15
5. At the Redbridge School, of pupils have a pet dog, & have a cat and have a
The rest have no pets.
A)What fraction of the children has no pet? Write this in its simplest form.
B)There are 250 pupils at the School. How many children have no pets?
C)At the Bluebell School, 23% have no pets. What percentage increase of pupils at
the Bluebell School have no pets compared to the Redbridge School?

Answer:

g = -0.5

Step-by-step explanation:

Divide both sides by the numeric factor on the left side, then solve.

Answer:

g = -0.5 or -1/2

Step-by-step explanation:

-10(g – 6) = 65 : apply distributive property

(-10 • g) + (-10 • -6) = 65 : multiply

-10g + 60 = 65 : apply property of equal subtraction

-10g + 60 – 60 = 65 – 60 : subtract

-10g = 5 : apply property of equal division

-10g ÷ -10 = 5 ÷ -10 : divide

g = -0.5

The number of chickens and sheeps on the Old MacDonals farm is 39 and 16 respectively.

let

x + y = 55

2x + 4y = 142

Multiply (1) by 2

2x + 2y = 110

2x + 4y = 142

4y - 2y = 142 - 110

2y = 32

y = 32/2

y = 16

substitute y = 16 into (1)

x + y = 55

x + 16 = 55

x = 55 - 16

x = 39

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Answer:

Loan A will have the lowest total paycheck because the total of the principal than the same total for loan B

The correct answer is : of those planning to not to go on vacation, more of them are women.

The correct answer is (C)

Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event

Given:  total people asked =145

The first option is not correct because plan to go on vacation women are more than men.

The second option is also incorrect because more women were asked than men.

The third option is correct because out of planning who were not going any vacation, women are more than of men.

The fourth option is incorrect because More men plan to go on vacation.

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Answer:

0+4

Step-by-step explanation:

Answer:

Step-by-step expiation : 5575.3

Answer:

4/27 ezzzzzzzzzzzzzzzzzzzz

a) The time taken by the rocket to reach the maximum height is 2.75 seconds.

b) The maximum height of the rocket will be 127 feet.

Speed is defined as the ratio of the time distance travelled by the body to the time taken by the body to cover the distance.

Given that:-

a) The time taken by the rocket will be calculated as:-

We have a function h(t)=−16t²+88t+6. Now by taking the derivative of the function with respect to time and equating it to zero we will get the maximum time in which the rocket will reach the maximum height.

h(t)  =  -16t² + 88t + 6

h'(t)    =     0

-32t   +   88   =  0

32t      =      88

t     =    88 / 32    =  2.75 seconds

So the maximum time is 2.75 seconds.

b) Maximum height will be calculated as:-

Put the value of maximum time in the function h(t)  =  -16t² + 88t + 6 we will get the maximum height.

h(t)  =  -16t² + 88t + 6

h(2.75) = -16(2.75)² + ( 88 x 2.75 ) + 6

h(2.75) = 127 feet

Therefore the time taken by the rocket to reach the maximum height is 2.75 seconds and the maximum height of the rocket will be 127 feet.

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Answer:

A ≈ 25.1 cm²

Step-by-step explanation:

the area (A) of the sector is calculated as

A = area of circle × fraction of circle

  = πr² × [tex]\frac{\frac{4\pi }{9} }{2\pi }[/tex]

  = π × 6² × [tex]\frac{\frac{4\pi }{9} }{2\pi }[/tex]

  = 36π × [tex]\frac{\frac{4\pi }{9} }{2\pi }[/tex]

  = 18 × [tex]\frac{4\pi }{9}[/tex]

  = 2 × 4π

  = 8π

  ≈ 25.1 cm² ( to the nearest tenth )

Real life scenarios of acute angles are:

Acute angles are angles that have a measure less than 90 degrees.

To create a real life scenario to illustrate acute angles, the following must be put in place:

Real life scenarios that satisfy the above highlights are:

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Step-by-step explanation:

any rational number can be expressed as a/b, with a and b being integer numbers, and b different to 0.

