please help asap only question i don’t know

Answer: a. 1

b. 12

c. 2x+11

Step-by-step explanation:

f(x) = 2x+5

a. f(-2)=

2(-2)+5=

-4+5=1

b. f(1)+5=

2(1)+5+5=

2+5+5=12

c. f(x+3)=

2(x+3)+5=

2x+6+5=2x+11

Joy was charged the labor of 11 hours .

Joy was charged a total of $1215 from the landscaper for listed parts and labor.

He was charged $445 for listed parts

He was charged the rest for labor

The hourly rate for labor is $ 70

Let the number of hours worked by labor be x

so, we can form a linear equation here as 445 + 70x = 1215

Now we have to solve this linear equation

445 + 70x = 1215

70x= 1215 - 445

70x = 770

x = 770/ 70

x = 11

Therefore, 11 hours of labor were needed to complete the job.

The correct question is Joy recently hired a landscaper to do some necessary work. On the final bill, Joy was charged a total $1215.  $445 was listed for parts and the rest for labor. If the hourly rate for labor was $70, how many hours of labor were needed to complete the job?

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A composite function exists as a function that depends on another function. The values for f(g(x)) = 4x-7 and g(f(x)) = 4x -28

In mathematics, function composition exists as an operation ∘ that takes two functions f and g, and has a function h = g ∘ f such that h(x) = g. In this operation, the function g exists used to the result of applying the function f to x.

Function composition is a mathematical procedure where two functions, f, and g, are taken and a function, h = g ∘ f, is produced such that h(x) = g(f(x)). The function g exists used in this operation on the outcome of applying the function f to x.

Let the two functions be f(x) = -x-7 and g(x) = 4 x

f(g(x)) = f(4x)

f(4x) = 4x-7

g(f(x)) = g(x-7)

simplifying the above equation, we get

g(x-7) = 4 (x - 7) = 4x -28

g(x-7) = 4x - 28

Therefore, the value of f(g(x)) = 4x-7 and g(f(x)) = 4x -28

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Answer:

h(x) = 3x^2 - x^3 + 10x - 7 == 2

Step-by-step explanation:

Step-by-step explanation:

BQS=78

BQN=48

BQS+BQN=78+48=126

180-126=54°

NQS=54°

comfirm=BQN+BQS+NQS=180

48+78+54=180

Answer:

30°

Step-by-step explanation:

∠NQS + ∠BQN = ∠BQS

Knowing this information, we use the known measures to find ∠NQS

∠NQS + 48° = 78°

∠NQS = 78° - 48°

∠NQS = 30°

Thus, angle NQS is 30 degrees.

Hope this helps!

x = 5 is not an extraneous solution of the equation.

✓(2x + 6) - ✓(x - 1) = 2

✓(2x + 6) = 2 + ✓(x - 1)

Squaring both the sides of equation, we get

(✓(2x + 6))² = (2 + ✓(x - 1))²

2x + 6 = (2)² + (✓(x - 1))² + 2(2)(✓(x - 1))

2x + 6 = 4 + x - 1 + 4(✓(x - 1))

x + 3 = 4✓(x - 1)

Squaring again on both the side of the equation, we get

(x + 3)² = (4✓(x - 1))²

x² + 3² + 2×3×x = 16(x - 1)

x² + 9 + 6x = 16x - 16

x² - 10x + 25 = 0

By applying middle term splitting on equation (1), we get

x² - 5x - 5x + 25 = 0

x(x - 5) -5(x - 5) = 0

(x - 5) (x - 5) = 0

x = 5 and 5.

Check for extraneous solutions by putting value of x in original equation -

✓(2×5 + 6) - ✓(5 - 1) = 2

✓16 - ✓4 = 2

4 - 2 = 2

2 = 2

LHS = RHS

x = 5 is not an extraneous solution of the equation.

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Answer:

D= (-3,1]

R= [0,-4]

Step-by-step explanation:

[,] <-- used for when the point is included in the answer (closed circle or greater than/less than or equal to sign)

(,) <-- used for when the answer does not include point (open circle or greater than/less than sign   not equal to)

Plug in the points where shown on each line.

