Based on the graph, what is the initial value of the linear relationship?

Lynn is making custom bricks. She combines 39 pounds of water and 48 pounds of brick dust in a mixing bucket. She spills 13 poun

ehidna [41]

Lynn mixes 39 pounds of water with 48 pounds of brick dust, so the mixture has a total weight of 39 + 48 = 87 pounds.

13 pounds are spilled during mixing, so she's left with 87 - 13 = 74 pounds to turn into bricks.

7 pounds are needed for 1 brick, so she can make 10 bricks because 7 • 10 = 70 and 7 • 11 = 77, and 70 < 74 < 77.

If she uses the mixture to make 10 bricks, she uses up 70 pounds of it, which leaves 4 pounds to be washed out.

5 0

3 years ago

Y=-1/4x-5 and 2x+8y=56 determine if paralllel ,nethier or perpendicular

Dominik [7]

Answer:

Parallel

Step-by-step explanation:

In the slope-intercept form (y=mx +c), the coefficient of x tells us the slope of the line.

2x +8y= 56

Let's rewrite this equation into the slope-intercept form.

8y= -2x +56

Dividing both sides by 8:

$y= - \frac{2}{8} x + \frac{56}{8}$

$y = - \frac{1}{4} x + 7$

Slope= -¼

y= -¼x -5

Slope= -¼

Since both lines have the same slope, they are parallel to each other.

3 0

3 years ago

Four times a number w minus 3.5 is less than or equal to -1.5

DIA [1.3K]

Answer:

This is your answer ☺️☺️

6 0

3 years ago

if we add 150 to the minuend of a subtraction and we remove 220 from the subtrahend. How much does the difference vay?

Alex17521 [72]

Answer:

Step-by-step explanation:

(m+150) - (s-200) = m-s + 350

The difference increases by 350

7 0

3 years ago

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0

4 years ago