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Eight randomly selected customers at a local grocery store spent the following amounts on groceries in a single visit: $216, $18

chubhunter [2.5K]

Answer:

a) 910

b) 828100

c) 144320

Step-by-step explanation:

a. ∑y

This is the sum of all values of y. So

$\sum y = 216 + 184 + 35 + 92 + 142 + 175 + 9 + 57 = 910$

b. (∑y)^2

This is the square of the sum, so the square of 910, that we found in the previous item.

910² = 828100

c. ∑y^2

This is the sum of the squares of each amount spent. So

$\sum y^2 = 216^2 + 184^2 + 35^2 + 92^2 + 142^2 + 175^2 + 9^2 + 57^2 = 144320$

8 0

3 years ago

A house on the market was valued at $245,000. After several years, the value increased by 9%. By how much did the house's value

aniked [119]

We are given that a house had a value of $245000 and its value was increased by 9%. To determine how much is this value increase in dollars we need to multiply the value of the house by 9/100. We get this:

$245000\times\frac{9}{100}=22050$

Therefore, the increase in dollars is $22050. Now, to determine the current value we need to add the increase to the previous value, we get:

$245000+22050=267050$

Therefore, the current value of the house is $267050.

6 0

11 months ago

Mikaela friends brought a big box of crackers to the movies. 53 crackers are whole wheat and 29 are cheese flavored. The remaini

Elena-2011 [213]

Answer: 123 crackers

Step-by-step explanation:

If 1/3 of the box contains sesame seed crackers, then, 2/3 of the box contains whole wheat and cheese flavored crackers.

Total number of whole wheat and cheese flavored crackers = 53 + 29 = 82

Let x be the total number of crackers in the box, so we will have the equation

2/3 x= 82

x = 82 ÷ 2/3; we will get

X =82 x 3/2

X = 123 crackers

8 0

3 years ago

Use Cramer's Rule to solve the following system: –2x – 6y = –26 5x + 2y = 13

andriy [413]

$\bf \begin{cases}
-2x-6y&=-26\\
\quad 5x+2y&=13
\end{cases}\stackrel{\textit{determinant of the coefficients}}{D=
\begin{bmatrix}
-2&-6\\5&2
\end{bmatrix}}\implies (-4)-(-30)
\\\\\\
D=-4+30\implies \boxed{D=26}\\\\
-------------------------------\\\\$

$\bf x=\cfrac{D_x}{D}\implies x=\cfrac{
\begin{bmatrix}
\boxed{-26}&-6\\\\ \boxed{13}&2
\end{bmatrix}}{D}\implies x=\cfrac{(-52)-(-78)}{26}
\\\\\\
x=\cfrac{-52+78}{26}\implies x=\cfrac{26}{26}\implies \boxed{x=1}\\\\
-------------------------------\\\\
y=\cfrac{D_y}{D}\implies y=\cfrac{
\begin{bmatrix}
-2&\boxed{-26}\\\\ 5&\boxed{13}
\end{bmatrix}}{D}\implies y=\cfrac{(-26)-(-130)}{26}
\\\\\\
y=\cfrac{-26+130}{26}\implies y=\cfrac{104}{26}\implies \boxed{y=4}$

4 0

3 years ago

Consider the following equation. cos x = x3 (a) Prove that the equation has at least one real root. f(x) = cos x − x3 is continu

skelet666 [1.2K]

Answer:

b. 0.86, 0.87

Step-by-step explanation:

a. Find attached solution to a

4 0

3 years ago