Suppose that a certain college class contains 51 students. Of these, 23 are freshmen, 30 are mathematics majors, and 7 are neither. A student is selected at random from the class.
(a) What is the probability that the student is both a freshman and a mathematics major?
(b) Given that the student selected is a freshman, what is the probability that he is also a mathematics major

Based on the current price of gas per gallon, and the fact that it was cheaper the year before, the percent decrease in price is 31%

First, find the price of gas last year:

=Current price of gas + Decrease in price

= 3.20 + 1.43

= $4.63

The percent of decrease in price can be found by the formula:

= Change in gas price / Price of gas last year x 100%

= 1.43 / 4.63 x 100%

= 0.30885529157667386609071274298056 x 100%

= 31%

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Using the Fundamental Counting Theorem, there is a 0.7143 = 71.43% probability that the code either starts with an 'A' or ends with an even digit.

It is a theorem that states that if there are n things, each with [tex]n_1, n_2, \cdots, n_n[/tex] ways to be done, each thing independent of the other, the number of ways they can be done is:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

In this problem, we have that the code has:

Hence the number of codes is:

N = 14 x 5 x 14 x 5 x 14 x 5 = 343,000.

The number of codes starting with an A is:

Na = 5 x 14 x 5 x 14 x 5 = 24,500.

The number of codes ending with an even digit(0, 2 or 4) is:

Ne = 14 x 5 x 14 x 5 x 14 x 3 = 205,800.

The number of codes starting with an A and ending with an even digit is:

Nae = 5 x 14 x 5 x 14 x 3 = 14,700.

The number of desired outcomes is:

24500 + 205800 + 14700 = 245,000.

Hence the probability is:

245,000/343,000 = 0.7143.

0.7143 = 71.43% probability that the code either starts with an 'A' or ends with an even digit.

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Answer:

[tex] \: \: \: \: \: { \rm{2 \: \: \: 3 \: \: \: 4}} \\ { \underline{ \times \: \: 0. \: \: 3 \: \: 5}} \\ { \rm{ 1 \: \: 1 \: \: 7\: \: 0}} \\ { \underline{ + 7\: \: 0\: \: 2 \: \: 0 }} \\ { \underline{ \: \boxed{ 8\: \: 1 \: .\: \: 9 \: \: 0}}}[/tex]

Answer: 81.9

The domain of this function is all the real numbers except - 7 / 3 and 23 / 3.

In accordance with function theory, the domain of a function is a set of x-values such that y-values exist. Herein we find that the given function is a rational function whose denominator includes an absolute value function. The numerator is a linear equation.

Therefore, the domain of the function is all real numbers except those x-values such that denominator is equal to zero. We need to solve the following expression in two cases:

|3 · x - 15| = 8

For x ≥ 5:

3 · x - 15 = 8

3 · x = 23

x = 23 / 3

For x < 5:

- 3 · x + 15 = 8

- 3 · x = - 7

x = - 7 / 3

Then, the domain of this function is all the real numbers except - 7 / 3 and 23 / 3.

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The number is 5.67

Let the number be 'x'

The difference of four times a number and seven = 4x - 7

Ten more than the number = 10 + x

According to the condition given in the question,

The difference between four times a number and seven is equal to ten more than the number.

4x - 7 = 10 + x

4x - x = 10 + 7

3x = 17

x = 17/3 = 5.67

We can verify the answer,

Consider the LHS,

=> 4x - 7 = 4(5.67) - 7 = 22.68 - 7 = 15.68

Consider the RHS,

=> 10 + x = 5.67 + 10 = 15.67

Both LHS and RHS are equal.

Therefore the number that satisfies the given condition is 5.67

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Answer:

Option D) 6.8 × 10⁻²

Answer:

D

Step-by-step explanation:

Answer:

[tex]\frac{12}{4} = \frac{3}{1}[/tex]

Step-by-step explanation:

[tex]\frac{12\div4}{4\div4}=\frac{3}{1}[/tex]

Hope this helps!

