -x+12

Add x to both sides

2x-8+12

Add 8 to both sides

2x=20

Divide both sides by 2

x=10

The answer is (C) 10.

Here is the full question .

We are interested in finding an estimator for [tex]Var (X_i )[/tex] and propose to use :

[tex]\hat {V} = \bar {X}_n (1- \bar {X} )_n[/tex]

Now; we are interested in the basis of [tex]\hat V[/tex]

Compute :

[tex]E \ \ [ \bar V] - Var (X_i) =[/tex]

Using this; find an unbiased estimator [tex][ \bar V][/tex] for [tex]p(1-p) \ if \ n \geq 2[/tex]

Write [tex]bar \ x{_n} \ for \ X_n[/tex]

Answer:

Step-by-step explanation:

[tex]\bar X_n = \dfrac{1}{n} {\sum ^n _ {i=1} } \\ \\ E(X_i) = - \dfrac{1}{n=1} \sum p \dfrac{1}{n}*np = \mathbf{p}[/tex]

[tex]V(\bar X_n) = V ( \dfrac{1}{n_{i=1} } \sum ^n \ X_i )} = \dfrac{1}{n^2} \sum ^n_{i=1} Var (X_i) \\ \\ = \dfrac{1}{n^2} \ \sum ^n _{i=1} p(1-p) \\ \\ = \dfrac{1}{n^2}*np(1-p) \\ \\ = \dfrac{p(1-p)}{n}[/tex]

[tex]E( \bar X^2 _ n) = Var (\bar X_n) + [E(\bar X_n)]^2 \\ \\ = \dfrac{p(1-p)}{n}+ p \\ \\ = p^2 + \dfrac{p(1-p)}{n} \\ \\ \\ \hat V = \bar X_n (1- \bar X_n ) = \bar X_n - \bar X_n ^2 \\ \\ E [ \hat V] = E [ \bar X_n - \bar X_n^2] \\ \\ = E[\bar X_n ] - E [\bar X^2_n] \\ \\ = p-(p^2 + \dfrac{p(1-p)}{n}) \\ \\ = p-p^2 -\dfrac{p(1-p)}{n}[/tex]

[tex]=p(1-p)[1-\dfrac{1}{n}] = p(1-p)\dfrac{n-1}{n}[/tex]

[tex]Bias \ (\bar V ) = E ( \hat V) - Var (X_i) \\ \\ = p(1-p) [1-\dfrac{1}{n}] - p(1-p) \\ \\ - \dfrac{p(1-p)}{n}[/tex]

Thus; we have:

[tex]E [\hat V] = p(1-p ) \dfrac{n-1}{n}[/tex]

[tex]E [\dfrac{n}{n-1} \ \ \bar V] = p(1 -p)[/tex]

[tex]E [\dfrac{n}{n-1} \ \ \bar X_n (1- \bar X_n )] = p (1-p)[/tex]

Therefore;

[tex]\hat V ' = \dfrac{n}{n-1} \bar X_n (1- \bar X_n)[/tex]

[tex]\mathbf{ \hat V ' = \dfrac{n \bar X_n (1- \bar X_n)} {n-1}}[/tex]

Answer:

z=12500

Step-by-step explanation:

Of means multiply and is means equals

85% *z = 106250

Change to decimal form

.85z = 106250

Divide each side by .85

.85z/.85 = 106250 /.85

z=12500