Can someone help me please?
A. 0.43, because 13 of the 30 days were cloudy.
B.0.35, because 13 of the 37 days were cloudy.
C.0.22, because 13 of the 60 days were cloudy.
D.0.80, because 24 of the 30 days were cloudy.

Answer:

  (2)  A=4, B=0, C=9 . . . (4x = 9)

Step-by-step explanation:

A linear equation in standard form has mutually prime integer coefficients with a positive leading coefficient.

The values that would be the appropriate values for an equation to be in standard form (Ax + By = C) is A = 4, B = 0 and C = 9.

The linear equation is the equation which has the coefficient as the mutual prime numbers and the coefficient must be the positive leading coefficient.

From the choices,

Therefore, the values that would be the appropriate values for an equation to be in standard form as it lead to positive leading coefficient and mutual prime is A = 4, B = 0 and C = 9.

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Answer:

w = 12

Step-by-step explanation:

We are given the equation:

[tex]\displaystyle{\dfrac{w-2}{4} = \dfrac{2w+1}{10}}[/tex]

First, clear out the denominators by multiplying both sides by LCM. In this scenario, our LCM is 40. Thus, multiply both sides by 40:

[tex]\displaystyle{\dfrac{w-2}{4} \cdot 40= \dfrac{2w+1}{10} \cdot 40}[/tex]

Simplify the expressions/equations:

[tex]\displaystyle{(w-2)\cdot 10= (2w+1)\cdot 4}\\\\\displaystyle{10w-20= 8w+4}[/tex]

Isolate w-variable:

[tex]\displaystyle{10w-20= 8w+4}\\\\\displaystyle{10w-8w=4+20}\\\\\displaystyle{2w=24}\\\\\displaystyle{w=12}[/tex]

Hence, the solution is w = 12

If you have any questions regarding my answer or explanation, do not hesitate to ask away in comment!

Answer:

0.772

Step-by-step explanation:

Original equation:

[tex]18 * 2^{5t}=261[/tex]

Divide both sides by 18

[tex]2^{5t} = 14.5[/tex]

Rewrite in logarithmic form ([tex]b^x=c \implies log_bc=x[/tex])

[tex]log_214.5 = 5t[/tex]

Divide both sides by 5

[tex]\frac{log_214.5}{5}=t[/tex]

Rewrite the equation so that base is 10 using change of base formula: [tex]log_ba = \frac{log_na}{log_nb}[/tex]

[tex]\frac{(\frac{log14.5}{log2})}{5}=y[/tex]

Keep, change, flip

[tex]\frac{log14.5}{log2}*\frac{1}{5} = \frac{log14.5}{5 * log2}[/tex]

Use a calculator to approximate log14.5 and log2

[tex]\frac{1.161368}{5 * 0.301029996} = t[/tex]

Multiply in denominator

[tex]\frac{1.161368}{1.505149978} = t[/tex]

Divide two values

[tex]t\approx 0.771596[/tex]

Round to nearest thousandth

[tex]t\approx 0.772[/tex]

Answer:

A = 72, P = 38

Step-by-step explanation:

Area can be found by getting the area of the overall figure, and then subtracting the blank space in the corner

12 * 7 = 84

3 * 4 = 12(invisible rectangle in top right)

84 - 12 = 72

Perimeter can be found by adding all of the sides

12 + 7 + (12-4) + 3 + 4 + (7-4)

The numbers in brackets are the unlabeled sides on the top and left.

Answer:

[tex]P=38\\ A=72[/tex]

Step-by-step explanation:

To find the perimeter (P), add up all of the side lengths of the figure:

(numbers in parenthesis are the unlabeled values in the figure).

[tex]12+7+(8)+3+4+(4)=38[/tex]

To find the area of the figure (A), multiply the length and width of the original rectangle [tex](12*7=84)[/tex]. Then, subtract this amount (84) minus the area of the missing rectangle in the top left corner [tex](4*3=12)[/tex].

Subtract the required values: [tex]84-12=72[/tex].

Therefore, the perimeter of the figure is 38 units, and the area of the figure is 72 units.

