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2 years ago

There are 24 people using the gym, tha ratio of men to women is 2;1, hoiw many men are using the gym

Mathematics

2 answers:

melamori03 [73]2 years ago

5 0

2 men for every 1 woman

24/3 =8

8*2 = 16

there are 16 men and 8 women

wlad13 [49]2 years ago

4 0

So if I understand the question ratio is 2 women for every men so that would men. So i think it would be 8 men

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Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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2 years ago

An estimated regression equation that was fit to estimate ductility in steel using its carbon content was found to be significan

AnnZ [28]

Answer:

At a level of 95%, it is expected that the interval [0.45; 11.59] contains the value of the ductility in steel when its carbon content is 0.5%.

Step-by-step explanation:

Hello!

Considering the dependent variable:

Y: Ductility in steel.

And the independent variable:

X: Carbon content of the steel.

The linear regression was estimated and a prediction interval was calculated.

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In this case 95% prediction interval for Y/X=0.5

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I hope it helps!

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2 years ago

The camera shop wants to use the measure of central tendency that expresses how successful they are in selling camaers which mea

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2 years ago

Directions: Calculate the area of a circle using 3.14x the radius

Leokris [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

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2 years ago

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egoroff_w [7]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers