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lisabon 2012 [21]

3 years ago

7

There are 32 student,nine of those are women,what percent are men,round to the nearest tenth?

1 answer:

12345 [234]3 years ago

8 0

First you have to find out how many guys you have-1.) 32-9=23, then you divide the number of men by the number of students-2.) 23÷32=0.718, so the answer is 0.7

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0.146 divided by 0.04

laiz [17]

3.65 is the correct answer to 0.146 divided by 0.04

6 0

3 years ago

Can someone help with this math question

Tcecarenko [31]

Based on corresponding angles, the upper left angle will be the supplement of 130, which is 50 degrees. The upper right angle is perpendicular, so it is 90 degrees. Since the sum of all internal angles of a triangle is 180, the remaining angle (lower angle) is 40 degrees. This is choice G.

6 0

3 years ago

2. The sum of two numbers is 67. The

DaniilM [7]

Answer:

below

Step-by-step explanation:

l = large no

s = small no

l=s+13

l+s = 67

s+13+s = 67

2s+13 = 67

2s = 54

s=27

l= 40

5 0

2 years ago

The value of y varies directly as the square of x and y=27 when x=9. Find the equation represents this relationship. Report the

skad [1K]

Answer:

The relation between x and y is

$y=\frac13 x^2$

Step-by-step explanation:

Given that,

The value of y varies directly as the square of x.

$y\propto x^2$

$\Rightarrow y=rx^2$

where r is constant of proportional.

Given that,

y= 27 and x=9

Plug y= 27 and x=9 in above relation

$y=rx^2$

$\therefore 27=r(9)^2$

$\Rightarrow 27 = r.81$

$\Rightarrow r=\frac{27}{81}$

$\Rightarrow r=\frac{1}{3}$

The relation between x and y is

$y=\frac13 x^2$

8 0

3 years ago

A pile of coins, consisting of quarters and half dollars, is worth $11.75. If there are 2 more quarters than half dollars, how m

Black_prince [1.1K]

The pile contains 17 quarters and 15 half-dollars.

Let <em>x</em> = the number of quarters and <em>y</em> = the number of half-dollars.

We have two equations:

(1) $0.25<em>x</em> + $0.50<em>y</em> = $11.75

(2) <em>x</em> = <em>y</em> +2

Substitute the value of <em>x</em> from Equation (2) into Equation (1).

0.25(<em>y</em>+2) + 0.50<em>y</em> = 11.75

0.25<em>y</em> + 0.50 + 0.50<em>y</em> = 11.75

0.75<em>y</em> = 11.75 – 0.50 = 11.25

<em>y</em> = 11.25/0.75 = 15

Substitute the value of <em>y</em> in Equation (2).

<em>x</em> = 15 + 2 = 17

The pile contains 17 quarters and 15 half-dollars.

<em>Check</em>: 17×$0.25 + 15×$0.50 = $4.25 + $7.50 = $11.75.

4 0

3 years ago