Answer: 1. A

2. A

Step-by-step explanation:

The value of 'x' from the expression is 1/3

Given the expression;

[tex]x + \frac{x}{3} = \frac{4}{9}[/tex]

Find the LCM of the left side, we have

[tex]\frac{3x + x}{3} = \frac{4}{9}[/tex]

Cross multiply

[tex]9(4x) = 4 *3[/tex]

[tex]36x = 12[/tex]

Make 'x' the subject

[tex]x = \frac{12}{36}[/tex]

x = [tex]\frac{1}{3}[/tex]

Thus, the value of 'x' from the expression is 1/3

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Hello !

Answer:

[tex]\boxed{\sf x=\dfrac{1}{3} }[/tex]

Step-by-step explanation:

Our aim is to find the value of x that verifies the following equation :

[tex]\sf x+\frac{x}{3} =\frac{4}{9}[/tex]

Let's isolate x :

Multiply both sides by 3 :

[tex]\sf 3(x+\frac{x}{3} )=\frac{4}{9}\times 3\\ \sf 3x+x=\frac{4}{3}[/tex]

Now we can combine like terms :

[tex]\sf 4x=\frac{4}{3}[/tex]

Finally, let's divide both sides by 4 :

[tex]\sf \frac{4x}{4} =\frac{4}{3} \times \frac{1}{4} \\\boxed{\sf x=\dfrac{1}{3} }[/tex]

Have a nice day ;)

The sum of the series is: (c) lim n⇒∝ 1/3(1 - (1/4)^n)

The complete question is added as an attachment

The given series is a geometric series. So, we first calculate the common ratio (r) using:

r = T2/T1

From the series, we have:

T1 = 1/4 and T2 = 1/16

Substitute the known values in the above equation

r = (1/16)/(1/4)

So, the equation becomes

T = 1/16 / 1/4

Rewrite as product

T = 1/16 * 4

Evaluate the product

r = 1/4

The formula to calculate the sum of a geometric series of is:

Sn = a(1 - r^n)/(1 - r)

Where

a = 1/4 -- the first term

r = 1/4 --- the common ratio

Substitute the known values in the above equation

Sn = 1/4 * (1 - (1/4)^n)/(1 - 1/4)

Simplify the denominator

Sn = 1/4 * (1 - (1/4)^n)/(3/4)

Divide 1/4 by 3/4

Sn = 1/3 * (1 - (1/4)^n)

Take the limit to infinity of the series

Sn = lim n⇒∝ 1/3(1 - (1/4)^n)

We can conclude that, the sum of the series is: (c) lim n⇒∝ 1/3(1 - (1/4)^n)

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Answer:

Step-by-step explanation:

B

The percentage of tardiness among the 102 pupils is 2.94.

According to the query,

Course A: Total number of students = 102

                   Number of tardy students = 3

Percent of students tardy in course A = (number of tardy students/total  

                                                                   number of students) * 100

  =(3/102)*100

  = 2.941

  ≈2.94

In case of course B: Total number of students = 85 & tardy student = 5

So, the percent of tardy student = (5/85)*100 = 5.88 %

Therefore, it is concluded that the percentage of student tardiness in course A is 2.94.

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Answer:

there should have been names of every line and joining parts

Similarities and differences between solving an absolute value equation and solving an absolute value inequality are shown below.

Similarities and differences between solving an absolute value equation and solving an absolute value inequality:

1) The absolute value equation of a number is simply the number's distance from zero.

2) Absolute value inequalities and linear inequalities share the fact that they both have two variables and so require a second equation to obtain the variables.

Therefore, the similarities and differences between solving an absolute value equation and solving an absolute value inequality are shown.

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The solution to the given inequality is y < 5

Inequality are equations  not separated by an equal sign. Given the following parameters

15 > y + 10

Subtract 10 from both sides

15 - 10 > y + 10 - 10

5 > y

Swap

y < 5

Hence the solution to the given inequality is y < 5

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The answers have been shown below.

Questions regarding the time spent by Jack jogging and bike riding are needed to be answered.

