answer.
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
ehidna [41]
2 years ago
10

The graph shows the linear relationship between the height of a plant (in centimeters) and the time (in weeks) that the plant ha

s been growing. Which statements are correct? Check all that apply. The rate of change is 4. The rate of change is 1. The rate of change is . The plant grows 4 cm in 1 week. The plant grows 1 cm in 4 weeks

Mathematics
2 answers:
notka56 [123]2 years ago
8 0

Answer:

A. The rate of change is 4,  C. The rate of change is 4/1,  and D. The plant grows 4 cm in 1 week.

Step-by-step explanation:

It was right on edge

bija089 [108]2 years ago
5 0
I found a graph with the same problem, so I guess this is the graph to be based on. To find the rate, just determine the slope between any two points along the line. Suppose these points are: (30,10) and (50,20).

Slope = (20 - 10)/(50 - 30) = 1/2

<em>The correct answer should be: the rate of change is 1 cm per 2 weeks, or 1/2.</em>

You might be interested in
1. In Suri's coin purse she has 6 dimes and
ollegr [7]

Answer:

The ratios are not equal

Step-by-step explanation:

Let

x ----> the number of dimes

y ----> the number of quarters

we know that

<em> Suri's coin purse</em>

x=6, y=4

The ratio of dimes  to quarters is

\frac{x}{y} =\frac{6}{4}

<em> Martha's coin purse</em>

x=5, y=3

The ratio of dimes  to quarters is

\frac{x}{y} =\frac{5}{3}

Compare the ratios

\frac{6}{4}=\frac{5}{3}\\6(3)=4(5)\\18\neq 20

therefore

The ratios are not equal

3 0
2 years ago
Alexi divided 256 by 16. Which could he have been converting? 256 ounces to pounds 256 pounds to ounces 256 pounds to tons 256 t
dedylja [7]
Pounds to ounces. This is because there are 16 ounces in a pound
6 0
2 years ago
Read 2 more answers
If f(x)=13x3−4x2+12x−5 and the domain is the set of all x such that 0≤x≤9 , then the absolute maximum value of the function f oc
Fittoniya [83]

The correct format of the question is

If f(x)= \frac{1}{3}x^3 -4x^2+12x -5 and the domain is the set of all x such that 0≤x≤9 , then the absolute maximum value of the function f occurs when x is

Answer:

The absolute maximum value of the function F(x) occurs when x is 9

Step-by-step explanation:

F'(x) = x^2 -8x +12= 0

      = (x-6)(x-2) = 0

      x = 2,6

so we have boundary points 0 , 9 and 2,6

The value of function at these four points

  x =     0    2       6     9

F(x) =  -5    17/3   -5     22

So the absolute maximum value of the given function is x = 9 and F(x) is 22.

8 0
2 years ago
find the probability that a randomly selected automobile tire has a tread life between 42000 and 46000 miles
maria [59]
Given that in a national highway Traffic Safety Administration (NHTSA) report, data provided to the NHTSA by Goodyear stated that the mean tread life of a properly inflated automobile tires is 45,000 miles. Suppose that the current distribution of tread life of properly inflated automobile tires is normally distributed with mean of 45,000 miles and a standard deviation of 2360 miles.

Part A:

Find the probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is between two numbers, a and b is given by:
P(a \ \textless \  X \ \textless \  b) = P(X \ \textless \  b) - P(X \ \textless \  a) \\  \\ P\left(z\ \textless \  \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \  \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles is given by:
P(42,000 \ \textless \ X \ \textless \ 46,000) = P(X \ \textless \ 46,000) - P(X \ \textless \ 42,000) \\ \\ P\left(z\ \textless \ \frac{46,000-45,000}{2,360} \right)-P\left(z\ \textless \ \frac{42,000-45,000}{2,360} \right) \\  \\ =P(0.4237)-P(-1.271)=0.66412-0.10183=\bold{0.5623}


b. Find the probability that randomly selected automobile tire has a tread life of more than 50,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, a, is given by:
P(X \ \textgreater \  a) = 1-P(X \ \textless \ a)  \\  \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of more than 50,000 miles is given by:
P(X \ \textgreater \  50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\  \\ =1-0.98294=\bold{0.0171}


Part C:

Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, a, is given by:
P(X \ \textless \  a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of less than 38,000 miles is given by:
P(X \ \textless \  38,000) = P\left(z\ \textless \ \frac{38,000-45,000}{2,360} \right) \\  \\ =P(z\ \textless \ -2.966)=\bold{0.0015}


d. Suppose that 6% of all automobile tires with the longest tread life have tread life of at least x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, x, is given by:
P(X \ \textgreater \ x) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at least x miles is 6%.

Thus:
P(X \ \textgreater \ x) =0.06 \\  \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\  \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\  \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\  \\ \Rightarrow x=3,669.8+45,000=48,669.8
Therefore, the value of x is 48,669.8


e. Suppose that 2% of all automobile tires with the shortest tread life have tread life of at most x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, x, is given by:
P(X \ \textless \ x) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at most x miles is 2%.

Thus:
P(X \ \textless \ x)=0.02 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.02=0.98 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 2.054) \\ \\ \Rightarrow \frac{x-45,000}{-2,360}=2.054 \\ \\ \Rightarrow x-45,000=-2,360(2.054)=-4,847.44 \\ \\ \Rightarrow x=-4,847.44+45,000=40,152.56
Therefore, the value of x is 40,152.56
4 0
2 years ago
Ms. Patel invested a total of $825 in two stocks. At the time of her investment, one share of Stock A was valued at $12.41 and a
Rom4ik [11]

Answer:

D

Step-by-step explanation:

a+b=79(compares the number of stocks purchased)

12.41a+8.62b=825(compares the prices of the stocks purchases)

a=38, b=41

7 0
2 years ago
Other questions:
  • In the diagram, transversal t cuts across the parallel lines a and b. Match the pairs of angles with the relationship that shows
    6·1 answer
  • Identify the volume of a cone with base area 64π m2 and a height 4 m less than 3 times the radius, rounded to the nearest tenth.
    9·1 answer
  • Which equations represent the data in the table? Check all that apply.
    15·2 answers
  • Two standardized​ tests, a and​ b, use very different scales of scores. the formula upper a equals 40 times upper b plus 50a=40×
    5·1 answer
  • A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15.
    15·1 answer
  • What integer represents a rise in temperature of 19°?
    11·1 answer
  • Advertisers often refer to CPM (Cost per thousand people reached) when quoting advertising rates. If a website
    15·2 answers
  • A coach purchases 47 hats for his players and their families at a total cost of $302. The cost of a small hat is $5.50. A medium
    5·1 answer
  • Scarlett is designing a package for a candy her company makes. She has cut several cardboard equilateral triangles, squares, rec
    5·1 answer
  • Solve 7x- c= k for x. <br>A. X = 7(k+C) <br>B. x = 7(K-C) <br>C. X = k+c/7 <br>D. X= k-c/7​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!