Hope that helps a bit!

The probability that 3 numbers chosen with replacement contain both even and odd is 0.75 while the probability that 3 numbers chosen without replacement contain both even and odd is 0.8.

The numbers include 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16

Case 1: with replacement

Let P (E) = probability of chosing even number is

P (E) = (Required outcome)/(Total outcome)

Required outcome = numbers of even number

Total outcome = total numbers

P (E) = 8/16

P (E) = 1/2

Let P (o) = probability of chosen odd number is

P (o) = (Required outcome)/(Total outcome)

Required outcome = numbers of odd number

Total outcome = total numbers

P (o) = 8/16

P (o) = 1/2

Since 3 numbers are chosen, there are 6 possible combination which are OEE, EOE, OEO, OOE EEO, and EOO

P (chosen 3 numbers) = P (OEE) + P (EOE) + P (OEO) + P (OOE) + P (EEO) + P (EOO)

P (OEE) =P(O)×P(E)×P(E)  

P (OEE) =1/2×1/2×1/2

P (OEE) =1/8

P (EOE) =P(E)×P(O)×P(E)  

P (EOE) =1/2×1/2×1/2

P (EOE) =1/8

P (OEO) =P(O)×P(E)×P(O)  

P (OEO) =1/2×1/2×1/2

P (OEO) =1/8

P (OOE) =P(O)×P(O)×P(E)  

P (OOE) =1/2×1/2×1/2

P (OOE) =1/8

P (EEO) =P(E)×P(E)×P(O)  

P (EEO) =1/2×1/2×1/2

P (EEO) =1/8

P (EOO) =P(E)×P(O)×P(O)  

P (EOO) =1/2×1/2×1/2

P (EOO) =1/8

P (chosen 3 numbers)=1/8+1/8+  1/8+1/8+1/8+1/8

P (chosen 3 numbers)=6(1/8)

P (chosen 3 numbers)=3/4

P (chosen 3 numbers)=0.75

Case 2: without replacement, when the first number is chosen there is a reduction in the amount of the remaining numbers.

Let P (E) = probability of chosen even number is

P (E) = (Required outcome)/(Total outcome)

Required outcome = numbers of even number

Total outcome = total numbers

P (E) = 8/16

P (E) = 1/2

Let P (o) = probability of chosing odd number is

P (o) = (Required outcome)/(Total outcome)

Required outcome = numbers of odd number

Total outcome = total numbers

P (o) = 8/16

P (o) = 1/2

Since 3 numbers are chosen, there are 6 possible combination which are OEE, EOE, OEO, OOE EEO, and EOO

P (chosen 3 numbers) = P (OEE) + P (EOE) + P (OEO) + P (OOE) + P (EEO) + P (EOO)

P (OEE) =P(O)×P(E)×P(E)  

P (OEE) =8/16×8/15×7/14

P (OEE) =1/2×8/15×1/2

P (OEE) =8/60

P (OEE) =2/15

P (EOE) =P(E)×P(O)×P(E)  

P (EOE) =8/16×8/15×7/14

P (EOE) =1/2×8/15×1/2

P (EOE) =8/60

P (EOE) =2/15

P (OEO) =P(O)×P(E)×P(O)  

P (OEO) =8/16×8/15×7/14

P (OEO) =1/2×8/15×1/2

P (OEO) =8/60

P (OEO) =2/15

P (OOE) =P(O)×P(O)×P(E)  

P (OOE) =8/16×7/15×8/14

P (OOE) =1/2×7/15×4/7

P (OOE) =28/210

P (OOE) =2/15

P (EEO) =P(E)×P(E)×P(O)  

P (EEO) =8/16×7/15×8/14

P (EEO) =1/2×7/15×4/7

P (EEO) =28/210

P (EEO) =2/15

P (EOO) =P(E)×P(O)×P(O)  

P (EOO) =8/16×8/15×7/14

P (EOO) =1/2×8/15×1/2

P (EOO) =8/60

P (EOO) =2/15

P (chosen 3 numbers)=2/15+2/15+  2/15+2/15+2/15+  2/15

P (chosen 3 numbers)=6(2/15)

P (chosen 3 numbers)=4/5

P (chosen 3 numbers)=0.8

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Answer:

23 and 26? which numbers would you like to know?

