Divide £50.00 in the ratio of 1:4 SOMEBODY HELP!!!!

Answers

Answer 1

Answer: 1+4=5
50.00/5=10
1 unit is 10
1 unit to 4 unit

10 times 1=10
10 times 4=40

1:4
10:40

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Lily has 3 dog toys that are red. one fourth of all her dog toys are red. how many dog toys does lily have? first draw _ circles to show _ equal parts.

Answers

so she has 12 dogs toys because if she has three dogs and they are a fourth of the total you can just multiply 3 four times.

do you get it?

i hope i helped

Answer:

Step-by-step explanation:

The following data shows rainfall in a forest, in inches, on consecutive days of a month:2.7, 2.6, 2.4, 2.5, 2.8, 2.6, 2.7, 2.8, 2.4, 2.3, 2.5
What would be the values where each line is in a box plot?? Like the median, maximum, minimum

Answers

Data set: 2.7, 2.6, 2.4, 2.5, 2.8, 2.6, 2.7, 2.8, 2.4, 2.3, 2.5

arrange in ascending order:
2.3, 2.4, 2.4, 2.5, 2.5, 2.6, 2.6, 2.7, 2.7, 2.8, 2.8

The 5 number summary that is presented in a box plot
1) minimum value - 2.3
2) 1st quartile - 2.4
3) median - 2.6
4) 3rd quartile - 2.75
5) maximum value - 2.8

Answer:

The answer is the First graph

Step-by-step explanation:

Write this equation in standard form
9x^2-4y^2-24y-72=0

Answers

$9x^2-4y^2-24y-72=0\n \n 9x^2-4y^2-24y=72\n \n 9x^2-(4y^2+24y+36)=72-36\n \n 9x^2-4(y+3)^2=36\n \n (9x^2)/(36)-(4(y+3)^2)/(36)=(36)/(36)\n \n \boxed{(x^2)/(4)-((y+3)^2)/(9)=1}$

Which equation is true for x = –6 and x = 2?2x2 – 16x + 12 = 0
2x2 + 8x – 24 = 0
3x2 – 4x – 12 = 0
3x2 + 12x + 36 = 0

Answers

we know that

using a graph tool

let's proceed to graph each case to determine the roots

case 1) $2x^(2) - 16x + 12 = 0$

the roots are

$x1=0.8\ x2=7.2$ ------> is not the solution

see the attached figure N $1$

case 2) $2x^(2) +8x -24 = 0$

the roots are

$x1=-6\ x2=2$

see the attached figure N $2$ --------> is the solution

case 3) $3x^(2) - 4x - 12 = 0$

the roots are

$x1=-1.4\ x2=2.8$ ------> is not the solution

see the attached figure N $3$

case 4) $3x^(2) +12x +36 = 0$

the graph does not have x-intercepts

see the attached figure N $4$ ------> is not the solution

therefore

the answer is

the solution is

$2x^(2) +8x -24 = 0$

Second statement 2x2 + 8x – 24 = 0 Is true for the given conditions. When x = -6 2x2 + 8x – 24 = 0 Becomes 2(-6)2 + 8(-6) – 24 = 0 2(36) - 48 - 24 = 0 72 - 48 - 24 = 0 0 = 0 Which is true. When x = 2 2x2 + 8x – 24 = 0 Becomes 2(2)2 + 8(2) – 24 = 0 2(4) + 16 - 24 = 0 8 + 16 - 24 = 0 0 = 0 Which is true. So 2x2 + 8x – 24 = 0 will be answer.

Which of the following is the equation of a line parallel to the line y=-x+1 passing through the point (4,1)a. x+y=5
b. x-y=-5
c. x+y=-5
d. -x-y=5

Answers

y=-x+b

plug in x = 4 and y = 1 to find b

1 = -4 + b and we know that if you add 5 and -4 you get 1. So ...

5 = b so if we plug that in

y= -x + 5

which is the same as y + x = 5, So it would be A

1. Each of 9 friends chooses her favorite positive integera. The median of the chosen number is 91, what is the smallest the average of the 9 chosen numbers could be?

b. The median of the chosen number is 91, is there an limit to how large the aerge of the chosen numbers can be? If so, what is the largest the average can be?

c. The average of the chosen number is 91, what is the smallest the median of the 9 chosen numbers could be?

d. The average of the chosen numbers is 91. What is the largest the median of the chosen numbers could be?

Answers

Answer:

a) 1

b) There is no limit to which the largest number can be because we are only given information about the median.

c) 1

d) 90

Final answer:

The smallest average is 49 and the largest average is 91. The smallest median is 91 and the largest median is also 91.

Explanation:

a. Since the median is 91, at least 5 friends must choose numbers greater than or equal to 91, and at most 4 friends can choose numbers less than 91. To minimize the average, let's assume the four friends choose the smallest possible numbers less than 91 (1, 2, 3, and 4). The remaining five friends can then choose 91, 91, 91, 91, and 91. The average of the nine chosen numbers is (1 + 2 + 3 + 4 + 91 + 91 + 91 + 91 + 91)/9 = 49.

b. There is no limit to how large the average of the chosen numbers can be. The nine friends can all choose the same number, such as 91, which would make the average 91.

c. Since the average is 91, let's assume the eight friends choose the smallest possible numbers less than 91 (1, 2, 3, ..., 8). The remaining friend can then choose a number greater than or equal to 91. To minimize the median, the friend can choose the smallest possible number greater than or equal to 91, which is 91. So, the smallest median would be 91.

d. Since the average is 91, let's assume the eight friends choose the largest possible numbers less than 91 (84, 85, ..., 91). The remaining friend can then choose a number greater than or equal to 91. To maximize the median, the friend can choose the largest possible number greater than or equal to 91, which is 91. So, the largest median would also be 91.

Learn more about median and average of chosen numbers here:

brainly.com/question/33899535

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