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3 years ago

5

On a sunny day, cindy notices that her shadow is 21 inches long. if it is 75 inches from the top of her head to the end of the s

hadow, how tall is cindy?

1 answer:

Ray Of Light [21]3 years ago

5 0

Cindy is 72 inches long. Imagine it like a right triangle. Her shadow is the base, she is the side, and the distance between her head and the end of the shadow is the diagonal line connecting the two. Use the Pythagorean theorem to find how tall cindy is.

a²+b²=c²        a=21, the length of the shadow
                       b= ? cindy's height
                     c= 75, the distance from her head to the edge of the shadow
21²+ b²=75²
441 +b² = 5625
-441         -441
------         -------
0             5184
b² = 5184
√b² =√5184
b= 72

hope it helps

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Cos 0 = 12/13. Find sin 0.<br> A. 5/12<br> B. 12/5<br> C. 13/12<br> D. 5/13

andreyandreev [35.5K]

Answer:

Sine θ = 5/13

Step-by-step explanation:

From the question given above, the following data were obtained:

Cos θ = 12/13

Sine θ =?

Next, we shall determine the opposite. This can be obtained as follow:

Cos θ = Adjacent /Hypothenus

Cos θ = 12/13

Adjacent = 12

Hypothenus = 13

Opposite =?

Hypothenus² = Opposite² + Adjacent²

13² = Opposite² + 12²

169 = Opposite² + 144

Collect like terms

169 – 144 = Opposite²

25 = Opposite²

Take the square root of both side

Opposite = √25

Opposite = 5

Finally, we shall determine the Sine θ. This can be obtained as follow:

Opposite = 5

Hypothenus = 13

Sine θ =.?

Sine θ = Opposite / Hypothenus

Sine θ = 5/13

8 0

2 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

A Ski resort tracks the proportion of seasonal employees who are rehired each season. Rehiring a seasonal employee is beneficial

DaniilM [7]

The test statistic z will be equal to -0.946 and it shows that there is no significant difference in the proportion of rehires between full time and part time.

Given sample sizes of 833 and 386 and result of samples 434 and 189.

$n_{1} =833, n_{2}=386$

Proportion of full time=434/833=0.52

Proportion of part time=189/386=0.49.

Difference in proportion =0.52-0.49

$H_{0}:$ TTF- i∈ rho=0

$H_{1}:$ TTF+i∈ rho≠0.

Mean of difference=0.03

Z=(X-μ)/σ

σ= $\sqrt{0.4517/447}$

=0.0317

σ=0.0317

z=(0-0.03)/0.0317

=-0.03/0.0317

=-0.317

p value will be =0.1736.

Because p value is greater than 0.01 so we will accept the null hypothesis which shows that there is no significant difference in the proportions.

Hence there is no significant difference in the proportion of rehires between full time and part time.

Learn more about z test at brainly.com/question/14453510

#SPJ4

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2 years ago

Mark had 4 members in his family each of the person get 1/6 what Is the fraction of the tomato use fraction strips

umka21 [38]

If each person got 1/6 then there is still 2/6 left over, =1/3. In fraction strips, we would do

1

1/3 1/3 1/3 1/3

1/6 1/6 1/6 1/6 1/6 1/6 for the overall tomato and then 1/6 1/6 for the remaining piece of the tomato.

Hope this helps, if you need any more help just comment :)

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3 years ago

What is this shape which is 75 mm in diameter and 112 mm in height.<br> This shape has to be a prism

klasskru [66]

The shape you are looking for is a pyramyde

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3 years ago