1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Montano1993 [528]
1 year ago
9

Write an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the line described by y=1/2

+5

Mathematics
1 answer:
Marat540 [252]1 year ago
8 0

Given:

The point, (4, -3)

The line,

y=\frac{1}{2}x+5

To find an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the given line:

The slope of the line is,

m=\frac{1}{2}

Since the given line is parallel to the new line, so the slope will be same for the both.

Using the point-slope formula,

y-y_1=m(x-x_1)

Substitute the point and slope we get,

\begin{gathered} y-(-3)=\frac{1}{2}(x-4) \\ y+3=\frac{1}{2}x-2 \\ y=\frac{1}{2}x-2-3 \\ y=\frac{1}{2}x-5 \end{gathered}

Hence, the equation in slope-intercept form for the line is,

y=\frac{1}{2}x-5

You might be interested in
Choose the number sentence that illustrates the distributive property of multiplication over addition.
patriot [66]

Answer:

  • Option A

Step-by-step explanation:

A.

  • 5 × (4 + 9) = (5 × 4) + (5 × 9), correct

B.

  • 5 × (4 + 9) = (5 × 4) + 9, incomplete, missing 9 x 5

C.

  • 5 × (4 + 9) = (5 + 4) × (5 + 9), wrong
5 0
2 years ago
Help with #1 please show work!
horrorfan [7]
1. 4n=g
;
{24*4=96
23*4=92
19*4=76
15*4=60}
2. Each question is worth 4 points. For every dependent variable, multiply by 4 to get the independent variable.

Hope this helped☺☺

6 0
3 years ago
A study was designed to investigate the effects of two variables ​(1) a​ student's level of mathematical anxiety and​ (2) teachi
kicyunya [14]

Answer:

Solution using the Empirical Rule, which is explained here.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

The question is incomplete, but you solve this using the Empirical Rule.

8 0
2 years ago
A certain reaction was run several times using each of two catalysts, A and B. The catalysts were supposed to control the yield
S_A_V [24]

Answer:

a) See the code and the figure attached below

b) See the code and the figure attached below

c) We see significant differences on the boxplots. For Catalyst A the variation is less than the variation for Cat B. For both categories we see that we don't have a symmetric distributions. The maximum for Cat A is a little higher than the maximum for Cat B. The median for Cat A is a little higher than Cat B. And the third quartile for Cat A is lower than the third quartile for Cat B.

Step-by-step explanation:

Data given

Catalyst A: 4.4,3.4,2.6,3.8,4.9,4.6,5.2,4.7,4.1,2.6,6.9,0.8,3.6,2.9,2.6,4.0,4.3 .9,4.8,4.5,4.4,3.1,5.7,4.5

Catalyst B: .4,1.1,2.9,5.5,6.4,5.0,5.8,2.5,3.7,3.8,3.1,1.6, 3.5,5.9,6.7,5.2, 6.3, 2.6,4.3, 3.8

Solution to the problem

We are going to use R in order to do the plots required, we put the code and the output its attached to the solution.

Part a

R Code:

> catA<-c(4.4,3.4,2.6,3.8,4.9,4.6,5.2,4.7,4.1,2.6,6.9,0.8,3.6,2.9,2.6,4.0,4.3,0.9,4.8,4.5,4.4,3.1,5.7,4.5 )

> length(catA)

[1] 24

> catB<-c(.4,1.1,2.9,5.5,6.4,5.0,5.8,2.5,3.7,3.8,3.1,1.6,3.5,5.9,6.7,5.2,6.3,2.6,4.3,3.8 )

> length(catB)

[1] 20

> hist(catA)

> hist(catB)

Part b

R code:

> par(mfrow=c(1,2))

> boxplot(catA,main="Boxplot for CatA")

> boxplot(catB,main="Boxplot for CatB")

The result is attached.

Part c

We see significant differences on the boxplots. For Catalyst A the variation is less than the variation for Cat B. For both categories we see that we don't have a symmetric distributions. The maximum for Cat A is a little higher than the maximum for Cat B. The median for Cat A is a little higher than Cat B. And the third quartile for Cat A is lower than the third quartile for Cat B.

8 0
3 years ago
Before the summer season begins, you need to decide how many postcards to stock in your shop. Historically, the number of postca
Vikentia [17]

Answer:

You should stock 59 thousands of postcards

Step-by-step explanation:

In the standard normal distribution table, you want to find the Z with probability 0.4, which is (see figure attached) 1.28. Z is calculated as follows:

Z = (x - μ)/σ

where μ is the mean and σ is the standard deviation. Replacing with data:

1.28 = (x - 50)/7

x = 1.28*7 + 50 = 59 thousands of postcards

So, approximately 80% of the demand would be satisfied with a number of postcards between 41000 and 59000 (notice the curve is symmetrical).

8 0
3 years ago
Other questions:
  • Please help thank you very much
    12·1 answer
  • Given g(x) is a linear function and passes through the points (3,2) and (5,-1),
    8·1 answer
  • 12 - 8 x 9 +16<br> Order of operation
    14·1 answer
  • Michael is 6 feet tall and is standing on top of
    9·1 answer
  • 10 POINTS!!! PLEASE ANSWER SOON! THANK YOU!
    14·1 answer
  • If a number is divisible by 2,5,9 what is one digit of the number
    11·1 answer
  • Simplify the expression -2m-6(m+1)​
    14·2 answers
  • How many tiles each 30cm by 30cm will be needed for floor​
    13·2 answers
  • A-15=4a-3<br> if u dont want spam then dont answer
    12·1 answer
  • Simply the following expression: 8x + 5y - 4z + 9y - 3x + 13z
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!