Answer:

12 = 3

4      1

Step-by-step explanation:

hope the helps

Based on the balances in Reese Corp's trial balance, the current assets total can be found to be B. $5,555,000

The total of the current assets can be found by the formula:

= Inventory + Accounts Receivable + Cash

Solving then gives:

= 775,000 + 2,695,000 + 2,085,000

= $5,555,000

In conclusion, the current assets for Reese Corp is $5,555,000 .

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Answer:

-22 I arrived as fast as I could

Answer:In his equation height is given by h(t) and t represents time. The shape of this equation is that of a downward facing parabola because the coefficient of the first term is negative. Therefore, the maximum height is found by determining the vertex of the parabola which is given by b/2a or 50/(16*2). Thus after, 1.5625 seconds the ball reaches it maximum height of 44 feet (plug 1.5625 into formula.)

Notice that the formula has a constant of 5 in it meaning that the ball was throw up at a 5 foot starting position above the ground. Thus, for the ball to hit the ground h(t) = -5. So, what is the value of t that gives that value? Here we just use the quadratic formula and find that t = 3.31 seconds. That makes sense doesn't it? 2*1.5625 would get the ball back to the starting position ( 3.125 sec) and the ball needs another .185 sec to fall the last 5 feet. What speed is that in MPH? That would be 5/.185*(3600/5280) = 18.4 MPH. That is a lot slower than a 95 MPH fast ball.

Step-by-step explanation:

The value of the x in the given geometry figure is  108° .

The labelled diagram of the figure is given below.

From the figure , we can see that ∠OEA = ∠BOF as they are vertically opposite angles .

Hence ∠BOF = x

∠OFT and ∠BOF are linear pair , so on adding them , we will get sum 180°

∠OFT + ∠BOF = 180°

 72° + x = 180°

        x = 108°

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Answer:

  x° = 22°

Step-by-step explanation:

You want the measure of the angle marked x° in the figure with two parallel lines and a "bent" transversal.

Extending the upward-sloping line in the figure as shown in the attachment results in a triangle against the top horizontal line.

The angle on the right side of that triangle corresponds to the one marked 50°, so has the same measure. It is one of two "remote interior angles" whose sum is the measure of the exterior angle marked 72°. The other "remote interior angle" making up that sum is the vertical angle to the one marked x°.

So, we have ...

  x° +50° = 72°

  x° = 22° . . . . . . . . subtract 50°

__

Additional comment

You can do a "sanity check" on this answer by considering some extreme cases. If the 72° angle is reduced to 50°, the top leg of the "72°" angle would be horizontal. That would effectively reduce x to 0. Similarly, if the 72° angle were made to be 180°, then the value of x would be supplementary to the 50° angle. In other words, x is the difference between the one marked 72° and 50°, or 22°.

Answer:

25 quarters and 4 dimes. Louisa used 21 more quarters than she did dimes.

Step-by-step explanation:

q=number of quarters

d=number of dimes

6.65=.25q+.10d   6.65 (the total) is equivalent to the amount of quarters times .25 (the value of a quarter) plus the amount of dimes times .10 (the value of a dime)

Now divide 6.65 by .25.

26 quarters can be used.

26 times .25 = 6.50

26 quarters is $6.50

29-26=3

3 times .10 = .30

.30+6.50=6.80

This is too much. Let's try taking off one quarter.

Now we have 25 quarters. .25 times 25 = 6.25

29-25=4 (This means there will be 4 dimes if she used 25 quarters, but let's see)

4 times .10 = .40

.40 + 6.25=6.65

6.65 is the total amount she had to pay, so we did it. Solving this showed us that she used 25 quarters and 4 dimes. This equals $6.65, the amount she paid, uses only quarters and dimes, and uses a total of 29 coins.

I hope this helps you! Please mark as Brainliest answer! I worked really hard to type all of this out for you.

Answer:

15m⁹

Step-by-step explanation:

-3m² × -5m⁷ multiply bases (-3×-5=15). when multiplying bases you add exponents (2+7=9)

= 15m⁹

Answer:

.9 inches

Step-by-step explanation:

Set up and solve this proportion:

[tex] \frac{1.2}{400} = \frac{x}{300} [/tex]

[tex]400x = 360[/tex]

[tex]x = .9[/tex]

Answer:

Step-by-step explanation:

Let Matt's age be x.

