Set up an equation that can be used to solve the problem. Solve the equation and answer the question asked.

A pool measuring 16 meters by 22 meters is surrounded by a path of uniform width, as shown in the figure. If the area of the po

lesantik [10]

Let X be the width of the path.

The width of the path would be the width of the pool plus 2x, so 2x +16

The length of the path would be the length of the pool plus 2x, so 2x+22

The total area is 832 square feet.

Area is found by multiplying the length by the width.

So you have:

(2x+16) (2x+22) = 832

Use the Foil Method:

(2x+16) (2x+22) = 4x^2 +76x+352

Now you have:

4x^2 +76x+352 = 832

Subtract 832 from both sides:

4x^2 +76x+352-832 = 0

Simplify:

4x^2 +76x -480

Factor the left side:

4(x-5)(x+24) = 0

Divide both sides by 4:

(x-5)(x+24) = 0

Set both parenthesis to equal 0 and solve for x:

x=5 and x = -24

The width of the path cannot be a negative number, so the width is 5 meters.

8 0

3 years ago

Help !!!!!!!!!!!!!!!

Ostrovityanka [42]

Answer:

x = 29

Step-by-step explanation:

Follow the directions: substitute 13 for y in the first equation.

x + 13 = 42

Subtract 13 from both sides of the equation.

x +13 -13 = 42 -13

x = 29

4 0

3 years ago

Given m||n, find the value of x.

kkurt [141]

As angle x° and angle 76° are alternate interior angles their measure will be equal.

Which means :-

$x° = 76° \:$

Therefore , the value of x = 76° .

7 0

3 years ago

A monarchy has one leader how many terms do you think a monomial has

Sholpan [36]

I would think i has one term because the prefix "mono" means one

3 0

3 years ago

Read 2 more answers

Bodytemperatures ofhealthy koalasarenormally distributed with a mean of 35.6°C and a standard deviation of 1.3°C. a.What is the

Licemer1 [7]

Answer:

a

$P(X < 35) = 32.3 \%$

b

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

c

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 35.6^oC$

The standard deviation is $s = 1.3^oC$

The sample size is n = 30

Generally the probability that a health koala has a body temperature less than 35.0°C is mathematically represented as

$P(X < 35) = P(\frac{X - \mu }{s} < \frac{35 - 35.6}{1.3} )$

Here $(\frac{X - \mu }{s} = Z (The \ standardized \ value \ of \ X )$

So

$P(X < 35) = P(Z < -0.46)$

From the z-table P(Z < -0.46) = 0.323

So

$P(X < 35) = 0.323$

Converting to percentage

$P(X < 35) = 0.323 * 100$

$P(X < 35) = 32.3 \%$

considering question b

The sample mean is $\= x = 35$

Generally the standard error of the mean is mathematically represented as

$\sigma_{\= x} = \frac{s}{\sqrt{n} }$

=> $\sigma_{\= x} = \frac{1.3}{\sqrt{30} }$

=> $\sigma_{\= x} = 0.2373$

Generally the probability of the mean body temperature of koalas being less than 35.0°C is mathematically represented as

$P(\= X < 35) = P(\frac{\= X - \mu }{\sigma_{\= x }} < \frac{35 -35.6}{0.2373 } )$

$P(\= X < 35) = P(Z< -2.53 )$

From the z-table we have that

$P(Z< -2.53 ) = 0.006$

So

$P(\= X < 35) = 0.006 /tex] Converting to percentage [tex]P(\= X < 35) = 0.006 * 100$

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

4 0

4 years ago