Step-by-step explanation:

It should be noted that the graph which will correctly show the number of pages which Louise will have left to read after the number of days is D. A graph titled How much Louise has left to read has days on the x-axis and pages on the y-axis. A line goes through (0, 300) and (2, 270).

It should be noted that he a graph is a diagram that shows the relationships between two or more things.

In this case, it can be seen that Louise is reading a 300-page book and she reads 15 pages every day.

Therefore, the graph that correctly displays the number of pages which Louise has left to read after the number of days is a graph titled How Much Louise Has Left to Read has days on the x-axis and pages on the y-axis. A line goes through (0, 300) and (2, 270).

In conclusion, the correct option is D.

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The cost per cup of milk as required in the task content is; $0.14375.

The cost per fluid ounce of milk as required in the task content is; $0.018.

The cost per quart of milk as required in the task content is; $0.575.

It follows from the task content that the cost of each unit measure of milk is to be determined. Therefore, since the cost per gallon of milk is; $2.30.

1). Since there are 16 cups of milk per gallon, it follows that the cost per cup of milk is; 2.30/16 = $0.14375

2). Also, Since there are 128 fluid ounces of milk per gallon, it follows that the cost per fluid ounce of milk is; 2.30/128 = $0.018

3). Also, Since there are 4 quarts of milk per gallon, it follows that the cost per quart of milk is; 2.30/4 = $0.575

Ultimately, the cost per unit measure of milk in each case scenario has been determined and is as represented above.

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After solving the expression 3-√(x-3)=x algebraically, we get the result as  3.

Isolate the square root from the left hand side √x-3 = 3-x

Eliminate the radical on the left hand side by squaring on both sides-

Both sides must be raised to the second power.    

(√x-3)² = (3-x)²  

After squaring we get       

x-3 = x²-6x+9

Solving the quadratic equation-

We get rearranged equation as -

x²  - 7x  + 12 = 0

(x-4)(x-3)=0

x=4,3

Check if the first solution is correct by putting it in original equation-

putting x=3 in 3-√(x-3)=x  we get 3-0=3

Hence 3 is correct.

Check if the second solution is correct by putting it in original equation-

putting x=4 in  3-√(x-3)=x  we get 3-1=4 which is not true

hence 4 is rejected.

Therefore, after solving the expression 3-√(x-3)=x algebraically, we get the result as  3.

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Step-by-step explanation:

so, MN + NO = MO

MO = 50 miles

MN = 45 miles

therefore,

45 + NO = 50

NO = 50 - 45 = 5 miles

The probability that the student will answer at least 10 questions correctly from a multiple-choice examination that has 15 questions, each with five possible answers and only one of which is correct is 0.00

Let P (C) = probability of a student picking a correct option is

P (C) = (Required outcome)/(Total outcome)

Required outcome = numbers of correct option

Total outcome = total numbers of option

P (C) = 1/5

Recall that

P (C) + P (C1) = 1

 Where P (C1) is the probability of picking wrong option

P(C)+P(C^1 )=1

1/5+P(C^1 )=1

P(C^1 )=1-1/5

P(C^1 )=4/5

The probability of answering at least 10 questions correctly can be solved by using Binomial probability which states that

P(x)=(nCx) (p)^x (q)^(n-x)

P(x) is the probability of an event

n is the number of trial (15)

x number of specific outcome

p is the probability that the event will happen (1/5)

q is the probability that the event will not happen (4/5)

      Probability of answering at least 10 questions correctly will be

P(10) + P(11) + P(12) + P(13) +P(14) + P(15)

P(x)=(nCx) (p)^x (q)^(n-x)

P(10)=(15C10) (1/5)^10 (4/5)^(15-10)

P(10)=(15C10) (1/5)^10 (4/5)^5

15C10=15!/((15-10)!(10!))

