Mr. Coelho and Ms. Steiner are grading math and science tests of sixth-grade and seventh-grade students, respectively,
Mr. Coelho graded 25 science tests in 50 minutes and 30 math tests in 40 minutes. Ms. Steiner graded 20 science tests in
50 minutes and 12 math tests in 30 minutes.
Whose grading shows a proportional relationship between the number of tests graded and the time spent grading?

The answer is only ms.Steiner’s grading

You just need to integrate 3 times:

[tex]f'''(t)=t^{1/2}-9\cos t[/tex]

[tex]f''(t)=\displaystyle\int f'''(t)\,\mathrm dt=\frac23 t^{3/2}-9\sin t+C[/tex]

[tex]f'(t)=\displaystyle\int f''(t)\,\mathrm dt=\frac4{15} t^{5/2}+9\cos t+Ct+D[/tex]

[tex]f(t)=\displaystyle\int f'(t)\,\mathrm dt=\frac8{105} t^{7/2}+9\sin t+\frac C2 t^2+Dt+E[/tex]

Answer:

[see below]

Step-by-step explanation:

The figure is made out of two rectangular prisms.

V = w * l * h

Volume for the left prism:

V = 9 * 5 * 24

V = 1080 m³

Volume for the right prism:

V = 9 * 18 * 10

V = 1620 m³

Combine both volumes:

1620 + 1080 = 2700

2700 cubic meters should be your answer.

Hope this helps.

Answer:

Commutative Property of Addition: a + b = b + a

Step-by-step explanation:

The Commutative Property of Addition implies that even on changing the order of addition the final result (i.e. the sum) remains the same.

a + b = b + a

Suppose a = 5 and b = 6, then:

a + b = 5 + 6 = 11

b + a = 6 + 5 = 11.

Thus, a + b = b + a.

a + b + c= a + c + b = b + a + c = c + a + b

Suppose a = 4, b = 3 and c = 6, then:

a + b + c = 4 + 3 + 6 = 13

a + c + b = 4 + 6 + 3 = 13

b + a + c = 3 + 4 + 6 = 13

c + a + b = 6 + 4 + 3 = 13.

Thus, a + b + c= a + c + b = b + a + c = c + a + b.

Answer: about 73 %

Step-by-step explanation:

because 88 of 120 is about 73 %

The percent of change in the price of this coat is 26.66 %.

To find the percent of change in the price of this coat.

A part of a whole expressed in hundredths a high percentage of students attended. Also the result obtained by multiplying a number by a percent the percentage equals the rate times the base.

Given that:

Cost price of coat(C.P)= 120$

Selling price of coat(S.P)=88$

We know that ,

Percent change=change in price /C.P*100

Percent change=120$-88$/120$*100

Percent change=32/120*100

Percent change=26.66 %

Therefore, the percent of change in the price of this coat is 26.66 %.

Learn more about percentage here:

https://brainly.com/question/17198381

#SPJ2

Answer:

= 5 ( 12d + 23f )

Step-by-step explanation:

-2(5d-9f)+7f-10(-9f-7d)

Open parenthesis

= -10d + 18f + 7f + 90f + 70d

Collect like terms

= -10d + 70d + 18f + 7f + 90f

= 60d + 115f

Factorise

= 5 ( 12d + 23f )

Therefore,

-2(5d-9f)+7f-10(-9f-7d) in its simplest form is 5 ( 12d + 23f )

Answer:

11n

Step-by-step explanation:

The expression is 7n + 4n. Since 7n and 4n are like terms (they both are variables with n), we can combine them so the expression becomes 7n + 4n = 11n.

Answer:

234252464375675734

Step-by-step explanation:

addition...

Answer:

here 234252464375734

Greetings from Brasil...

We notice 2 dashes on the NO and NP line. This means that both are the same size. Since NO = 17 then OP is also 17 in length.

So

NP = NO + OP

NP = 17 + 17

NP = 34

As already said

NP = 5X - 6 = 34

5X - 6 = 34

Answer:

a. 7(3) - 20 = 1°

b. 9 cm

c. Aly is correct

Step-by-step explanation:

Answer:

a. ZWY

b. 9cm

c. Aly's solution is correct

Step-by-step explanation:

a. YW bisects the whole angle, thus angle XWY and angle ZWY are same

NOTE: REMEMBER TO WRITE THE LETTER ACCORDINGLY

b. the two triangles are congruent by AAS (angle-angle-side) thus the two legs of the the triangles are also congruent. when one is 9cm, the other is also 9cm.

c. Since the triangles are congruent, their sides also congruent.

 7x-20=2x-5

 7x-2x=-5+20

     5x=15

       x=3

This is same as Aly's solution

Answer: Hi!

The equation for inverse proportion is x y = k or x = k/ y.