(A) The equations are [tex]x+y=75, y=15+x[/tex]

Time spent jogging is 30 minutes

The total time would be [tex]45+60=105[/tex] minutes which is not equal to 75 minutes.

(B) Let [tex]x[/tex] be the time spent jogging.

[tex]y[/tex] be the time spent bike riding.

[tex]x+y=75\\y=15+x\\x+15+x=75\\2x+15=75\\x=\frac{75-15}{2} =30[/tex]

Time spent jogging is 30 minutes.

[tex]y=60\\x+y=75[/tex]

(C) If he rides his bike 15 minutes longer than he jogs then he would have to jog [tex]60-15 = 45[/tex] minutes.

Therefore, the total time would be [tex]45+60=105[/tex] minutes which is not equal to 75 minutes.

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The statement that would most likely have a negative linear correlation coefficient is amount of money spent on baby food as a child ages

Correlation coefficient is the value that relates two variables in question. Correlation can be positive, negative or no correlation

Note that the correlation coefficients are only values from 0 and 1. The statement that would most likely have a negative linear correlation coefficient is amount of money spent on baby food as a child ages

The amount of money spent on a baby is never a function of its age. The baby will age regardless.

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Combining the like terms, the simplified polynomials are given as follows:

a) 4x² - 14x + 17

b) -5x² - 20x + 8

Polynomials are simplified combining the like terms, that is, adding these numbers with the same variable.

Item a:

4(x - 2)(x + 1) - 5(2x - 5)

Applying the distributive property:

4(x² - x - 2) - 10x + 25

4x² - 4x - 8 - 10x + 25

Combining the like terms:

4x² - 4x - 10x - 8 + 25

4x² - 14x + 17

Item b:

-5(x + 2)² + 28

-5(x² + 4x + 4) + 28

-5x² - 20x - 20 + 28

-5x² - 20x + 8

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he has 231 pennies..............

Step-by-step explanation:

1. Not neccesarily true

2. Not necessarily true

3. True because angles in a triangle add to 180°

4. Not necessarily true

5. True because the medians are being constructed, and the three medians of a triangle are always concurrent.

6. Not necessarily true

The statements that are true about the triangle are:

Option C: m∠P + m∠Q + m∠R = 180°

Option E: PT, QU, and RS intersect at the same point.

1. m∠Q = m∠R: This is not true because there is no indication that the angles are equal.

2.The length of QU is half the length of RP: This is not true because there is no length given to show that measurement.

3. m∠P + m∠Q + m∠R = 180°:

This is true because the sum of angles in a triangle add to 180°

4. QU ≅ RS:

This is not true because we are not told that they are congruent

5. PT, QU, and RS intersect at the same point:

This is true because the medians are being constructed, and the three medians of a triangle are always concurrent.

6. The sum of the lengths of QU and RS is equal to the length of PT: This is not true because we are not told that.

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Answer:

See below

Step-by-step explanation:

To pay the loan off more quickly.

To pay less overall interest.

Often get a lower interest rate for a shorter term loan.

Answer:

y = -⅔x + 11/3

Step-by-step explanation:

The slope is

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

Substituting into point-slope form,

y - 3 = -⅔(x - 1)

y - 3 = -⅔x + ⅔

y = -⅔x + 11/3

The expression that must be true for a to be parallel to b are mAngle2 = 58Degrees and <1 = 4x - 10

A line is the shortest distance between two points and the point where two lines meet is known as an angle

From the figure shown

m<2 = 58 degrees (alternate exterior angle)

Since the sum of angles on a straight line is 180 degrees, hence;

3x-1+<2 + 4x - 10= 180

3x-1 + 58 + 4x - 10 =180

7x + 47 = 180

7x = 180- 47

7x = 133

x = 19

Hence the expression that must be true for a to be parallel to b are mAngle2 = 58Degrees and <1 = 4x - 10

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Answer:

angle 2=58 degrees

angle 1=4x-10 degrees

Step-by-step explanation:

The polynomial approximate values are

a. The linear approximating polynomial is p1(x) = 2 - x

b.The quadratic approximating polynomial is  p2(x) = x² - 3*x + 3

c. The approximating polynomial value is p1(0.97) = 1.03; p2(0.97) = 1.0309

According to the statement

we have given that the f(x) = 1/x

And we have to find that polynomial approximate values which are written below

The linear approximating polynomial And quadratic approximating polynomial And approximate the given quantity of polynomials obtained in parts a. and b.

