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3 years ago

Cindy is 5 feet tall with a shadow of 5 feet. The flagpole has a shadow of 12.5 feet. What is the height of the flagpole?

2 answers:

7 0

I am guessing cindy is standing inside the shadow of the flagpole. So the answer would be 12.5. To answer the question you have to set up proportions. 12.5/5=x/5, so you would cross multiply 12.5 and 5 then 5 and x so 62.5=5x then divide by 5 for 12.5.

Black_prince [1.1K]3 years ago

3 0

I don't want to give you the answer so here is a thorough explanation and an answer if you want it. Set this up as a proportion. Set it up as 5/5=x/12.5. Then multiply 5 by x to get 5x and 12.5 by 5. Once you have this answer, divide it by five to isolate the x. Once you have finished this, you have your answer.

I got the answer 12.5. Make sure you check this so you understand how to solve these kinds of problems.

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How did i work this out?

bogdanovich [222]

knowing the answer is between 3 & 4 plug those as x and see what you get for answers than adjust the number accordingly

3^3+2(3) = 33 so its higher than 3

4^3 + 2(4) = 72 so it is lower than 4

try the middle so try 3.5

3.5^3 + 2(3.5) = 49.875

question states use 1 decimal place so now try 3.6

3.6^3+2(3.6) = 53.856

3.5 is closest

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3 years ago

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goldenfox [79]

Answer:

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Step-by-step explanation:

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3 years ago

$49.95 CD player; 5% tax

jekas [21]

Answer:

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Step-by-step explanation:

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3 years ago

Please could you help me with this. 132 people took a driving test. 80 people predicted they would pass. 64 people

GenaCL600 [577]

You have to do 64 x 3 and that equals 192 the answer is 192 if that was what you was asking! Respond if not!

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2 years ago

The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ric

Papessa [141]

Let $X$ denote the number of minutes spent brushing his teeth, and let $\mu$ be the mean and $\sigma$ the standard deviation for this distribution.

$\mathbb P(X$

The z-score corresponding to this probability is approximately $z=-0.2533$ , which means

$\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1$

Next, (note the sign change)

$\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98$

The corresponding z-score is approximately $z=2.0538$ , so you have

$\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2$

Solving the two equations for $\mu$ and $\sigma$ , you'll find that the mean is approximately $\mu=1.1098$ and the standard deviation is approximately $\sigma=0.4334$ .

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3 years ago