When finding the value of the constant k, you can use the known values and then use this formula to calculate all of the unknown values.

Hope this helps!

Answer:

-2s is ur answer

hope it helps u

Answer:

Step-by-step explanation:

2s+(−4s)

Combine 2s and −4s to get −2s.

−2s

Answer:

Step-by-step explanation:

Surface area of a pyramid =

area of base + area of triangular faces

Since it's a square based pyramid

It's surface area is

area of base + 4( area of one triangular face)

Since the square has equal sides

For square base

Area of a square = l²

where l is the length

From the question l = 5

So we have

Area of square base = 5² = 25ft²

For one of the triangular face

Area of a triangle = ½ × base × height

base = 5

height = 6

Area = ½ × 5 × 6 = 15ft²

So the surface area of the pyramid is

25 + 4(15)

= 25 + 60

We have the final answer as

Hope this helps you

Answer:

yes

Step-by-step explanation:

20x10= 200 therefor its 10times as much

I hope u will get help frm it.....

Answer:

-2

Step-by-step explanation:

To find the slope of the line you have to use the equation,

(y2-y1)/(x2-x1)

In this case it is, (-4-2)/7-4)

This simplifies to -2 and this is the slope of the line

Answer:

-8/5

hope this help!

Answer:

3(x^y(5x^2y-x³)+25x⁴))

Step-by-step explanation:

15x^2y²-3x^3y+75x⁴

From 15x^2y²-3x^3y only, 3x^y is the common factor

=> 3x^y(5x^2y-x³)+75x⁴

Taking the common factor of the latter expression, 3 shows to be the common factor of all the expression.

=> 3(x^y(5x^2y-x³)+25x⁴)

Answer:

[tex]x = 2a[/tex]

Step-by-step explanation:

Required

Represent twice a number is x as an algebra

Given that the number is a;

Then

[tex]Twice\ a\ number = 2 * a[/tex]

[tex]Twice\ a\ number = 2a[/tex]

Also,

[tex]Twice\ a\ number = x[/tex]

So, we have that

[tex]x = 2a[/tex]

Hence, the algebraic representation of the given parameters is

[tex]x = 2a[/tex]

Answer: 9 : 10  or 9/10.

Step-by-step explanation:

The ratio of  muffins to scones   is 300 : 270.   Now to reduce it to the lowest term divide each by 30.

You will then get  9: 10

Answer:

Step-by-step explanation:

Given that:

A set of numbers is transformed by taking the log base 10 of each number. The mean of the transformed data is 1.65. What is the geometric mean of the untransformed data.

To obtain the geometric mean of the untransformed data,

X = set of numbers

N = number of observations

Arithmetic mean if transformed data = 1.65

Log(Xi).... = transformed data

Arithmetic mean = transformed data/ N

Log(Xi) / N = 1.65

(Πx)^(1/N), we obtain the antilog of the aritmétic mean simply by raising 10 to the power of the Arithmetic mean of the transformed data.

10^1.65 = 44.668359

Answer:

(a) Yes, the data suggest that females are more likely to graduate from high school than males.

(b) A 95% confidence interval for the difference in the graduation rates between females and males is [0.024, 0.404] .

Step-by-step explanation:

We are given that a survey of over 25,000 Americans aged between 18 and 24 years revealed the following: 88.1% of the 12,678 females and 84.9% of the 12,460 males had high school diplomas.

Let [tex]p_1[/tex] = population proportion of females who had high school diplomas.

[tex]p_2[/tex] = population proportion of males who had high school diplomas.

(a) So, Null Hypothesis, [tex]H_0[/tex] : [tex]p_1\leq p_2[/tex]    {means that females are less or equally likely to graduate from high school than males}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]p_1 > p_2[/tex]     {means that females are more likely to graduate from high school than males}

The test statistics that will be used here is Two-sample z-test statistics for proportions;

                        T.S.  =    ~  N(0,1)

where, [tex]\hat p_1[/tex] = sample proportion of females having high school diplomas = 88.1%

[tex]\hat p_2[/tex] = sample proportion of males having high school diplomas = 84.9%

[tex]n_1[/tex] = sample of females = 12,678

= sample of males = 12,460

So, the test statistics =  

                                    =  7.428  

The value of the standardized z-test statistic is 7.428.

Now, at a 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 7.428 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that females are more likely to graduate from high school than males.

(b) Firstly, the pivotal quantity for finding the confidence interval for the difference in population proportion is given by;

                        P.Q.  =  [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex]  ~  N(0,1)

where, [tex]\hat p_1[/tex] = sample proportion of females having high school diplomas = 88.1%

[tex]\hat p_2[/tex] = sample proportion of males having high school diplomas = 84.9%

[tex]n_1[/tex] = sample of females = 12,678

[tex]n_2[/tex] = sample of males = 12,460

Here for constructing a 95% confidence interval we have used a Two-sample z-test statistics for proportions.