Then Pat is x + 2, Dave is x + 4 and Dan is x + 6.

Dan's age is 36 years less than twice Pat's age:

Solve it for x:

So Matt is 38, Pat is 40, Dave is 42 and Dan is 44.

the linear equation tangent at x = -2 is:

y = 123.33*x + 226.74

To get the slope of the line we need to evaluate the first derivative in x = -2, we get:

f'(x) = 3√(2 - 4x) + (-12/2)*3x*(2 - 4x)^(-1/2)

Evaluating in x = -2 we get:

f'(-2) = 3√(2 - 4*-2) + (-12/2)*3*-2*(2 - 4*-2)^(-1/2) = 123.33

Then the line is something like:

y = 123.33*x + b

And the line must pass through:

f(-2) = 3*(-2)*√(2 - 4*(-2)) = -18.97

So the line must pass through the point (-2,  -18.97)

Replacing that in the linear equation:

-18.97 = 123.33*2 + b

-19.92 + 123.33*2 = b = 226.74

Then we conclude that the linear equation tangent at x = -2 is:

y = 123.33*x + 226.74

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We are given a table for input and output values and the relation between them.

Consider the table,

Input           Output = Input × itself + (-5)           Output

0                                        

1

6    

                                                                              95

                                                                              59

                                                                              76

For Input = 0 :

Output = Input × itself + (-5)

Output = 0 × 0 - 5

Output = -5

Now, for Input = 1,

Output = Input x Itself + (-5)

Output = (1) × 1 + (-5)

Output = 1 - 5

Output = - 4

For Input = 6,

Output = Input x Itself + (-5)

Output = (6) × 6 + (-5)

Output = 36 - 5

Output = 31

For Output = 95,

Output = Input x Itself + (-5)

95 = (Input)²+ (-5)

95 + 5 = (Input)²

(Input)² = 100

Input = √100

Input = 10

For Output = 59,

Output = Input x Itself + (-5)

59 = (Input)²+ (-5)

59 + 5 = (Input)²

(Input)² = 64

Input = √64

Input = 8

For Output = 76,

Output = Input x Itself + (-5)

76 = (Input)²+ (-5)

76 + 5 = (Input)²

(Input)² = 81

Input = √81

Input = 9

Therefore the complete table will be:

Input           Output = Input × itself + (-5)              Output

0                     Output = 0 × 0 + (-5)                          -5

1                      Output = 1 × 1 + (-5)                             -4

6                     Output = 6 × 6 + (-5)                          31

10                  95 = Input × itself + (-5)                      95

8                    59 = Input x Itself + (-5)                      59

9                    76 = Input x Itself + (-5)                      76

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The dimensions of the rectangular field with area 120 square meters is 10 meters by 12 meters.

The measurement that expresses the size of a region on a plane or curved surface is called area. Surface area refers to the area of an open surface or the boundary of a three-dimensional object, whereas the area of a plane region or plane area refers to the area of a form or planar lamina.

Given perimeter of the rectangular plot is 44 meters.

let the length is x , and width = y

Therefore perimeter = 2 (x + y)

or, 2 (x + y) = 44

or, x + y = 22

or, y = 22 - x

Now the area of the field is given as 120 square meters.

Area of the fields in terms of x is given by:

x ( 22 - x ) = 22 x - x²

Therefore  22 x - x² = 120

or,   x² -22x +120 = 0

or,  x² - 10x - 12x + 120 = 0

or, x (x -10) -12(x-10)=0

or, (x-10)(x-12) = 0

Now either the value of x is 10m or the value of x is 12m

Therefore the dimensions of the rectangular field is 10 meters by 12 meters.

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Answer: 54

Step-by-step explanation:

3x=72-x       ==> 3x and 72-x are both congruent angles, so they're equal

3x+x=72-x+x

4x=72

x=18

3x=

3(18)=54 degrees

The length of EC will be 154 units.

We have the data that:

BD is the bisector of AC at point E.