15C10=15!/((5)!(10!))

15C10=1307674368000/((120)(3628800))

15C10=1307674368000/43545600

15C10=3003

P(10)=(3003) (0.2)^10 (0.8)^5

P(10)=(3003)(1.024×〖10〗^(-7))(0.32768)

P(10)=0.0001

P(11)=(15C11) (1/5)^11 (4/5)^(15-11)

P(11)=(15C10) (1/5)^11 (4/5)^4

15C11=15!/((15-11)!(11!))

15C11=15!/((4)!(11!))

15C11=1307674368000/((24)(39916800))

15C11=1307674368000/958003200

15C11=1365

P(11)=(1365) (0.2)^11 (0.8)^4

P(11)=(1365)(2.048×〖10〗^(-8))(0.4096)

P(11)=0.000011

P(12)=(15C12) (1/5)^12 (4/5)^(15-12)

P(12)=(15C12) (1/5)^12 (4/5)^3

15C12=15!/((15-12)!(12!))

15C12=15!/((3)!(12!))

15C12=1307674368000/((6)(479001600))

15C12=1307674368000/2874009600

15C12=455

P(12)=(455) (0.2)^12 (0.8)^3

P(12)=(455)(4.1×〖10〗^(-9))(0.512)

P(12)=0.0000009542

P(13)=(15C13) (1/5)^13 (4/5)^(15-13)

P(13)=(15C13) (1/5)^13 (4/5)^2

15C13=15!/((15-13)!(13!))

15C13=15!/((2)!(13!))

15C13=1307674368000/((2)(6227020800))

15C13=1307674368000/12454041600

15C13=105

P(13)=(105) (0.2)^13 (0.8)^2

P(13)=(105)(8.19×〖10〗^(-10))(0.64)

P(13)=0.00000005505

P(14)=(15C14) (1/5)^14 (4/5)^(15-14)

P(14)=(15C14) (1/5)^14 (4/5)^1

15C14=15!/((15-14)!(14!))

15C14=15!/((1)!(14!))

15C14=1307674368000/((1)(87178291200))

15C14=1307674368000/87178291200

15C14=15

P(14)=(15) (0.2)^14 (0.8)^1

P(14)=(15)(1.638×〖10〗^(-10))(0.8)

P(14)=0.00000000197

P(15)=(15C15) (1/5)^15 (4/5)^(15-15)

P(15)=(15C15) (1/5)^15 (4/5)^0

15C15=15!/((15-15)!(15!))

15C15=15!/((0)!(15!))

15C15=1307674368000/((1)(1307674368000))

15C15=1307674368000/1307674368000

15C15=1

P(15)=(1) (0.2)^15 (0.8)^0

P(15)=(1)(3.277×〖10〗^(-13))(1)

P(15)=3.277×〖10〗^(-13)

Probability of answering at least 10 questions correctly will be

P(10) + P(11) + P(12) + P(13) +P(14) + P(15)

P (at least 10)=0.0001+0.000011+ 0.0000009542+0.00000005505+0.00000000197+ 3.277×〖10〗^(-13)

P (at least 10)=0.00011250667

Approximately

P (at least 10)=0.0001

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Answer:

Probably puberty

The values of x and y of the parallelogram PQRS are x = 2 and y = 11

We're given the dimensions of parallelogram PQRS as;

PT = y

TR = 5x + 1

QT = 2y

TS = 6x + 10

T is the intersection of the two diagonals PR and QS and so the diagonals bisect each other.

The diagonal PR is cut into PT and TR, both of which are congruent or equal. Thus, PT = TR.

Similarly, the other diagonal QS is split in half as well. The two equal pieces are QT and TS. So QT = TS.