So, 95% confidence interval for the difference in population proportions, ([tex]p_1-p_2[/tex]) is;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                      of significance are -1.96 & 1.96}    

P(-1.96 < [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] < 1.96) = 0.95  

P( [tex]-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] < [tex]{(\hat p_1-\hat p_2)-(p_1-p_2)}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ) = 0.95  

P( [tex](\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] < ([tex]p_1-p_2[/tex]) < [tex](\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ) = 0.95

95% confidence interval for ([tex]p_1-p_2[/tex]) = [[tex](\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] , [tex](\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ]

= [ [tex](0.881-0.849)-1.96 \times {\sqrt{\frac{0.881(1-0.881)}{12,678}+\frac{0.849(1-0.849)}{12,460}} }[/tex] , [tex](0.881-0.849)+1.96 \times {\sqrt{\frac{0.881(1-0.881)}{12,678}+\frac{0.849(1-0.849)}{12,460}} }[/tex] ]

= [0.024, 0.404]

Therefore, a 95% confidence interval for the difference in the graduation rates between females and males is [0.024, 0.404] .

(c) The assumptions and conditions necessary for the above inferences to hold are;

Answer:

Step-by-step explanation:

In the tossing of a fair coin, there are equal probabilities of getting a HEAD and getting a TAIL.

Total probability is always 1 and a coin has 2 faces - Head & Tail.

The probability of getting a Head is 1/2 = 0.5

The probability of getting a Tail is 1/2 = 0.5

E1 is the event that TAIL comes up when the coin is tossed the first time

E2 is the event that HEAD comes up when the coin is tossed the second time

The probability value for EVENT 1 is 0.5

The probability value for EVENT 2 is 0.5

Answer: Joseph Kabila

Step-by-step explanation:

Answer:

Joseph Kabila

Step-by-step explanation:

Answer:

14 mph   ( average speed during the second part of the trip )

Step-by-step explanation:

Let´s call  "x"  the average speed during the first part then

t = 5 hours

t  =  t₁  +  t₂        t₁   and  t₂   times during part 1 and 2 respectively

l = t*v           (  distance is speed by time )      t =  l/v

First part

t₁  = 16/x        and      t₂  = 42 / ( x + 6)

Then

t =   5   =  16/x   +   42 /(x + 6)

5 = [ 16 * ( x  +  6 ) +  42 * x ] / x* ( x + 6 )

5 *x * ( x + 6 )  =  16*x  + 96 +  42 x

5*x² + 30*x  - 58*x - 96  =  0

5*x²  -  28*x  -  96  =  0

We obtained a second degree equation, we will solve for x and dismiss any negative root since negative time has not sense

x₁,₂  =   [28 ± √ (28)² + 1920  ] / 10

x₁,₂  = ( 28  ± √2704 )/ 10

x₁  = 28  -  52 /10        we dismiss that root

x₂  = 80/10

x₂  =  8 mph       average speed during the first part, and the average speed in the second part was 6 more miles than in the firsst part. then the average spedd dring the scond part was 8 + 6 = 14 mph

Answer:

Step-by-step explanation:

Hello !

Using the cross multiplication method, which of the following is the solution to 21/(x+4) = 7/(x+2)?

21(x + 2) = 7(x + 4)

21x + 42 = 7x + 28

21x - 7x = 28 - 42

14x = -14

x = -14/14

x = -1

Answer:

3/(x+2)= x+1

Step-by-step explanation:

21/7=3

x/x= x

4/2=2

Answer:

[tex] \boxed{ \bold{ \huge{\boxed{ \sf{ - 54w + 45}}}}}[/tex]

Step-by-step explanation:

[tex] \sf{3 + 6( - 9w + 7)}[/tex]

Distribute 6 through the parentheses

⇒[tex] \sf{3 - 54w + 42}[/tex]

Add the numbers : 42 and 3

⇒[tex] \sf{ - 54w + 45}[/tex]

Hope I helped!

Best regards!!

Answer:

10x.

Step-by-step explanation:

It is given that,

JK = 3x

LM = 12x

JM = 25x

Let as consider the line as shown in the below figure.

From the figure it is clear that,

[tex]JK+KL+LM=JM[/tex]

[tex]3x+KL+12x=25x[/tex]

[tex]KL+15x=25x[/tex]

[tex]KL=25x-15x[/tex]

[tex]KL=10x[/tex]

Therefore, the length of KL is 10x.

Answer:

We are given order pairs  (–9, –8), (–{(8, 4), (0, –2), (4, 8), (0, 8), (1, 2)}.