So, the given function is f(x) = 1/x

And

f'(x) =  -1/x²

f''(x) = 2/x³

a = 1

Now, we find the linear approximating polynomial

So,

a. The linear approximating polynomial is:

p1(x) = f(a) + f'(a)*(x - a)

p1(x) = 1/1 + -1/1² * (x - 1)

p1(x) = 1 - x + 1

p1(x) = 2 - x

And Now, we find the quadratic approximating polynomial

So,

b. The quadratic approximating polynomial is:

p2(x) = p1(x) + 1/2 * f''(a)*(x - a)²

p2(x) = 2 - x + 1/2 * 2/1³ * (x - 1)²

p2(x) = 2 - x + (x - 1)²

p2(x) = 2 - x + x² - 2*x + 1

p2(x) = x² - 3*x + 3

And Now, we find the approximating polynomial value

So,

c. approximate 1/0.97 using p1(x)

p1(0.97) = 2 - 0.97 = 1.03

approximate 1/0.97 using p2(x)

p2(0.97) = 0.97² - 3*0.97 + 3 = 1.0309

So, The polynomial approximate values are

a. The linear approximating polynomial is p1(x) = 2 - x

b.The quadratic approximating polynomial is  p2(x) = x² - 3*x + 3

c. The approximating polynomial value is p1(0.97) = 1.03; p2(0.97) = 1.0309.

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Disclaimer: This question was incomplete. Please find the full content below

Question:

A. Find the linear approximating polynomial for the following function centered at the given point a. b. Find the quadratic approximating polynomial for the following function centeredat the given point a. c. Use the polynomials obtained in parts a. and b. to approximate the given quantity.

f(x)=1/x, a=1; approximate 1/0.97

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The y-intercept of m is smaller than n of y-intercept.

The points where a line crosses an axis are known as the x-intercept and the y-intercept, respectively.

A function's y-intercept is the value of y when

x= 0

The cubic function represented by function n traverses the points (-1,0) and (0,2).

When, x=0 ,y=2 and y = 2 is the y-intercept.

Operation m:

When, x=0,y = -6 the y-intercept is y = -6, which is less than 2, meaning that the y-intercept of m is smaller than the y-intercept of n, and option A provides the correct response.So the y-intercept of m is smaller than n of y-intercept.

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[tex] \frac{y}{x} = \frac{4}{ - 3} = \frac{8}{ - 6} = \frac{12}{ - 9} \\ this \: relation \: holds \: so \: y \: varies \\ directly \: with \: x[/tex]

[tex] \frac{ 4}{3} m = \frac{8}{ - 6} \\ m = \frac{ \frac{ - 8}{6} }{ \frac{3}{ - 4} } = \frac{8 \times 4}{6 \times 3} = \frac{32}{18} = 2 \\ multipling \: by \: 2[/tex]

[tex] \frac{y}{x} = \frac{ - 40}{ - 0.5} = \frac{8}{2.5} = \frac{5}{4} \\ \frac{y}{x} = 80 = 3.2 = 1.25 \\ this \: is \: untrue \: so \: y \: is \: not \\ varying \: directly \: with \: x[/tex]

[tex]no \: constant \: of \: variation[/tex]

[tex] \frac{y}{x} = \frac{30}{6} = \frac{35}{7} = \frac{96}{8} \\ \frac{y}{x} = 5 = 5 = 12 \\ y \: does \: not \: vary \: directly \: with \: x[/tex]

In AON networks, "node" is a point where one or more lines (arrows) begin or terminate, commonly used for depicting an event or activity.

Scheduling logic diagrams of the most fundamental kind. Any activity on the schedule that has no float; Total Float = 0; is considered a critical activity.

Between the project's start and finish, there are vital activity(s) that must be completed continuously. This is known as the critical path.

