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3 years ago

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Ted researched the price of airline tickets and discovered a correlation between the price of a ticket and the number of miles t

raveled. After recording his data on a scatter plot, he determined the equation for the line of best fit is y = 300 + 0.45x. What does 0.45 represent in the equation? A) the lowest ticket price B) the number of miles traveled C) the expected change in the price of multiple tickets Eliminate D) the expected change in price of the ticket for each mile traveled

2 answers:

Nookie1986 [14]3 years ago

7 0

D) the expected change in price <span>of the ticket for each mile traveled
Because it is a coloration between the two </span>

andriy [413]3 years ago

4 0

D) the expected change in price <span>of the ticket for each mile traveled
Because it is a coloration between the two </span><span />

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drek231 [11]

QUESTION 1

We want to expand $(x-2)^6$ .

We apply the binomial theorem which is given by the formula

$(a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+^nC_2a^{n-2}b^2+...+^nC_na^{n-n}b^n$ .

By comparison,

$a=x,b=-2,n=6$ .

We substitute all these values to obtain,

$(x-2)^6=^6C_0x^6(-2)^0+^6C_1x^{6-1}(-2)^1+^6C_2x^{6-2}(-2)^2+^6C_3x^{6-3}(-2)^3+^6C_4x^{6-4}(-2)^4+^6C_5x^{6-5}(-2)^5+^6C_6x^{6-6}(-2)^6$ .

We now simplify to obtain,

$(x-2)^6=^nC_0x^6(-2)^0+^6C_1x^{5}(-2)^1+^6C_2x^{4}(-2)^2+^6C_3x^{3}(-2)^3+^6C_4x^{2}(-2)^4+^6C_5x^{1}(-2)^5+^6C_6x^{0}(-2)^6$ .

This gives,

$(x-2)^6=x^6-12x^{5}+60x^{4}-160x^{3}(-2)^3+240x^{2}-1925x+64$ .

Ans:C

QUESTION 2

We want to expand

$(x+2y)^4$ .

We apply the binomial theorem to obtain,

$(x+2y)^4=^4C_0x^4(2y)^0+^4C_1x^{4-1}(2y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(2y)^3+^4C_4x^{4-4}(2y)^4$ .

We simplify to get,

$(x+2y)^4=x^4(2y)^0+4x^{3}(2y)^1+6x^{2}(2y)^2+4x^{1}(2y)^3+x^{0}(2y)^4$ .

We simplify further to obtain,

$(x+2y)^4=x^4+8x^{3}y+24x^{2}y^2+32x^{1}y^3+16y^4$

Ans:B

QUESTION 3

We want to find the number of terms in the binomial expansion,

$(a+b)^{20}$ .

In the above expression, $n=20$ .

The number of terms in a binomial expression is $(n+1)=20+1=21$ .

Therefore there are 21 terms in the binomial expansion.

Ans:C

QUESTION 4

We want to expand

$(x-y)^4$ .

We apply the binomial theorem to obtain,

$(x-y)^4=^4C_0x^4(-y)^0+^4C_1x^{4-1}(-y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(-y)^3+^4C_4x^{4-4}(-y)^4$ .

We simplify to get,

$(x+2y)^4=^x^4(-y)^0+4x^{3}(-y)^1+6x^{2}(-y)^2+4x^{1}(-y)^3+x^{0}(-y)^4$ .

We simplify further to obtain,

$(x+2y)^4=x^4-4x^{3}y+6x^{2}y^2-4x^{1}y^3+y^4$

Ans: C

QUESTION 5

We want to expand $(5a+b)^5$

We apply the binomial theorem to obtain,

$(5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{5-1}(b)^1+^5C_2(5a)^{5-2}(b)^2+^5C_3(5a)^{5-3}(b)^3+^5C_4(5a)^{5-4}(b)^4+^5C_5(5a)^{5-5}(b)^5$ .

We simplify to obtain,

$(5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{4}(b)^1+^5C_2(5a)^{3}(b)^2+^5C_3(5a)^{2}(b)^3+^5C_4(5a)^{1}(b)^4+^5C_5(5a)^{0}(b)^5$ .

This finally gives us,

$(5a+b)^5=3125a^5+3125a^{4}b+1250a^{3}b^2+^250a^{2}(b)^3+25a(b)^4+b^5$ .

Ans:B

QUESTION 6

We want to expand $(x+2y)^5$ .

We apply the binomial theorem to obtain,

$(x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{5-1}(2y)^1+^5C_2(x)^{5-2}(2y)^2+^5C_3(x)^{5-3}(2y)^3+^5C_4(x)^{5-4}(2y)^4+^5C_5(x)^{5-5}(2y)^5$ .

We simplify to get,

$(x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{4}(2y)^1+^5C_2(x)^{3}(2y)^2+^5C_3(x)^{2}(2y)^3+^5C_4(x)^{1}(2y)^4+^5C_5(x)^{0}(2y)^5$ .

This will give us,

$(x+2y)^5=x^5+^10(x)^{4}y+40(x)^{3}y^2+80(x)^{2}y^3+80(x)y^4+32y^5$ .

Ans:A

QUESTION 7

We want to find the 6th term of $(a-y)^7$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=7,b=-y$

We substitute to obtain,

$T_{5+1}=^7C_5a^{7-5}(-y)^5$ .

$T_{6}=-21a^{2}y^5$ .

Ans:D

QUESTION 8.

We want to find the 6th term of $(2x-3y)^{11}$

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=11,a=2x,b=-3y$

We substitute to obtain,

$T_{5+1}=^{11}C_5(2x)^{11-5}(-3y)^5$ .

$T_{6}=-7,185,024x^{6}y^5$ .

Ans:D

QUESTION 9

We want to find the 6th term of $(x+y)^8$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=8,a=x,b=y$

We substitute to obtain,

$T_{5+1}=^8C_5(x)^{8-5}(y)^5$ .

$T_{6}=56a^{3}y^5$ .

Ans: A

We want to find the 7th term of $(x+4)^8$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=6,n=8,a=x,b=4$

We substitute to obtain,

$T_{6+1}=^8C_5(x)^{8-6}(4)^6$ .

$T_{7}=114688x^{2}$ .

Ans:A

4 0

3 years ago

A clock has a face with a radius of 11.5 centimeters. What is the approximate circumference of the clock?

antiseptic1488 [7]

C = 2 pi r
C = 2 x 3.14 x 11.5
C = 72.22
answer is <span>
A. 72.22 cm</span>

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3/4 x 6 I need Help With This

scZoUnD [109]

Answer:

4.5

Step-by-step explanation:

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Can two numbers have the same absolute value

xxMikexx [17]

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no they can not

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Help Me Pls Thanks <3

dybincka [34]

If there are 150 total instruments in said Orchestra, we'll need to divide that number in half.
75, simple enough. Now we know that half of 150 is 75.
Start by writing your problem like this: 150= 75w+37.5b
That equation is all I can think of, because 37.5 is half of 75, which means that's roughly the number you're looking for. Sorry if this doesn't help much<span />

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