We need to remove in order to make the relation a function.

Step-by-step explanation:

Note: A relation is a function only if there is no any duplicate value of x coordinate for different values of y's of the given relation.

In the given order pairs, we can see that (0, –2) and (0, 8) order pairs has same x-coordinate 0.

So, we need to remove any one (0, –2) or (0, 8) to make the relation a function. hope this helps you :) god loves you :)

Answer:

1372cm²

Step-by-step explanation:

If this were a complete rectangle, it would be 50x40=2000cm².  So we need to take that number and subtract the half-circle.  A=[tex]\pi[/tex]r²  A=3.14*20*20=1256

1256*1/2 = 628

2000-628=1372cm²

Answer:

1372 cm²

Step-by-step explanation:

First, find the area of the rectangle:

A = lw

A = 40(50)

A = 2000 cm²

Next, find the area of the semicircle:

A =([tex]\pi[/tex]r²) / 2

A = (3.14)(20)²

A = 1256/2

A = 628 cm²

Then, subtract the semicircle area from the rectangle's area:

2000 - 628

= 1372 cm²

Answer:

The probability the die chosen was green is 0.9

Step-by-step explanation:

From the information given :

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has faces numbered 1, 2, 3, 4, 4, and 4.

SO, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) =  [tex]\dfrac{1}{6}[/tex]

P (4 | green dice) =  [tex]\dfrac{3}{6}= \dfrac{1}{2}[/tex]

A die is selected at random and rolled four times.

When the die is selected randomly; the probability of the first die must be equal to the probability of the second die =  [tex]\dfrac{1}{2}[/tex]

The probability of two 1's and two 4's in the first dice can be calculated as:

[tex]= \begin {pmatrix} \left\begin{array}{c}4\\2 \end{array}\right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4[/tex]

[tex]=\dfrac{4!}{2!(4-2)!}\times (\dfrac{1}{6})^4[/tex]

[tex]=\dfrac{4!}{2!(2)!}\times (\dfrac{1}{6})^4[/tex]

[tex]=\dfrac{4\times 3 \times 2!}{2!(2)!}\times (\dfrac{1}{6})^4[/tex]

[tex]=\dfrac{12}{2 \times 1}\times (\dfrac{1}{6})^4[/tex]

[tex]= 6 \times (\dfrac{1}{6})^4[/tex]

[tex]= (\dfrac{1}{6})^3[/tex]

[tex]= \dfrac{1}{216}[/tex]

The probability of two 1's and two 4's in the second  dice can be calculated as:

[tex]= \begin {pmatrix} \left\begin{array}{c}4\\2 \end{array}\right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]

[tex]= \dfrac{4!}{2!(4-2)!} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]

[tex]= \dfrac{4!}{2!(2)!} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]

[tex]= 6 \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]

[tex]= \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2[/tex]

[tex]= \dfrac{9}{216}[/tex]

∴  The probability of two 1's and two 4's in both dies

= P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's  = [tex](\dfrac{1}{216} \times \dfrac{1}{2} )+ ( \dfrac{9}{216} \times \dfrac{1}{2})[/tex]

The probability of two 1's and two 4's  = [tex]\dfrac{1}{432}+ \dfrac{1}{48}[/tex]

The probability of two 1's and two 4's  = [tex]\dfrac{5}{216}[/tex]

Using Bayes Theorem; the probability that the die was green can be computed as follows:  

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's)

P(second die (green) | two 1's and two 4's )  = [tex]\dfrac{\dfrac{1}{2} \times \dfrac{9}{216} }{\dfrac{5}{216}}[/tex]

P(second die (green) | two 1's and two 4's )  = [tex]\dfrac{\dfrac{1}{48} }{\dfrac{5}{216}}[/tex]

P(second die (green) | two 1's and two 4's )  =[tex]\dfrac{1}{48} \times \dfrac{216}{5 }[/tex]

P(second die (green) | two 1's and two 4's )  = [tex]\dfrac{9}{10}[/tex]

P(second die (green) | two 1's and two 4's )  = 0.9

∴

The probability the die chosen was green is 0.9

Answer:

4, -2

Step-by-step explanation:

Hello, please consider the following.

[tex]\begin{aligned}4x^2-8x-32=0 &\text{ ***We divide by 4.***}\\\\x^2-2x-8=0 &\text{ ***We complete the square. ***}\\\\(x-1)^2-1-8=(x-1)^2-9=0 &\text{ ***We move the constant to the right.***}\\\\(x-1)^2=9=3^2 &\text{ ***We take the root.***}\\\\x-1=\pm3 &\text{ ***We add 1.***}\\\\x=1+3=\boxed{4} \ & or \ x=1-3=\boxed{-2}\end{aligned}[/tex]

Thanks