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2 years ago

Plzz what the answer I’m stuck

Mathematics

1 answer:

castortr0y [4]2 years ago

5 0

Answer:

the answer to your question is 576 square feet

Step-by-step explanation:

Step 1: 96 ÷ 4 = 24

Step 2: 24×24 =576

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Standar form of 1.204751131

densk [106]

1.204751131 × $10^{0}$

6 0

3 years ago

Given the functions, f(x) = 5x2 - 3x + 1 and g(x) = 2x2 + x - 2, perform the indicated operation. When applicable, state the dom

RUDIKE [14]

You plug in 1,2,3 and so on in the X and you graph , i think

3 0

3 years ago

Find the area of the circle with a circumference of

Anton [14]

Answer:

706.50 mm^2

Step-by-step explanation:

The formula to find the circumference is 2πr (r : radius)

if the circumference is 30π the the radius is = 15

to find the area we use the following formula :

π*r^2

r = 15 and π is given as 3.14

(3.14)*15^2 = 706.50 mm^2

8 0

2 years ago

Find the diameter of each circle.

Vedmedyk [2.9K]

Answer:

Diameter Length: ( About ) 5.4 km; Option B

Step-by-step explanation:

~ Let us apply the Area of the Circle formula πr^2, where r ⇒ radius of the circle ~

1. We are given that the area of the circle is 22.9 km^2, so let us substitute that value into the area of the circle formula, solving for r ( radius ) ⇒ 22.9 = π * r^2 ⇒ r^2 = 22.9/π ⇒ r^2 = 7.28929639361.... ⇒

radius = ( About ) 2.7

2. The diameter would thus be 2 times that of the radius by definition, and thus is: 2.7 * 2 ⇒ ( About ) 5.4 km

Diameter Length: ( About ) 5.4 km

5 0

3 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

2 years ago