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3 years ago

A homeowner uses a $10 bill to pay a neighbor for cutting his lawn. The

Mathematics

1 answer:

timama [110]3 years ago

7 0

Answer:

Step-by-step explanation:

It has a constant value.

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What does supplementary mean

murzikaleks [220]

Any two numbers, normally degrees, such as 10° and 170°, that add up to equal 180°

3 0

3 years ago

Which ordered pair is a solution of the equation -3x - y = 6

e-lub [12.9K]

Hello there!

The correct answer is option B (-3,3)

-3x - y = 6

-3(-3) - 3 = 6

9 - 3 = 6

6 = 6

Thus, the correct answer is option B

Good luck with your studied!

8 0

3 years ago

What are the coordinates of the point (-3,4) after a translation of 2 units down and 5 units left

marta [7]

Answer:

(-8, 2)

Step-by-step explanation:

Start with (-3, 4). Subtract 5 units from -3, obtaining -8, and subtract 2 units from 4, obtaining 2: (-8, 2)

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

3 years ago

Choose the solution for the equation 3(p+8) = 2(2p-1)

artcher [175]

Answer:

b) p=26

Step-by-step explanation:

3p + 3 · 8 = 2 ·2p - 2

3p + 24 = 4p - 2

p = 26

4 0

3 years ago