PT = TR

y = 5x + 1

and

QT = TS

2y = 6x + 10

Plugging the values gives;

2y = 6x + 10

2(y) = 6x + 10

2(5x + 1) = 6x + 10

2*5x + 2*1 = 6x + 10

10x + 2 = 6x + 10

10x + 2 - 6x = 6x + 10 - 6x

4x + 2 = 10

4x + 2 - 2 = 10 - 2

4x = 8

4x/4 = 8/4

x = 2

If x = 2, then y is...

y = 5x+1

y = 5*x+1

y = 5*2+1

y = 10+1

y = 11

Thus, the values of x and y of the parallelogram PQRS are x = 2 and y = 11

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Answer:

Step-by-step explanation:

Step-by-step explanation:

6 = 5/2x + k

6 = ( 5/2 . 2/1) + k

6 = 10/2 +k

6-5 = k

k = 1

Answer:

0.99 or 99%

Step-by-step explanation:

This is actually much more simple than it sounds

Take 95% and add 80% of 5 because 5% of the time the QK piece will not work. 80% of 5 is 4. 95+4=99

The expression (-1 ,-5) is not a solution to the inequality 4x + y < -3

The inequality is given as

4x + y < -3

The solution is given as

(-1, 5)

Substitiute (-1, 5) for (x, y) in the inequality expression 4x + y < -3

So, we have

4(-1) + 5 < -3

Open the bracket

-4 + 5 < -3

Evaluate the sum

1 < -3

The above inequality is false

Hence, (-1 ,-5) is not a solution to the inequality 4x + y < -3

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x = 512 is not an extraneous solution of the equation.

[tex] \sqrt[3]{x} [/tex] = 8

Cubing both the side of the equation, we get

[tex] { \sqrt[3]{x} }^{3} [/tex]

= 8³

x = 512

Check for extraneous solution by putting the value x = 512 in the original equation

[tex] \sqrt[3]{x} [/tex] = 8

[tex] \sqrt[3]{512} [/tex] = 8

8 = 8

⇒ LHS = RHS

∴ x = 512 is not an extraneous solution of the equation.

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We line up the decimal point so that the place value of other number does not changes.

Adding and subtracting decimal numbers is the same as adding and subtracting whole numbers. The most important thing to remember is to align the decimal places directly on top of each other when stacked. This stacks the digit values ​​so that 1 is above 1, , the tens on the tens, the tenths on the tenths,, and so on. Note that if a number does not have the same number of decimal places as other numbers, you can pad all the digits to the right of the decimal point with zeros. Numerically, this looks like 3.15 = 3.150 = 3.150000000.

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Answer: [tex]f\left(x\right)\le \:-4[/tex]

Step-by-step explanation:

[tex]\begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:-4\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-4]\end{bmatrix}[/tex]

Speed in still water= [tex]\sqrt{43}[/tex]miles

Let take the speed in still water be x and  the speed of current in the river be  y.

The speed of the current in a river= 5 miles

Let the distance be s which is 1.2 miles.

The time(t) difference is 40 minutes.

We have an equation,

[tex]\frac{s}{x-y}[/tex] - [tex]\frac{s}{x+y}[/tex] = t

[tex]\frac{1.2}{x - 5}[/tex] - [tex]\frac{1.2}{x + 5}[/tex] = [tex]\frac{40}{60}[/tex]

1.2 × 60(2 × 5) = 40([tex]x^{2}[/tex] - 25)

x =[tex]\sqrt{43}[/tex]

Therefore the speed in still water is [tex]\sqrt{43}[/tex].

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Answer:

15

Step-by-step explanation:

Answer:

The number is 1

Step-by-step explanation:

Let the number be x
Subtract 8 from x ==>   x - 8

Multiply by -3 ==>         -3(x - 8) = -3(x) -3(-8) = -3x - 24

This equals 21

So
-3x + 24 = 21

Subtract 24 from both sides:
-3x + 24 - 24 = 21 - 24

==> -3x + 0 = - 3

-3x = -3

Divide by -3

==> -3x/-3 = -3/-3

x = 1

Check

1 - 8 = -7

-3(1-8) = -3 x -7 = 21

The calorie serving in simplest form is 56.5

The rate in mentioned question will be number of calories present per serving.  

So writing the given information in the form of rate -

Rate of calorie per serving = 452/8

Now, to convert the rate of calories in simplest form, we will perform division of both numerator and denominator of fraction by a common number. We can directly divide the numerator and denominator of fraction with 8. Else, we can perform division by simpler numbers too.

