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3 years ago

given that the ratio of boys to girls in a choir group is 17:23, what is the relative frequency of the number of boys?

Mathematics

1 answer:

marusya05 [52]3 years ago

5 0

Answer:

50 Students

You can find the number of boys and girls by dividing the total, so there are a total of 20 boys and 30 girls in the choir of 50. The ratio 2 boys : 3 girls means we can divide the chorus into groups.

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4(2x + 2) + 12 > 100

Komok [63]

Answer:

x > 10

Step-by-step explanation:

4(2x + 2) + 12 > 100

4(2x + 2) > 100 -12

4(2x + 2) > 88

2x + 2 > 22

2x> 22-2

2x > 20

x > 10

3 0

3 years ago

Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary. B = 67°, a

kvasek [131]

Among many other formulas, the area of a triangle is

$S = \frac 1 2 a c \sin B$

where B is the angle between sides a and c.

$S=\frac 1 2(10)(20) \sin 67^\circ \approx 92.05 \textrm{ sq cm}$

Answer: 92.05 sq cm, first choice

Bonus. Here's a formula for the area S of a triangle your teacher doesn't know:

$16S^2 = 4a^2b^2 -(c^2-a^2-b^2)^2=(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$

7 0

3 years ago

-t=9t(t-10) Im stuck

viva [34]

T=0 is also a solution. Other than 0 here is a solution.

7 0

3 years ago

Can someone help me, please ?? im stuck

Tresset [83]

Part A

5x - 5 >= 10 or-3x + 1 > 13

1) This is a disjunction (OR)

5x - 5 >= 10

5x >= 15

x >= 3

-3x + 1 > 13

-3x > 12

x < -4

Answer

x <- 4 or x >= 3

Part C:

1) This is a conjunction (And)

5x+ 3 <= 18 and 4 - x < 6

5x <= 15 and - x < 2

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Answer

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8 0

3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago

given that the ratio of boys to girls in a choir group is 17:23, what is the relative frequency of the number of boys? ​

given that the ratio of boys to girls in a choir group is 17:23, what is the relative frequency of the number of boys?