-7 = -7/1 rational

sqrt(10) irrational (because there is no rational number multiplied by itself that results in 10)

sqrt(16) = sqrt(4×4) = 4 = 4/1 rational

52% = 52/100 rational

1.235 = 1235/1000 rational

Answer:

f(3) = 6

Step-by-step explanation:

you substitute 3 for x.

So you have 2(3) which is 2 multiplied by 3 which is 6

Answer:

  £850

Step-by-step explanation:

We can compute the left-over amount by comparing that to the amount spent.

__

Written as ratios, we have ...

  spent : left-over = 3/8 : 1 -3/8 = 3/8 : 5/8 = 3 : 5

Writing this using fractions, we have ...

  left-over / spent = 5 / 3

Solving for left-over gives ...

  left-over = spent × (5/3) = £510×5/3 = £850

Nate had £850 left over.

_____

Additional comment

Nate's money totaled £510 +850 = £1360.

The fraction spent is £510/£1360 = 3/8, as required.

Answer:

The correct answer is: "Option [D]".

Step-by-step explanation:

Hi student, let me help you out!

....................................................................................................................................

Let's use the acronym PEMDAS. With the help of this little acronym, we will not make mistakes in the Order of Operations!  :)

[tex]\dag\textsf{Acronym \: PEMDAS}[/tex]

P=Parentheses,

E=Exponents,

M=Multiplication,

D=Division,

A=Addition,

S=Subtraction.

Now let's start evaluating our expression, which is [tex]\mathsf{6-(\cfrac{2}{3})^2}[/tex]

According to PEMDAS, the operation that we should perform is "E-Exponents".

Notice that we have a fraction raised to a power. When this happens, we raise both the numerator (2 in this case) and the denominator (3 in this case) to that power, which is 2. After this we obtain  [tex]\mathsf{6-\cfrac{4}{9}}[/tex].

See, we raised both the numerator and the denominator to the power of 2.

Now what we should do is subtract fractions.

Note that 6 and -4/9 have unlike denominators. First, let's write 6 as a fraction: [tex]\mathrm{\cfrac{6}{1}-\cfrac{4}{9}}[/tex]. Now let's multiply the denominator and the numerator of the first fraction times 9: [tex]\mathrm{\cfrac{54}{9}-\cfrac{4}{9}}[/tex].

See, now the fractions have the same denominator. All we should do now is subtract the numerators: [tex]\mathrm{\cfrac{50}{9}}[/tex].

∴, the answer is Option D.

Hope this helped you out, ask in comments if any queries arise.

Best Wishes!

[tex]\star\bigstar\underline{\overline{\overline{\underline{\textsf{Reach \: far. Aim \: high. Dream \: big.}}}}}\bigstar\star[/tex]

[tex]\underline{\rule{300}{5}}[/tex]

Calculate 2/3, to the power of 3, therefore

[tex]\bf{\left(\dfrac{2}{3}\right)^{2}=\dfrac{2\times2}{3\times3}=\dfrac{4}{9} }[/tex]

[tex]\bf{6-\dfrac{4}{9} }[/tex]

Convert 6 to the fraction 54/9.

[tex]\bf{\dfrac{54}{9}-\dfrac{4}{9} }[/tex]

Since 54/9 and 4/9 have the same denominator, join their numerators to subtract them.

[tex]\bf{\dfrac{54-4}{9} \ \ \to \ \ \ Subtract }[/tex]

Subtract 4 from 54 to get 50.

[tex]\bf{\dfrac{50}{9} \ \ \to \ \ \ Answer }[/tex]

Therefore, the correct alternative is "D".

[tex]\red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}[/tex]

The answers to the question are

The median of the data set is gotten by arranging the values in order

3, 4, 5, 5, 7, 8

The median can be gotten by

5 + 5 / 2

= 10/2

Median = 5

This is the value under which 25 percent of the data are found in the group of data. The lower quartile is 4.

This is the point from which 75 percent of the data is found when it is in ascending order. The upper quartile is 7.