Indirect method -

Lets begin the division by 4

Rate of calorie per serving = 113/2

Further dividing by 2

Rate of calorie per serving = 56.5

Direct method will also give the same result, which is rate of calorie per serving = 56.5.

Hence, in simplest form (decimal) the rate of calorie per serving is 56.5.

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On multiplying and then simplifying 3∛2x² . 7 ∛32x⁴, we get, 84x⁶

1. Split the radicands in half. A number known as a radicand appears underneath the radical symbol.

2. To multiple radicands, multiply the integers as if they were whole numbers. Always keep the item beneath a single radical sign.

3. Use factoring to exclude any perfect squares from the radicand. To achieve this, check whether any perfect squares have the radicand as a factor.

4. Position the square root of the ideal square before the radical sign. Don't change the second component of the radical symbol.

5. Square the square's root. Sometimes it is necessary to multiply a square root by itself.

Given,

3∛2x² . 7 ∛32x⁴

⇒ [tex]21 (2^{1/3} ) x^{2+4} . 32^{1/3}[/tex]

⇒ [tex]21 (2^{1/3} ) x^{6} . 2^{5/3}[/tex]

⇒ [tex]21 (2^{1/3 + 5/3} ) x^{6}[/tex]

⇒ [tex]21 (2^{2} ) x^{6}\\84x^{6}[/tex]

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The solution for the compound inequality is 4 < b ≤ 5.

It is an expression mathematical that represents a non-equal relationship between a number or another algebraic expression. Therefore, it is common the use following symbols: ≤ (less than or equal to), ≥ (greater than or equal to), < (less than), and > (greater than).

The solutions for inequalities can be given by: a graph in a number line or numbers.

The question gives:

3b + 2 < 5b - 6 ≤ 2b + 9

You can write the previous compound inequality as:

3b + 2 < 5b - 6 (1)

5b - 6 ≤ 2b + 9 (2)

         3b + 2 < 5b - 6

         3b-5b < -6 -2

          -2b < -8

            2b >8

            b >4

         5b - 6 ≤ 2b + 9

         5b-2b  ≤ 9+6

          3b ≤ 15

          b ≤ 5

Combining the intervals, you have 4 < b ≤ 5.

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1.  The probability given as P(Y∩X) is; 2/25

2. The probability given as P(Y) * P(X) is; 4/15

We are given the following probabilities;

P(X) = 2/3 , P(Y) = 2/5 , and P(X|Y) = 1/5

1. We want to find P(Y ∩ X).

This means probability of the set intersection of Y and X

Since  P(X) = 2/3 , P(Y) =2/5 , and P(X|Y) =1/5, it means that we have a conditional probability.

Thus;  P(Y∩X) = P(Y) * P(X|Y)

P(Y∩X) = 2/5 × 1/5

P(Y∩X) = 2/25

2. We want to find P(Y) * P(X)

Plugging in the relevant values gives;

P(Y) * P(X) = 2/3 × 2/5

P(Y) * P(X) = 4/15

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Complete question is;

For two events X and Y, P(X) =2/3 , P(Y) =2/5 , and P(X|Y) =1/5 . Find the probabilities.

1.  P(Y∩X) = ____

2. P(Y)· P(X) =____

The new mean 101.5 is greater than the previous mean 94.

By adding up all the numbers in a data collection and dividing by the total number of values in the set, the mean (average) of the data set can be determined. The median is the middle when a data set is sorted from least to greatest.

The distribution of a basketball team's score with a mean of 94 points per game and a high of 109 for the first 20 games.

Mean = 94,

Points per game = 94, 110, 102, 93, 98, 99, 121, 96, 94, and 108 points.

Therefore,

The new mean = ( sum of observations) / Number of observations

New mean = ( 94 + 110 + 102 + 93 + 98 + 99 + 121 + 96 + 94 + 108) / 10

New mean = 1015 / 10

New mean = 101.5

101.5 > 94

Now, the new mean is greater than the previous mean.

Hence, the correct option is (A).

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