There are 420 students . The ratio from girls to boys is 4- 3 . How many more girls?

Answers

Answer 1
Answer: There are 60 more girls
Answer 2
Answer:

Final answer:

There are 420 students divided into 7 parts due to the ratio of girls to boys being 4:3. With one part totaling 60 students, there are 240 girls and 180 boys. Thus, there are 60 more girls than boys.

Explanation:

To solve this problem, you need to understand the concept of ratios. In this case, the ratio of girls to boys is 4:3. This means that for every 4 girls, there are 3 boys. If we add the two parts of the ratio together, we get 7 parts. This means that the total number of students, which is 420, is to be divided into 7 parts.

So, one part of this ratio is equal to 420 divided by 7, which equals 60 students. Since the ratio claims there are 4 parts of girls and 3 parts of boys, to find out the numbers of girls and boys, we multiply each part of the ratio by 60. Hence, the total number of girls is 4 multiplied by 60, which equals 240 and the number of boys is 3 multiplied by 60, which equals 180.

Therefore, the difference in number between girls and boys is 240 minus 180, which equals 60. So there are 60 more girls than boys.

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Given the formula below, solve for x. y - y1 = m(x-x1)

Answers

Answer:

y-3 = 2/9 (x-8)

Step-by-step explanation:

The half-life of radium-226 is 1620 yr. Given a sample of 1 g of radium-226, the quantity left Q(t) (in g) after t years is given by:Q(t)= 1/2^t/1620

Required:
a. Convert this to an exponential function using base e.
b. Verify that the original function and the result from part (a) yield the same result for Q(0), Q(1620), and Q(3240).

Answers

Answer:

(a)e^(-0.000428 t)

Step-by-step explanation:

We are given that

Half life of radium-226=1620 yr

The quantity left Q(t) after t years is given by

Q(t)=((1)/(2))^{(t)/(1620)}

a. We have to convert the given function into an exponential function using base e.

Q(t)=((1)/(2))^{(t)/(1620)}

=(((1)/(2))^t)^{(1)/(1620)

=e^(ln(1/2) t/1620)

=e^({(ln(1/2))/(1620)t)

=e^(-0.000428 t)

(b)

Q(0)=e^(-0.000428 * 0)

=1

From original function

Q(0)=1

Q(1620)=((1)/(2))^{(t)/(1620)}

Q(1620)=(1)/(2)=0.5

From exponential function

Q(1620)=e^(-0.000428 * 1620)

=0.499\approx 0.5

Q(3240)=((1)/(2))^{(3240)/(1620))=0.25

Q(3240)=e^(-0.000428 * 3240)

Q(3240)=0.249=\approx 0.25

Hence, verified.

A candy distributor needs to mix a 30% fat-content chocolate with a 50% fat-content chocolate to create 200 kilograms of a 46% fat-content chocolate. How many kilograms of each kind of chocolate must they use?

Answers

Step-by-step explanation:

If x is the kilograms of 30% chocolate, and y is the kilograms of 50% chocolate, then:

x + y = 200

0.30x + 0.50y = 0.46(200)

Solving the system of equations with substitution:

0.30x + 0.50(200 − x) = 0.46(200)

0.30x + 100 − 0.50x = 92

8 = 0.20x

x = 40

y = 200 − x

y = 160

The distributor needs 40 kg of 30% chocolate and 160 kg of 50% chocolate.

Final answer:

To obtain 200 kilograms of a 46% fat-content chocolate, the candy distributor needs to mix 40 kilograms of a 30% fat-content chocolate and 160 kilograms of a 50% fat-content chocolate.

Explanation:

This problem can be solved using a basic mixture problem method. Let's name the amount of the 30% fat-content chocolate as 'x' and the amount of the 50% fat-content chocolate as 'y'. The total weight of the resulting chocolate is provided in the problem, 200 kilograms, therefore we know that x + y = 200.

The total fat in the chocolates should be 46% of 200kg, or 92kg. This gives us another equation based on the fat content, 0.3x + 0.5y = 92.

Solving these two equations linearly, we find the values of x and y. The amount of 30% fat content chocolate (x) is 40 kilograms and the amount of 50% fat-content chocolate (y) is 160 kilograms.

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How can I calculate this expression?

Answers

Answer:

26 whole 3/40

Step-by-step explanation:

Multiply the denominator with the number in middle and add to top,

13/3+ 17/5+ 46/16

LCM& Changing into lowest terms,

(65+51)/15+23/8

116/5+23/8

(928+115)/40

1043/40

26 whole 3/40

Idek dog you should know honestly

Compact and oversized tires are produced on two different machines. The table shows the number of each type of tire produced, y, depending on the number of hours, x, the machines work.

Answers

Answer:

OPTION B: 57x - 3

Step-by-step explanation:

Total tires produced is the sum of number of over-sized tires and the number of compact tires.

So, we have when x = 1, tires, y = 23 + 31 = 54

When x = 2, total tires y = 48 + 63  = 111

When x = 3, total tires y = 73 + 95  = 168

When x = 4, total tires y = 98 + 127 = 225

We can substitute the options to check which one will be the correct representation.

Option A: 55x - 1

When x = 1, y = 55(1) - 1    = 54

When x = 2, y = 55(2) - 1  = 109    $ \ne $    111

So, this option is eliminated.

Option B: 57x - 1

When x = 1, y = 57(1) - 3    =   54

When x = 2, y = 57(2) - 3  =   111

When x = 3, y = 57(3) - 3  =   168

When x = 4, y = 57(4) - 3  =   225

These values exactly match with the values from the table. So, Option B is the right answer.

Option C: 110x - 2

When x = 1, y = 110(1) - 2 = 108  $ \ne $  54

Hence, this option is incorrect.

Option D: 114x - 6

When x = 1, y = 114(1) - 6 $ \ne $  54

Hence, this option is also eliminated.

Option B is the required answer.

Answer: D

Step-by-step explanation:

the person above me was on the right track but needed to multiply the sum of the y values by 2. Therefore the answer should be D

Pls help me someone this is annoying me

Answers

Answer:

They are both 42 cm

Step-by-step explanation: