All categories

3 years ago

Write down the equation of a line that is parallel to the line with equation y=7-4X

Mathematics

1 answer:

Leona [35]3 years ago

4 0

Answer:

y=8-4x

Step-by-step explanation:

You might be interested in

HELP QUICK ME QUICK PLEASE THANKS

GenaCL600 [577]

To solve this problem, we need to know the formula for the nth term of an arithmetic sequence. The formula is $a_{n}$ = t₁ + (n-1)d, with $a_{n}$ being the nth term, t₁ being the first term, and d being the distance between terms. Now we can substitute values to figure out what $a_{n}$ is. When substituting, the equation becomes:

$a_{n}$ = 12 + 29(-6)
$a_{n}$ = -162

That means that the 30th term in this arithmetic sequence is -162.

8 0

3 years ago

Read 2 more answers

3. What is the decimal equivalent of each

zaharov [31]

Answer:

a. 0.81818182

b. 0.8

c. -3.4

d. 0.55555556

Step-by-step explanation:

8 0

3 years ago

Have a nice day;) it’s ok to be tired but keep pushing. Just a lil while more

Savatey [412]

Answer:

thx

Step-by-step explanation:

6 0

3 years ago

Please help!! Will mark first answerer brainiest!! Thank you so much!

Mrac [35]

The answer is C i believe, always remember to add alike terms.

8 0

3 years ago

Read 2 more answers

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$

5 0

4 years ago