A local college is forming a seven-member research committee having one administrator, three faculty members, and three students. There are five administrators, 14 faculty members, and 18 students in contention for the committee. How many seven-member committees are possible

Answer:

r = 7

Step-by-step explanation:

This is how you solve the equation...

[tex]r^{3} = 49r\\r^{3} / r= 49r / r\\r^{2} = 49\\r = \sqrt{49} \\r = 7[/tex]

Answer:

r = 7 is the answer

Answer:

only 5 points

Step-by-step explanation:

Answer:

It's D

Step-by-step explanation:

100%

Answer:

First box: 35 Negative Result

Second box : 60 Positive Result

Step-by-step explanation:

Contains organic matter

Sample size : 700

95% positive means  [tex]\frac{95}{100} \times 700 = 665\\[/tex]

5% negative means [tex]\frac{5}{100}\times 700 = 35[/tex]

Does not contain organic matter

Sample size : 300

20% positive means [tex]\frac{20}{100} \times 300 = 60[/tex]

80% negative means [tex]\frac{80}{100}\times 300 = 240[/tex]

Answer:

Just subtract the less no by greater one .you get the answer.

Positive result of sample test gives a positive result than it is 665/1000*100*=66.5% likely that the soil sample contains organic matter.

Answer:

N'(-2,6)  

Noting Else To Put

Answer: The correct answer is B (-2,6)

Step-by-step explanation: it is show down below

Answer:

t(s) > t(c)  then t(s) is in the rejection region for H₀.

We reject H₀ we support the claim that the mean penetration is bigger than that of the specifications

Step-by-step explanation:

Sample Information:

sample size    n  =  35

sample mean    x  =  52.7

sample standard deviation  s  =  4.8

spec. true average  =  50

Test Hypothesis:

Null Hypothesis                             H₀              x  =  50

Alternative Hypothesis                 Hₐ              x  >  50

The standarddeviation of the population is unknown therefore even though we assume  distribution is normal we have to use t-student distribution

The alternative hypothesis indicates that the test is a one-tail test, we will test at significance level α  =  5 %     CI  =  95 %

for  α   =  0.05    t(c)  is from t-student -table   and with  df  = 35 - 1

df  =  34     t(c)  =  1.69

t(s)  =  (  x  -  μ ) / s/√n

t(s)  = ( 52.7  -  50 ) / 4.8/√35

t(s)  =  2.7 * 5.92 / 4.8

t(s)  =  3.33

Comparing   t(s)  and   t(c)  

t(s) > t(c)  then t(s) is in the rejection region for H₀.

We reject H₀ we support the claim that the mean penetration is bigger than that of the specifications

Answer:

it is 20

Step-by-step explanation:

well the angles looks the same

Answer:

x=20[vertically opposite angle are equal]

Answer:

C = 16 pi

Step-by-step explanation:

The circumference is given by

C = 2*pi*r

C = 2* pi*8

C = 16 pi

Answer:

Solution given:

radius [r]=8

Now

we have

circumference of circle=2πr=2*8*π=16π

16 is a required answer.

Answer:

3/10

Step-by-step explanation:

Answer:

8.333... Yards.

Step-by-step explanation:

To convert feet to yards, divide the feet by 3.

[tex]\frac{25}{3} = \frac{25}{3}/8.333...[/tex]

To check work

Multiply 8.333... by 3.

8.333 · 3 = 24.999...

It Is Not Exactly 25 Because The 3 Is Infinite But It Is Enough To Prove Your Answer.

Answer:

Slope = 1/4

Step-by-step explanation:

[tex]slope = \frac{y2-y1}{x2-x1} \\m=\frac{3-2}{3--1} \\m=1/4[/tex]

Answer:

(d) above [tex]74.1^o[/tex]

Step-by-step explanation:

Given

[tex]\tan w = 3.52[/tex]

Required

Determine w, using a calculator

We have:

[tex]\tan w = 3.52[/tex]

Take arctan of both sides

[tex]w = \tan^{-1}(3.52)[/tex]

Using a calculator, we have:

[tex]w \approx 74.1^o[/tex]

From the given options, (d) is true

Answer: 44 Us Cups

Step-by-step explanation:

Answer:

C)

Step-by-step explanation:

(n+4)(n+4) = n² + 4n + 4n + 16

this can be simplified to get:

n² + 8n + 16

now multiply the above by (n+4)

(n²+8n+16)(n+4) by multiplying each of the three terms in the first set of parentheses by each of the two terms in the second set to get:

n³ + 4n² + 8n² + 32n + 16n + 64  [Now look to combine 'like terms']

n³ + 12n² + 48n + 64 [this is option C]

=======================================================

Explanation:

There are...

We start with 6, and count down by 1 each time we fill up a slot. We stop once the third slot is filled or accounted for. The countdown is to ensure that we don't pick the same person twice. From here, multiply those values: 6*5*4 = 30*4 = 120

Interestingly, this is equal to 5! = 5*4*3*2*1 = 120 because the 3*2 becomes 6 and that *1 at the end doesn't affect things. Though usually results of permutation problems don't always end up like this. The order matters because a result like ABC is different from BAC, where A,B,C,D,E,F are the six school organizations.

