Find the inequality represented by the graph

Which lines have slope -4/5 and contains point (0,1)?

miv72 [106K]

ANSWER

The required line is BC

EXPLANATION

We want to find the line that passes through
$(0,1)$

and has slope
$m = - \frac{4}{5}$

Since the x-value of the given point is zero, the point

$(0,1)$

is the y-intercept.

From the above graph, the two lines that contains
$(0,1)$

are line CE and line BC.

Since, the slope

$m = - \frac{4}{5}$
is negative, line BC is the required line because it is sloping backwards.

We can confirm this by using the point,

$B(-5,5)$

and

$C(0,1)$

to find the slope and obtain

$m = \frac{5 - 1}{ - 5 - 0} = - \frac{4}{5}$

3 0

3 years ago

Connie Hargrave's tree fell in the last storm and damaged the car of

Nimfa-mama [501]

The answer to this question is D

5 0

3 years ago

The following graphs show the performance of two stocks over the first five months of the year. A graph titled Stock A B C has m

hichkok12 [17]

Answer:

C) Stock XYZ shows the best performance. The scale of Stock XYZ makes the graph appear to be rising more slowly.

Step-by-step explanation:

i did this question on my test, from reading the graph, i am almost certain this is the answer. correct me if i am wrong!

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

GIVING BRAINIEST IF CORRECT!!!

GREYUIT [131]

Answer:

C. 2/5 full

Step-by-step explanation:

$\frac{1}{\frac{1}{15} } :\frac{6}{y}$

y · 1 = 1/15 · 6

y = 0.4

$y=\frac{2}{5}$

6 0

3 years ago

Find the inequality represented by the graph​

Find the inequality represented by the graph