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3 years ago

Write standard form a.) 0.0000918 b.) 12.002×10-³

Mathematics

1 answer:

ad-work [718]3 years ago

8 0

Answer:

A. 9.18 x $10^{5}$

B.12002

Step-by-step explanation:

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Stanley is having trouble understanding how to find the GCF

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

All the factors of 8 and 16 are 1;2;4;8

3 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

3 years ago

Triangle ABC is dilated by a scale factor of 2 with a center of dilation at the origin. Finish the rule for this dilation.

Vitek1552 [10]

Abc is your answer

Step-by-step explanation:

you said ABC

4 0

3 years ago

Really need help on this!!

marin [14]

Answer:

Hey there!

There are 3+4+6+3, or 16 females.

3 of the females are seniors.

Thus, the percent is 3/16, or 19%

Let me know if this helps :)

4 0

3 years ago

In AVWX, XV<br> W X and mW<br> - 27°. Find m_X.

SashulF [63]

Answer:

m<X = $126^{o}$

Step-by-step explanation:

From the given isosceles triangle, we have;

<W and <V as the base angles

So that,

m<W = m<V = 27 (base angles of an isosceles triangle are equal)

Thus,

m<X + m<V + m<W = 180 (sum of angles in a triangle)

m<X + 27 + 27 = 180

m<X + 54 = 180

m<X = 180 - 54

= 126

m<X = $126^{o}$

8 0

2 years ago

Write standard form a.) 0.0000918 b.) 12.002×10-³​

Write standard form a.) 0.0000918 b.) 12.002×10-³