As a slightly longer way to do the problem, you can use the nPr formula which is [tex]_nP_r = \frac{n!}{(n-r)!}[/tex] where n = 6 and r = 3 in this case. The exclamation marks indicate factorial. If you go this route, you should find that one of the steps will involve 6*5*4.

Answer:

A.

[tex] {b}^{2} + 3b - 3 = 0[/tex]

Step-by-step explanation:

[tex] {b}^{2} + 3b - 3 = 0 \\ because \: \: {b}^{2} < 4ac \\ {b}^{2} = {3}^{2} = 9 \\ 4ac = 4 \times 1 \times - 3 = - 12 \\ hence : {b}^{2} < 4ac[/tex]

Answer:

175,760,000 different California license plates can be made.

Step-by-step explanation:

Possible outcomes for each number of letter:

10 numbers(from 0 to 9) and 26 letters.

Format of the plate:

Number - letter - letter - letter - number - number - number

The numbers of possible outcomes are:

10 - 26 - 26 - 26 - 10 - 10 -10

How many different plates?

We multiply these values. So

10*26*26*26*10*10*10 = 175,760,000

175,760,000 different California license plates can be made.

The Required equilibrium solution are (400, 40) (500, 50) (600,60).

To find the equilibrium solutions and to enter answer as a list of ordered pairs (A,L), where A is the number of aphids and L the number of ladybugs.

equation is the relationship between variable and represented as y =ax+b    is example of polynomial equation.

since the equilibrium solution given by the expression

L = 10A
where A is the number of aphids and L the number of ladybugs.

And ordered pairs with equilibrium solution  of one with 400 aphids and 40 ladybugs, one with 500 aphids and 60 ladybugs and one with 600 aphids and 60 ladybugs  are (400, 40) (500, 50) (600,60).

Thus, The Required equilibrium solution are (400, 40) (500, 50) (600,60).

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I guess you need to find the area :

Area of the figure = Square area + half of

circle area

A = ( length )^2 + 1/2 × pie × ( radius)^2

A = ( 10 )^2 + 1/2 × pie × ( 10/2 )^2

A = 100 + 1/2 × pie × 25

A = 100 + 12.5 pie

____________________________

If pie = 3 : | If pie = 3.14 :

|

A = 100 + 12.5 × 3 | A = 100 + 12.5 × 3.14

|

A = 100 + 37.5 | A = 100 + 39.25

|

A = 137.5 m^2 | A = 139.25 m^2

Answer:

y = x + 1

Step-by-step explanation:

Answer:

y = x + 1

Step-by-step explanation:

(1,2) and (3, 4)

y = mx + b

y = x + 1

thanks

Answer:

1/9 ÷ 2

= 1/9 × 1/2

= 1/18

Step-by-step explanation:

THAT IS MY ANSWER HOPE IT HELPS

Answer:

Cathy was driving at a speed of 67 miles per hour.

Step-by-step explanation:

Fined $15 for each mile per hour over the speed limit. Cathy was given a ticket for $105.

We solve this using a rule of three.

1 mile - $15

x miles - $105

Applying cross multiplication:

15x = 105

x = 105/5 = 7.

Cathy was 7 miles above the speed limit.

How fast was Cathy driving?

7 miles above the speed limit of 60 miles per hour, so 60 + 7 = 67 miles per hour.

Cathy was driving at a speed of 67 miles per hour.

Answer:

w = 6ft 8in

Step-by-step explanation:

the proportions will be the same

w/7.5 = 12/13.5

multiply both sides by 7.5

w = 12/13.5 * 7.5

w = 6.6666666667ft

w = 6ft 8in

The original truck was 6.67 feet wide.

"It is a comparison of two or more numbers that indicates their sizes in relation to each other."

"It is an equation in which two ratios are set equal to each other."

For given example,

A model truck is 13.5 inches long 7.5 inches wide.

The ratio of length to width of a model truck would be,

13.5 : 7.5                       ...........................(1)

The original truck was 12 feet long.

This means the original truck was 144 inches long.

Let 'x' be the width (in inches) of the original truck.

So, the ratio of the length to the width of the original truck would be,

144 : x                           .................................(2)

Also, the ratios given by (1) and (2) must be in proportion.

[tex]\Rightarrow \frac{13.5}{7.5} = \frac{144}{x} \\\\\Rightarrow 13.5 \times x = 144 \times 7.5\\\\\Rightarrow \bold{x=80~inches}\\\\\Rightarrow \bold{x=6.67~feet}[/tex]

Therefore, the original truck was 6.67 feet wide.

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Answer:

the formula for a cylinder is πr^2•h

Step-by-step explanation:

just plug in the numbers, and you can change π to 3.14 for an approximate answer!

hopefully this helps you!