The lowest value in the data set is 3

The highest value in the data set is 8

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The original volume of the gas would be 4.00 liters

Boyle's law is a gas law that states that the pressure of a gas is inversely proportional to it's volume with temperature kept constant.

It is represented thus;

P1V1 = P2V2

Where P1 = Original pressure = 6 atmospheres

V1 = Original volume =?

P2 = Final pressure =4 atmospheres

V2 = Final volume= 6 liters

Make the original volume V1, subject of formula

V1 = P2× V2 ÷ P1

V1 = 4× 6 ÷ 6 = 24 ÷ 6 = 4. 00 liters

Hence, the original volume of the gas is 4.00 liters

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Answer:

69

It wont always be easy, but always try to do what ' s right!

   In a world of worriers, be the warrior!

        When nothing goes right, go left.

Answer:

4/225

Step-by-step explanation:

Formula :

Probability (Event) = No. of times event has occurred / Total possibilities

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

Solving :

⇒ P (Joe or Sabine) = No. of times Joe or Sabine has entered / Total number of people

⇒ P (Joe or Sabine) = 3 + 5 / 450

⇒ P (Joe or Sabine) = 8/450

⇒ P (Joe or Sabine) = 4/225

Answer:

Point A:

(1, 3) ------> (3, -1)

Point B:

(3, 3) ------> (3, -3)

Point C:

(3, 2) ------> (2, -3)

Step-by-step explanation:

To find the coordinates of a 90° clockwise rotation, use the following:

[tex](x, y)[/tex] -----> [tex](y, -x)[/tex]

This means you will switch the x and y coordinates of the point and then take the opposite of the x coordinate (if it is positive make it negative and if it is negative make it positive).

Point A:

(1, 3) ------> (3, -1)

Point B:

(3, 3) ------> (3, -3)

Point C:

(3, 2) ------> (2, -3)

Answer:

[tex]3x+7=-3x+2+6x[/tex]

Combine the like terms on the right side of the equation.

[tex]3x+7=3x+2[/tex]

Eliminate 3x from the equation.

[tex]7\neq 2[/tex]

Seven will never equal 2.

Step-by-step explanation:

[tex]3x+7=-3x+2+6x[/tex]

Combine the like terms on the right side of the equation.

[tex]3x+7=3x+2[/tex]

Eliminate 3x from the equation.

[tex]7\neq 2[/tex]

Seven will never equal 2.

Check the picture below.

[tex]~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ Q(\stackrel{x_1}{-4}~,~\stackrel{y_1}{9})\qquad U(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) ~\hfill QU=\sqrt{[ 2- (-4)]^2 + [ 3- 9]^2} \\\\\\ ~\hfill \boxed{QU=\sqrt{72}} \\\\\\ U(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad A(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-2}) ~\hfill UA=\sqrt{[ -3- 2]^2 + [ -2- 3]^2} \\\\\\ ~\hfill \boxed{UA=\sqrt{50}}[/tex]

[tex]A(\stackrel{x_1}{-3}~,~\stackrel{y_1}{-2})\qquad D(\stackrel{x_2}{-9}~,~\stackrel{y_2}{4}) ~\hfill AD=\sqrt{[ -9- (-3)]^2 + [ 4- (-2)]^2} \\\\\\ ~\hfill \boxed{AD=\sqrt{72}} \\\\\\ D(\stackrel{x_1}{-9}~,~\stackrel{y_1}{4})\qquad Q(\stackrel{x_2}{-4}~,~\stackrel{y_2}{9}) ~\hfill DQ=\sqrt{[ -4- (-9)]^2 + [ 9- 4]^2} \\\\\\ ~\hfill \boxed{DQ=\sqrt{50}}[/tex]

now, let's take a peek at that above, DQ = UA and QU = AD, so opposite sides of the polygon are equal.