This means that AE = EC

Also, AE + EC = AC

AE + AE = AC

2 AE = AC

Now,

AE = 2x² - 8 and AC is 5x² - 25

Substituting the values, we get that:

2 (2x² - 8) = 5x² - 25

4x² - 16 = 5x² - 25

5 x² - 4 x² = 25 - 16

x² = 9

x = 3 ( as length cannot be negative)

AE = EC = 2x² - 8

EC = 2 (9)² - 8

EC = 162 - 8

EC = 154 units.

Therefore, we get that, the length of EC will be 154 units.

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Answer: Yes

Step-by-step explanation:

All you had/have to do is 19-11 and get 8

Answer:

Yes, you're correct.

Step-by-step explanation:

Because since the bases are the same you can set the exponents equal. Then move the constant to the right hand side and it changes the sign. Then you calculate the difference. Finally you change the signs on both sides of the equation. x = 8

Answer: 12,186.96

Step-by-step explanation:

The unit rate problem is discussed above and the unit rate of speed according to problem is 50 kilometers per hour.

What is the unit rate method ?

The unit rate method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

The real world situation can be as follows ;

A police officer who believes you were speeding stops you. You say that it took you exactly seven hours to travel 350 kilometers across the state. To determine if you were over speeding, the policeman must perform the necessary calculations. He requires to find out the unit rate.

Let's solve it to find out that at what unit rate of speed you were travelling.

We know that unit rate meant we need to find out the distance we have travelled in 1 hour.

So,

Unit speed = 350 / 7 km per hour

or

Unit speed = 50 kilometers per hour

Police man will check with the limit of speed and if he find that you were not over speeding , then he would allow you to go further.

Therefore , the unit rate problem is discussed above and the unit rate of speed according to problem is 50 kilometers per hour.

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[tex]\stackrel{\textit{\LARGE sides}}{\cfrac{x}{2}+(x-5)+(x-2)~}~ = ~~13\implies \cfrac{x}{2}+2x-7=13 \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( \cfrac{x}{2}+2x-7 \right)=2(13)}\implies x+4x-14=26\implies 5x-14=26 \\\\\\ 5x=40\implies x=\cfrac{40}{5}\implies \boxed{x=8} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(8)}{2}\implies 4\hspace{5em}(8)-5\implies 3\hspace{5em}\underset{Longest}{(8)-2\implies 6}[/tex]

[tex]5x+y=35 \implies 15x+3y=105 \\ \\ \implies (15x+3y)+(7x-3y)=105+27 \\ \\ 22x=132 \\ \\ \boxed{x=6} \implies 7(6)-3y=27 \\ \\ -3y=-15 \\ \\ \boxed{y=5}[/tex]

Step-by-step explanation:

first we need the mean value. this is the sum of all data points divided by the number of data points :

(3×9 + 5 - 3×13)/7 = -7/7 = -1

next, for the standard deviation, we need to sum the squares of the differences of the data points to the mean value, divide that by the number of data points again, and then pull the square root :

(3(-13 - -1)² + 3(9 - -1)² + (5 - -1)²)/7 =

= (3×196 + 3×100 + 36)/7 = (588 + 300 + 36)/7 =

= 924/7 = 132

the standard deviation is now the square root of 132 :

sqrt(132) = 11.48912529... ≈ 11.5

so, yes, as we can see, the numbers vary/deviate rather strongly from the mean value.

The number of large rolls that could be made from Rosa's dough is 4 numbers.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the number of large rolls that can be made from Rosa's dough, hence:

(3/4)x = 3 3/5

(3/4)x = 18/5

x = 4.8

The number of large rolls that could be made from Rosa's dough is 4 numbers.

The inequality that shows the number of shirts that Lisa buys is given as 15x + 35 ≤ 137

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Answer:

False

Step-by-step explanation:

This applies perfectly when x and y are negative.

What if both x and y are positive??

x=4, y=3.

4>3 which is true

but |4|<|3| = 4<3 ==> not true.

