Measure of Angle B round to the nearest hundredth

Plz answer this for 10 points

ki77a [65]

domain refers to the X values so option A would be correct :)

4 0

3 years ago

Evacuate the following using suitable identities (102)^3

snow_lady [41]

The value of $(102)^3$ exists 1061208.

<h3>How to estimate the value of $(102)^3$ ?</h3>

Let us rewrite $(102)^3$ as $(100+2)^3$

Now utilizing the identity $(a+b)^3=a^3+b^3+3ab(a+b)$ , we get

a = 100 and b = 2 then substitute the values of a and b then

$(100+2)^3=100^3+2^3+[(3\times100\times2)(100+2)]$

= 1000000 + 8 + (600 × 102)

= 1000000 + 8 + 61200

= 1061208

Hence, $(102)^3=1061208$

Therefore, the value of $(102)^3$ exists 1061208.

To learn more about cubic polynomial equation refer to:

brainly.com/question/28181089

#SPJ9

5 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

1 When the product of 4 and a number x is added to 12, the result is 16. What is the difference between x and 8?

Aleks [24]

Answer:

x = 1

Step-by-step explanation:

12 + 4 x = 16

Subtract 12 from both sides.

4x = 4

Divide 4 on both sides.

x = 1

3 0

3 years ago

Fair rickets for 2 adults and 3 children cost $34. An adult ticket costs $2 more than a child ticket. What is the price of an ad

Studentka2010 [4]

Answer:

Step-by-step explanation:

An adult ticket costs $2 more than a child ticket

Let x and (x + $2)represent the cost of the child and adult ticket respectively

2(x+ $2) + 3x = $34

now solving for x

2x+ $4 + 3x = $34

5x = $30

x = $6 , the cost of the child ticket.

the cost of adult ticket $6+2=$8

3 0

3 years ago