Now, let's check the slopes of DQ and QU

[tex]D(\stackrel{x_1}{-9}~,~\stackrel{y_1}{4})\qquad Q(\stackrel{x_2}{-4}~,~\stackrel{y_2}{9}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{9}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{(-9)}}} \implies \cfrac{9 -4}{-4 +9}\implies \cfrac{5}{5}\implies \cfrac{1}{1}\implies 1 \\\\[-0.35em] ~\dotfill[/tex]

[tex]Q(\stackrel{x_1}{-4}~,~\stackrel{y_1}{9})\qquad U(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{9}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{(-4)}}} \implies \cfrac{3 -9}{2 +4}\implies \cfrac{-6}{6}\implies \cfrac{-1}{1}\implies -1[/tex]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's notice that the reciprocal of 1/1 is just 1/1 and the negative of that is just -1/1 or -1, so QU has a slope that is really just the negative reciprocal of DQ, those two lines are perpendicular, thus making a 90° angle, and their congruent opposite sides will also make a 90°, that makes QUAD hmmm yeap, you guessed it.

Answer:

Step-by-step explanation:

1. 54 Students

2. 4 Students

3.) 33 Students

4.) 10 Students

Just justify the answer for 5 and 6 you can do it<<33

The solution for this expression is:[tex]9\sqrt{6}-15\sqrt{15}+6\sqrt{10}-50[/tex].

The properties of multiplication are:

Here, you should apply the distributive property for solving this expression.

[tex](3\sqrt{3}+2\sqrt{5} )*(3\sqrt{2}-5\sqrt{5})\\ \\[/tex]

Applying the distributive property, you have:

[tex]9\sqrt{6}-15\sqrt{15}+6\sqrt{10}-10*\sqrt{25} \\ \\ 9\sqrt{6}-15\sqrt{15}+6\sqrt{10}-10*5\\ \\ 9\sqrt{6}-15\sqrt{15}+6\sqrt{10}-50[/tex]

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Answer:

y = 2/3

Step-by-step explanation:

Assuming you are looking for "y":

15 * ( y - 4 ) - 2 * (y - 9 ) + 5 * (y + 6) = 0

15y - 60 - 2y + 18 + 5y + 30 = 0

15y - 2y + 5y -60 + 18 + 30 = 0

18y = 60 - 18 - 30

18y = 12

y = 12/18

y = 2/3

Step-by-step explanation:

hope you can send the questions again because your question is to long to answer everything. I answered 1,2,3,4

Answer:

a ) Electronically , by reviewing billing receipts

Step-by-step explanation:

The main purpose of controlling inventory is to make sure that the optimal amount of products are being kept.

Therefore, option A is correct.

Answer:

Step-by-step explanation:

The two relations can be translated to a system of linear equations in two variables. These can be solved in any of the usual ways.

__

Let d and m represent the amounts David and Mary saved last month, respectively.

The first relation tells us

  d/m = 3/5   ⇒   5d = 3m   ⇒   5d -3m = 0

The second relation tells us that adding 3 times what David had last month to 2 times what Mary had last month will give a total that is 154 greater than the sum of their funds last month.

  d +m + 154 = 3d +2m   ⇒   2d +m = 154

__

We can add 3 times the second equation to the first to eliminate m.

  (5d -3m) +3(2d +m) = (0) +3(154)

  11d = 462 . . . . . . . . simplify

  d = 42 . . . . . . . . . divide by 11

Using the second equation, we can find Mary's savings:

  m = 154 -2d

  m = 154 -2(420 = 70

David saved $42 last month; Mary saved $70.

Answer:

[tex]11\frac{7}{8}[/tex] ounces

Step-by-step explanation:

Since Ethan needs [tex]2 \frac{3}{8}[/tex] ounces of blue paint to make 1 batch of green paint, we can multiply the amount of blue paint needed by 5 to see how much blue paint he'll need for 5 batches of green paint.

[tex](2 \frac{3}{8})5 = 10 \frac{15}{8} \rightarrow \boxed{11\frac{7}{8}}[/tex]

This means he will need [tex]11\frac{7}{8}[/tex] ounces of blue paint to make 5 batches of green paint.