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3 years ago

Finding the Area of a Trapezoid

Mathematics

2 answers:

taurus [48]3 years ago

7 0

Answer:

the area of the rectangle is 44cm

the area of the triangle is 17/18?

and the area of the hexagon is 49/50?

Step-by-step explanation:

its a bit confusing

aalyn [17]3 years ago

5 0

Answer:

120, 24, 144

Step-by-step explanation:

1. 10 times 12=120

2. 4 times 12 divided by 2=24

3. add 24 and 120

Area of a rectangle is Length X Height

Area of a triangle is Length X Height / 2

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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

Pls help !!<br><br> Solve and CHECK.<br><br> 1 / 4 ( p - 7) = 1/6 ( p - 3)

victus00 [196]

Answer:

the answer would be p= 15

Step-by-step explanation:

3 0

3 years ago

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How many pounds of walnuts that cost $0.80 per pound must be mixed

jeyben [28]

Answer:

kt 5.20 yes

Step-by-step explanation:

6 0

3 years ago

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A person recently read that 84% of cat owners are women. How large a sample should the researcher take if she wishes to be 90% c

Lana71 [14]

Answer:

405

Step-by-step explanation:

To find sample size, use the following equation, where n = sample size, za/2 = the critical value, p = probability of success, q = probability of failure, and E = margin of error.

$n=\frac{(z_{\frac{\alpha }{2} })^{2} *p*q }{E^2}$

The values that are given are p = 0.84 and E = 0.03.

You can solve for the critical value which is equal to the z-score of (1 - confidence level)/2. Use the calculator function of invNorm to find the z-score. The value will given with a negative sign, but you can ignore that.

(1 - 0.9) = 0.1/2 = 0.05

invNorm(0.05, 0, 1) = 1.645

You can also solve for q which is 1 - p. For this problem q = 1 - 0.84 = 0.16

Plug the values into the equation and solve for n.

$n =\frac{(1.645)^2*0.84*0.16}{(0.03)^2}\\n= 404.0997333$