Which expression has the same value as -8x- (-5y)?

kiruha [24]

An irrational number cannot be expressed as a ratio between two numbers and it cannot be written as a simple fraction because there is not a finite number of numbers when written as a decimal. Instead, the numbers in the decimal would go on forever, without repeating.

8 0

3 years ago

For a science class, Sweet T counts clusters of bacteria growing in two Perri dishes. He records his finding as a linear functio

GREYUIT [131]

The correct answer is that "Dish A has a greater rate of change."

In Dish A, the values are going up by 2 each day. In Dish B, the values are staying the same.

Therefore, Dish A has a greater rate of change.

8 0

3 years ago

Read 2 more answers

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

The quotient of z and 8

kiruha [24]

Answer:

z/8

Step-by-step explanation:

cuz it says z and 8 in that order so z/8 please dont just ignore give me 5 star and thanks dont ignore if u do i will never help you

4 0

3 years ago

Help! this is in algebra 1

Scrat [10]

Answer: I have the pictures attached

Step-by-step explanation:

In order to graph this, we have to get it into this equation: y = mx + b, where m = slope and b = y intercept.

y = 3/2x - 4 is already in this form.

2y + 4 = 2 + 3x is not, so we have to isolate y

2y + 4 - 4 = 2 + 3x - 4

2y = -2 + 3x

2y/2 = -2/2 + 3x/2

y = -1 + 3/2x

y = 3/2x - 1

Okay, now graph it knowing your y intercepts and your slopes.

7 0

3 years ago