Hence, the answer is false

Answer:

Hello,

Answer 207

Step-by-step explanation:

[5, 29, 47, 61, 83, 225]

[5, 29, 41, 67, 83, 225]

[5, 23, 47, 61, 89, 225]

[5, 23, 41, 67, 89, 225]

[2, 5, 7, 43, 61, 89, 207]

[2, 3, 5, 47, 61, 89, 207]

[2, 3, 5, 41, 67, 89, 207]

import sys

#-------------------------------------------------------

def see(txt,arr):

   print (txt,"=",end =" ")

   for i in range(1,10 ):

       if arr[i]:

           print (i,end=" ")

   print ()

#-------------------------------------------------------

def libre(p):

   if p==0:

       return True

   if p<10:

       a=p

       if not(used[a]):

           return True

   else:

       a=p//10

       b=p%10

       if not(used[a]) and not(used[b] ):

           return True

   return False

#-------------------------------------------------------

def store(p):

# on ne stock pas 0

   global used

   #print ("in store ",p)

   if p==0:

       return

   if p<10:

       used[p]=True

   else:

       a=p//10

       b=p%10

       used[a]=True

       used[b]=True

   return            

def isSol():

   global mini,k,mini_set,all_mini_set

   for i in range(1,10):

       if not used[i]:

           return False

   s=i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9

   if s <= mini[0]:

       if s < mini[0]:

           k=k+1

       mini[0]=s

       mini[1]=i_1

       mini[2]=i_2

       mini[3]=i_3

       mini[4]=i_4

       mini[5]=i_5

       mini[6]=i_6

       mini[7]=i_7

       mini[8]=i_8

       mini[9]=i_9

       a_set=sorted(set([i for i in mini if i >0] ))

       if (a_set not in all_mini_set):

           mini_set=a_set.copy()

           all_mini_set.append(a_set)

           print (a_set)

       #print (mini,end="")

       """

       see("used ",used)

       see("sav_9",sav_9)

       see("sav_8",sav_8)

       see("sav_7",sav_7)

       see("sav_6",sav_6)

       see("sav_5",sav_5)

       see("sav_4",sav_4)

       see("sav_3",sav_3)

       see("sav_2",sav_2)

       see("sav_1",sav_1)

       """

       #if k>2:

       #    sys.exit(0)

   return True

#-------------------------------------------------------

#╚ 0 : ne pas utiliser

jeu=[

    [],

    [0,13, 17, 19, 31, 41, 61, 71],

    [0,2,23,29],

    [0,3,13,23,31,37,43,53,73,83],

    [0,41,43,47],

    [0,5,53,59],

    [0,61,67],

    [0,7,17,37,47,67,71,73,79,97],

    [0,83,89],

    [0,19,29,59,79,89,97]

    ]

print ("-------------------------------------------")

for i in range(10):

   print (jeu[i])

print ("-------------------------------------------")

mini=[1000]+[0 for i in range(1,10) ]

mini_set={}    

all_mini_set=[]

used=[False for i in range(10 )]

sav_1=used.copy()

k=0

for i_1 in jeu[1]:

   used=sav_1.copy()

   if libre(i_1):

       store(i_1)

       sav_2=used.copy()

       for i_2 in jeu[2]:

           used=sav_2.copy()

           if libre(i_2):

               store(i_2)

               sav_3=used.copy()

               for i_3 in jeu[3]:

                   used=sav_3.copy()

                   if libre(i_3):

                       store(i_3)

                       sav_4=used.copy()

                       for i_4 in jeu[4]:

                           used=sav_4.copy()

                           if libre(i_4):

                               store(i_4)

                               sav_5=used.copy()

                               for i_5 in jeu[5]:

                                   used=sav_5.copy()

                                   if libre(i_5):

                                       store(i_5)

                                       sav_6=used.copy()

                                       for i_6 in jeu[6]:

                                           used=sav_6.copy()

                                           if libre(i_6):

                                               store(i_6)

                                               sav_7=used.copy()

                                               for i_7 in jeu[7]:

                                                   used=sav_7.copy()

                                                   if libre(i_7):

                                                       store(i_7)

                                                       sav_8=used.copy()

                                                       for i_8 in jeu[8]:

                                                           used=sav_8.copy()

                                                           if libre(i_8):

                                                               store(i_8)

                                                               sav_9=used.copy()

                                                               for i_9 in jeu[9]:

                                                                   used=sav_9.copy()

                                                                   if libre(i_9):

                                                                       store(i_9)

                                                                       isSol()

                                                               used=sav_9.copy()