The importance of AON diagram are-

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Let [tex]C[/tex] be the set of all students in the classroom.

Let [tex]P[/tex] and [tex]M[/tex] be the sets of students that pass physics and math, respectively.

We're given

[tex]n(C) = 60[/tex]

[tex]n(P \cap M') = 17[/tex]

[tex]n(P \cap M) = 25[/tex]

[tex]n((P \cup M)') = n(P' \cap M') = 9[/tex]

i. We can split up [tex]P[/tex] into subsets of students that pass both physics and math [tex](P\cap M)[/tex] and those that pass only physics [tex](P\cap M')[/tex]. These sets are disjoint, so

[tex]n(P) = n(P\cap M) + n(P\cap M') = 25 + 17 = \boxed{42}[/tex]

ii. 9 students fails both subjects, so we find

[tex]n(C) = n(P\cup M) + n(P\cup M)' \implies n(P\cup M) = 60 - 9 = 51[/tex]

By the inclusion/exclusion principle,

[tex]n(P\cup M) = n(P) + n(M) - n(P\cap M)[/tex]

Using the result from part (i), we have

[tex]n(M) = 51 - 42 + 25 = 34[/tex]

and so the probability of selecting a student from this set is

[tex]\mathrm{Pr}(M) = \dfrac{34}{60} = \boxed{\dfrac{17}{30}}[/tex]

Answer:

Cookies = 44

Step-by-step explanation:

universal U=60

let the number of cookies be C,

n(C)=(3*b)-4

and the number of brownies be b,

n(b)=?

n(buc)=60

b+c=60

then, c=60-b

substitute for c=60-b.

60-b=3b-4

60+4=3b+b

64=4b

b=16

c=60-b

c= 60-16

c=44

The correct option is C.

The formula that could be used to represent a  function w is

= w(x)=v(x-7)+7

A polynomial divided by another polynomial can be used to represent a rational function. Since polynomials are defined everywhere, the set of all numbers excluding the zeros in the denominator constitutes the domain of a rational function.

Example 1. x = f(x) (x - 3). The denominator, x = 3, only contains one zero. When the denominator is zero, rational functions are no longer defined.

According to the given Information:

Rational functions whose point of discontinuity is at x=7.

When a point of discontinuity exists for a rational function,

It occurs when:

q(x) = r(x-a),  x=a

In this instance, we are required to notice the following relationship, which is a union of a parent rational function and a vertical translation:[tex]w(x)=v(x-a)+k\\ \forall k \in \mathbb{R}[/tex], (2)

If we are aware of a=7 and k=7 .

The equation that could be used to represent a  function w is. w(x)=v(x-7)+7

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I understand that the question you are looking for is:

Rational functions v and w both have a point of discontinuity at x = 7. Which equation could represent function w?

A. w(x)=v(x-7)

B. w(x)=v(x+7)

C. w(x)=v(x-7)+7

D. w(x)=v(x)+7

The information to fill one the box regarding the proof include:

It should be noted that when two triangles of each corresponding sides are equal, then it's said that they are similar.

Here AB = CD, and AD = CB as they illustrate the fact that they are parallel.

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Simplifying the expression gives 36x^2 - 82x - 10 = 0

Given the expression;

4x-5/3+2x=7+2/9x+2

[tex]\frac{4x - 5}{3 + 2x} = \frac{9}{9x + 2}[/tex]

Cross multiply

[tex]4x - 5( 9x + 2) = 9 (3 + 2x)[/tex]

Expand the bracket

[tex]36x^2 + 8x - 45x - 10 = 27x + 18x[/tex]

Collect like terms

36x^2 + 8x - 45x - 27x - 18x = 0

Add the like terms

36x^2- 82x - 10 = 0

Thus, simplifying the expression gives 36x^2- 82x - 10 = 0

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The error in the measurement of its volume is 3%.

The error in the measurement of its volume is:​

Volume[tex]=\frac{4\pi r^{3} }{3}[/tex]

Δv/v*100=3 ΔR/R*100

[tex]3*\frac{1}{100} *100[/tex]

=3

The error in the measurement of its volume is 3%.

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