1. Fill in the blank with the best regression model for each scm plot: Choose from:

A pack of gum costs 50 cents. That is 4 cents less than four times what the pack cost 15 years ago.

alisha [4.7K]

15 years ago pack
= 4 x 0.50 + 0.04
= 2.04

15 years ago, pack costed $2.04

Hope this helps. - M

7 0

3 years ago

X is a normally Distributed random variable with a standard deviation of 4.00. Find the mean of X when 64.8% of the area lies to

Cerrena [4.2K]

Answer:

7

Step-by-step explanation:

σ = 4 ; μ =?

8.52 to the left of X

.

P(X < 8.52) = 64.8%

P(X < 8.52) = 0.648

Using the Z relation :

(x - μ) / σ

P(Z < (8.52 - μ) / 4)) = 0.648

The Z value of 0.648 of the lower tail is equal to 0.38 (Z probability calculator)

Z = 8.52 - μ / 4

0.38 = 8.52 - μ / 4

0.38 * 4 = 8.52 - μ

1.52 = 8.52 - μ

μ = 8.52 - 1.52

μ = 7

5 0

3 years ago

Find all the angles listed. also list all of the lines that are secants, chords and tangents

Tanzania [10]

Answer:

Secants – 1. <CBG 2. <AGF 3. <ABD

Tangent – 1. <CDE

Chords – 1. <BD 2. <DF 3. <BG 4. <BF 5. <GD 6. <FG

Angles – 1. 45 2. 75 3. 35 4. 70 5. 75 6. 55 7. 50 8. 25 9. 35 10. 70 11. 50 12. 25 13. 70 14. 50. 15. 60 16. 85 17. 95 18. 85 19. 95

Step-by-step explanation:

4 0

3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

Divide the decimal 134.55 by the decimal 0.5

lina2011 [118]

Answer:

269.1

Step-by-step explanation:

8 0

3 years ago