[tex]\pi {r}^{2} h \\ 22 \div 7 \times {r}^{2} h[/tex]

Answer:

Solution given:

perpendicular [P]=15m

base[B]=8m

hypotenuse [H]=17m

rate :4 plants per square metre

no of plants=?

we have

Area of triangular garden: ½(P*B)=½(15*8)=60m²

Now

Total no of plants =rate ×Area =4×60=240

Michael needs 240 plants in the garden

Answer:

97,655

Step-by-step explanation:

5(5)^(n-1) = 78,125

5^n = 78,125

n = 7

=> S7 = 5(5^7 -1) / (5-1)

= 5/4 (78, 125 -1)

= 5/4 (78 124) = 97,655

angle B is teh answer

100% sure !

Plz mark me brainliest

Answer:

Step-by-step explanation:

(a) We will prove by induction that if n is a multiple of 3 then Bob has a winning strategy.

Let n=3

It is given that Alice always plays first.

Then, for the first move, Alice has a choice of removing 1 or 2 stones.

Case 1:

Alice removes 1 stone. Then it is now Bob's turn. There are 2 stones left. Bob has the choice of removing 1 or 2 stones. Then Bob's winning strategy should be to remove 2 stones. Then Bob removes the last stone and wins.

Case 2:

Alice removes 2 stones. Then it is Bob's turn. There is exactly 1 stone left, which Bob removes in his turn. Since he removes the last stone, he wins.

Thus, in either case, for n=3, Bob has a winning strategy. ____ (A)

Now, let us assume that Bob has a winning strategy for n=3p.

We will now check if Bob has a winning strategy for n=3p+3

It is given that Alice plays first.

Case 1:

Alice starts the game by removing 1 stone. Then, Bob has a choice of removing 1 or 2 stones. If he chooses to remove 2 stones, then the number of stones left is 3p+3-3=3p and it is now Alice's turn to play. This is exactly the game when n=3p and we have already assumed that Bob has a winning strategy for n=3p. Thus, in this case Bob has a winning strategy.

Case 2:

Alice starts the game by removing 2 stones. Then, Bob has a choice of removing 1 or 2 stones. If he chooses to remove 1 stone, then the number of stones left is 3p+3-3=3p and it is now Alice's turn to play. This is again exactly the game when n=3p and we have already assumed that Bob has a winning strategy for n=3p. Thus, in this case also, Bob has a winning strategy.

Thus, in either case Bob has a winning strategy for n=3p+3 if he has a winning strategy for n=3p. _______ (B)

Thus, from (A) & (B), using induction, we can say that Bob has a winning strategy if n is a multiple of 3.

(b) We will now prove that if n is not a multiple of 3, then Alice has a winning strategy.

If n is not a multiple of 3, then n can have either of the forms of 3m+1 or 3m+2, m ∈ W.

We will prove the given fact for both the forms simultaneously

Let m=0, i.e., n=1 or n=2

Since Alice starts first, she removes 1 stone, if n=1 or 2 stones if n=2 and thus wins. Thus Alice has a winning strategy if m=0. ______ (A)

Let us assume that Alice has a winning strategy for m=k, i.e., for n=3k+1 & n=3k+2

Now, we will check if Alice has a winning strategy for m=k+1, i.e., for n=3(k+1)+1=3k+4 or n=3(k+1)+2=3k+5

Let n=3k+4

Since Alice plays first, she has a choice to remove 1 or 2 stones.

Note that, if Alice removes 2 stones and in turn Bob removes 2 stones, then the number of stones becomes a multiple of 3 such that it is Alice's turn to play. In that case, Bob will have a winning strategy as shown in the previous part.

Then, Alice should remove 1 stone. Then, it is now Bob's turn and he has a choice of removing 1 or 2 stones.

Case 1:

Bob removes 1 stone. Then there are 3k+4-1-1=3k+2 stones remaining and it is Alice's turn. This is identical to the game where n=3k+2. We have already assumed that Alice has a winning strategy in this case.

Case 2:

Bob removes 2 stones. Then there are 3k+4-1-2=3k+1 stones remaining and it is Alice's turn. This is identical to the game where n=3k+1. We have already assumed that Alice has a winning strategy in this case.

Thus, in either case, Alice has a winning strategy.

Let n=3k+5

Since Alice plays first, she has a choice to remove 1 or 2 stones.

Note that, if Alice removes 1 stone and in turn Bob removes 1 stone, then the number of stones becomes a multiple of 3 such that it is Alice's turn to play. In that case, Bob will have a winning strategy as shown in the previous part.

Then, Alice should remove 2 stones. Then, it is now Bob's turn and he has a choice of removing 1 or 2 stones.

Case 1:

Bob removes 1 stone. Then there are 3k+5-2-1=3k+2 stones remaining and it is Alice's turn. This is identical to the game where n=3k+2. We have already assumed that Alice has a winning strategy in this case.

Case 2:

Bob removes 2 stones. Then there are 3k+5-2-2=3k+1 stones remaining and it is Alice's turn. This is identical to the game where n=3k+1. We have already assumed that Alice has a winning strategy in this case.

Thus, in either case, Alice has a winning strategy.

Then, we can say that Alice has a winning strategy for m=k+1 if she has a winning strategy for m=k. _____ (B)

Then, by induction, from (A) & (B), we can say that if n is not a multiple of 3, then Alice has a winning strategy.