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
GREYUIT [131]
3 years ago
7

Write the equation 2x + 4y = 12 in slope intercept form.

Mathematics
1 answer:
Gwar [14]3 years ago
7 0
4y=12-2x
y=(12-2x)/4
-------------
You might be interested in
Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean
Natali [406]

Answer:

The percentage of snails that take more than 60 hours to finish is 4.75%

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

There is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours.

This means that \mu = 50 and \sigma = 6.

The percentage of snails that take more than 60 hours to finish is %

The pvalue of the zscore of X = 60 is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 50}{6}

Z = 1.67

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

The relative frequency of snails that take less than 60 hours to finish is

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is:

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

X = 67

Z = \frac{X - \mu}{\sigma}

Z = \frac{67 - 50}{6}

Z = 2.83

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

p = .9977 - 0.9525 = .0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 100% subtracted by the pvalue of the Zscore of X = 76.

Z = \frac{X - \mu}{\sigma}

Z = \frac{76 - 50}{6}

Z = 4.33

The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most hours.

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

Z = \frac{X - \mu}{\sigma}

-1.29 = \frac{X - 50}{6}

X - 50 = -7.74

X = 42.26

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical 80% of snails take between and hours to finish.

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 50}{6}

X - 50 = 7.68

X = 57.68

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

5 0
3 years ago
What is the answerrrrr to this
Alex_Xolod [135]
Area = 140

16 x 8 = 128

16-12 = 4

(6x4)/2 = 12

128 + 12= 140
8 0
2 years ago
Write an equation for a polynomial function that has zeros of {3,2}
REY [17]

Answer:

  • f(x) = x² - 5x + 6

Step-by-step explanation:

<u>There are two roots:</u>

  • x = 3 and x = 2

<u>The function is:</u>

  • f(x) = (x - 3)(x - 2)
  •      = x² - 5x + 6
8 0
2 years ago
What is 178to the nearest hundred
Kay [80]
800 is to the nearest hundred
7 0
3 years ago
Read 2 more answers
Identify the percent change <br><br> 28 seconds to 23 seconds
Zarrin [17]
Percent change=change/original times 100
chane=28-23=5
oriiginal=28

percent change=5/28 times 100
0.178 times 100
17.8

about 18% change
4 0
3 years ago
Other questions:
  • PLZ NEED HELP WILL MARK BRAINLIEST IF RIGHT!!!!!
    15·2 answers
  • The quesiton is in the picture
    10·1 answer
  • From a rope 15m long, 4 1/3 m is cut off and 3/5 of the remaining is cut off again. Find the length of the remaining part of the
    14·1 answer
  • If r = 12, what is the value of t?
    12·2 answers
  • A boy throws a rock off a cliff. The function shown below models the flight of the rock, where x equals the distance the rock is
    12·1 answer
  • What does this equation mean?
    9·2 answers
  • ♦HURRY PLEASE!!! ♦<br><br> ↓<br> Don't worry about the answer choice that is highlighted : )
    14·1 answer
  • (-2,1); perpendicular to y= - 2/5x -4<br><br> Help explain step by step?
    15·1 answer
  • Which function is continuous at x = 5?
    14·2 answers
  • 35-{15-(19+5)÷3 simplify​
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!