All categories

3 years ago

A class has 6 boys and 15 girls. What is the ratio of boys to girls

Mathematics

2 answers:

Gnoma [55]3 years ago

4 0

$the \ ratio \ is : \\ \\ boys \ to \ girls = 6:15= 2:5$

Brut [27]3 years ago

3 0

we know that

Ratio is the quantitative relation between two numbers

Let

x-------> number of boys in the class

y-------> number of girls in the class

$ratio=\frac{x}{y}$

in this problem we have

$x=6\ boys\\y=15\ girls$

<u>Find the ratio</u>

substitute the values in the formula

$ratio=\frac{6}{15}$

Divide by $3$ both numerator and denominator

$ratio=\frac{2}{5}$

therefore

<u>the answer is</u>

the ratio of boys to girls is $\frac{2}{5}$

You might be interested in

How do the angle measures of polygon ABCD compare with the angle measures in the reflected image? How do the side lengths compar

Tresset [83]

The measures of the angles ABCD are the same as the measures of the angles that is corresponding based on the reflected images.

<h3>How to show that the angles are equal</h3>

Based on the fact that this is a reflection, the angles would be equal measures.

What this means is that there is a preservation of the of the reflected image. This is the same logic for the length of the sides.

The reflection is known to be a rigid type of transformation that is known to have its angles and sides preserved.

Read more on reflection in mathematics here:

brainly.com/question/1908648

#SPJ1

8 0

2 years ago

write the equation of the line that passes through (2, 4) and has a slope of –1 in point-slope form. x – 4 = –1(y – 2) x – 2 = –

Kobotan [32]

The point-slope form: $y-y_0=m(x-x_0)$

m = -1; (2; 4) $\to x_0=2;\ y_0=4$

subtitute

y - 2 = -1(x - 4)

6 0

3 years ago

The population of a town is 58,000 and is growing by 23% per year. If the trend continues, what will the population be in 8 year

mina [271]

The population in 8 yrs would be 164,720

6 0

3 years ago

Henry and his family are eating at a rotating restaurant at the top of a tower. The restaurant rotates clockwise 90° every 15 mi

frutty [35]

Answer:

The table of Henry's family will be at (4, -4)

Step-by-step explanation:

The problem states that the initial coordinates of Henry's family is at the coordinates (-4, 4). Each 15 minutes the restaurant will rotate 90° every 15 minutes. If we follow the Cartesian plane and the center of the restaurant being at (0, 0) we know that the table will be situated at the (4, 4) coordinates of the restaurant after the first rotation. After the 30 minute mark, the restaurant rotates one more time and moves the table of Henry's family at the (4, -4) coordinates of the restaurant as the center of the restaurant will never move.

4 0

3 years ago

Moose Drool Makes Grass More Appetizing Different species can interact in interesting ways. One type of grass produces the toxin

antiseptic1488 [7]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: level of toxin ergovaline on the grass after being treated with Moose saliva.

>The parameter of interest is the population mean of the level of ergovaline on grass after being treated with Moose saliva: μ

The best point estimate for the population mean is the sample mean:

>X[bar]=0.183 Sample average level of ergovaline on the grass after being treated with Moose saliva.

Assuming that the sample comes from a normal population.

>There is no information about the sample size takes to study the effects of the Moose saliva in the grass. Let's say that they worked with a sample of n=20

Using the Student-t you can calculate the CI as:

X[bar] ± $t_{n-1;1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

Where $t_{n-1;1-\alpha /2}= t_{19;0.975}= 2.093$

0.183 ± 2.093 * (0.016/√20)

[0.175;0.190]

Using a 95% confidence level you'd expect that the interval [0.175;0.190] will contain the true average of ergovaline level of grass after being treated with Moose saliva.

The information of a study that shows how much does the consumption of canned soup increase urinary BPA concentration is:

Consumption of canned soup for over 5 days increases the urinary BPA more than 1000%

75 individuals consumed soup for five days (either canned or fresh)

The study reports that a 95% confidence interval for the difference in means (canned minus fresh) is 19.6 to 25.5 μg/L.

>This experiment is a randomized comparative experiment.

Out of the 75 participants, some randomly eat canned soup and some randomly eaten fresh soup conforming to two separate and independent groups that were later compared.

>The parameter of interest is the difference between the population mean of urinary BPA concentration of people who ate canned soup for more than 5 days and the population mean of urinary BPA concentration of people that ate fresh soup for more than 5 days.

>Using a 95% confidence level you'd expect that the interval 19.6 to 25.5 μg/L would contain the value of the difference between the population mean of urinary BPA concentration of people who ate canned soup for more than 5 days and the population mean of urinary BPA concentration of people that ate fresh soup for more than 5 days.

> If the sample had been larger, then you'd expect a narrower CI, the relationship between the amplitude of the CI and sample size is indirect. Meaning that the larger the sample, the more accurate the estimation per CI is.

> The parameter of interest is Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later of people with an average age of 26 years old.

> The only way of finding the true value of the parameter is if you were to use the information of the whole population of interest, this is making all swedes with an average age of 26 increase calorie intake by 70% (mostly by eating fast food) and limit their daily activity to a maximum of 5000 steps per day and then measure their weight gain over two years. since this es virtually impossible to do, due to expenses and size of the population, is that the estimation with a small but representative sample is conducted.

n= 18

X[bar]= 6.8 lbs

S= 1.2

X[bar] ± $t_{n-1;1-\alpha /2}$ * (S/√n)

6.8 ± 2.101 * (1.2/√18)

$t_{n-1;1-\alpha /2}= t_{17;0.975}= 2.101$

[6.206;7.394]

With a 95% confidence level, you'd expect that the interval [6.206;7.394] will contain the true value of the Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later of people with an average age of 26 years old.

The margin of error of the interval is

$t_{n-1;1-\alpha /2}$ * (S/√n)= 2.101 * (1.2/√18)= 0.59

With a 95% it is expected that the true value of the Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later of people with an average age of 26 years old. will be 0.59lbs away from the sample mean.

I hope it helps

8 0

3 years ago