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3 years ago

Which of the following equations best models the data plotted below?

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

6 0

The key with these problems is to find which function has the closest y-intercept to the graph, and then try to figure out which one best approximates the slope.
Here are our options:
A. y = x + 4
B. y = 4x + 9
C. y = x + 18
D. y = 3x + 22

Which has the closest approximation of the y-intercept?
The y-intercept is not directly given, but we can assume it is less than 10.
That leaves us with A and B.

Which has the closest approximation of the slope?
The graph, on average, seems to move up about 60 and over about 15.
Slope = rise/run = 60/15 = 4. Although the slope isn't exactly 4, it's much closer to 4 than 1, which is slope for option A.

Therefore, the answer is
B) y= 4x + 9

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CaHeK987 [17]

Answer:

He must sell 800 bushels.

Step-by-step explanation:

Divide 2400 by 3, giving you 800.

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3 years ago

Read 2 more answers

Which equation would have real zero(s) corresponding to the x-intercept(s) of the graph below?

MA_775_DIABLO [31]

Answer:

Choice A.

$y = - {2}^{x} + 4$

Step-by-step explanation:

Use graphing calculator

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2 years ago

Which expression is equivalent to r to the 9th over r to the 3rd

alisha [4.7K]

Answer:

$r^{6}$

Step-by-step explanation:

To make this a bit easier to see, we'll expand that expression.

R to the 9th is just r times itself 9 times. R to the third is r times itself 3 times.

(excuse my bad formatting)

The expression:

r • r • r • r • r • r • r • r • r • 1

r • r • r • 1

Remember, anything and everything has a coefficient or denominator of 1.

So, we can cancel 3 r's from the numerator and denominator.

r • r • r • r • r • r • 1

Simplify....

$\frac{1r^{6}}{1}$

That just equals $r^{6}$ .

3 0

2 years ago

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

3 years ago

What is y+2=-4(X-6) in Ax+By=C form

PSYCHO15rus [73]

Y+2=-4x+24
2=-4x-y+24
-22=-4x-y

7 0

3 years ago