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3 years ago

5

There are 21 students in a class. The ratio of the number of boys to the number of girls is 4:3. How many girls are in the class

?

2 answers:

ZanzabumX [31]3 years ago

8 0

Answer:

Step-by-step explanation:

4:3 total is 7

8:6 total is 14

12:9 total is 21

So there are 9 girls.

k0ka [10]3 years ago

4 0

Answer: 9

Step-by-step explanation:

Given : There are 21 students in a class.

It is given that the ratio of the number of boys to the number of girls is 4:3.

Let 4x be the number of boys and 3x be the number of girls in class.

According to the question , we have

$3x+4x=21\\\\\Rightarrow\ 7x=21\\\\\Rightarrow\ x=\dfrac{21}{7}=3$

Now, the number of girls in class = 3x= 3(3)=9

Hence, there are 9 girls in class.

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4 years ago

70=(4x+36)/2 figure this out please

adell [148]

Answer:

$\boxed{\bold{\huge{\boxed{x = 26}}}}$

Step-by-step explanation:

$70 = \frac{4x+36}{2}$

<u>Multiplying both sides by 2</u>

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$104 = 4x$

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3 years ago

Read 2 more answers

Six times a number is greater than 20 more than that number. What are the possible values of that number? On-4 On>4 20 On>

Vaselesa [24]

Answer:

a > 4

Step-by-step explanation:

$6a > 20 + a \\ collect \: like \: terms \\ 6a - a > 20 \\ then \: add \\ 5a \: > 20 \\ divide \: both \: sides \: by \: 5 \\ \frac{5a}{5} > \frac{20}{5} \\ a > 4$

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2 years ago

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

melamori03 [73]

Answer:

We conclude that the mean wedding cost is less than $30,000 as advertised.

Step-by-step explanation:

We are given the following data set:(in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{236100}{8} = 29512.5$

Sum of squares of differences = 3408750

$S.D = \sqrt{\frac{3408750}{7}} = 697.82$

Population mean, μ = $30,000

Sample mean, $\bar{x}$ = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $ 697.82

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars}$ We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975$

Now,

$t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it.

We conclude that the mean wedding cost is less than $30,000 as advertised.

8 0

4 years ago