Solve each of the quadratic equations 3x=0.5x2

Amanda is a cabinetmaker. She makes cabinets using one of three types of wood: birch, maple, or cherry. She can finish each type

PIT_PIT [9117]

Assuming that every material in their respective categories has an equal likelihood of selection.

Categories:
(a) Wood types for cabinets: birch, maple, cherry.
Each wood type has a selection probability of 1/3.
(b) Finish types: transparent, semi-transparent.
Each finish type has a 1/2 chance of being selected.
(c) Knob types: bronze, steel, wood.
Each knob material carries a probability of 1/3.

The selection of materials for the cabinet, finish, and knobs is independent of one another.

Consequently,
Probability(birch wood AND bronze knob) = (1/3)*(1/3) = 1/9
Probability(wood knob) = 1/3
Probability(transparent stain) = 1/2
Probability(cherry wood AND semi-transparent stain) = (1/3)*(1/2) = 1/6

6 0

16 days ago

Turkey sandwiches cost $2.50 and veggie wraps cost $3.50 at a snack stand. Ben has sold no more than $30 worth of turkey sandwic

tester [8842]

Answer:

Ben could have sold a maximum of 6 turkey sandwiches.

Step-by-step explanation:

Turkey sandwiches are priced at $2.50, while veggie wraps cost $3.50 at the snack stand.

Our goal is to determine the largest number of turkey sandwiches Ben might have sold.

$2. 5 0 x + 3. 5 0 y \leq 3 0$

4 veggie wraps were sold (y).

Thus, the inequality is: 2.50x + 3.50(4) < 30

2.50x + 14 < 30

- 14 - 14

2.50x < 16

$x$

Ultimately, Ben could sell a maximum of 6 turkey sandwiches.

4 0

1 month ago

Read 2 more answers

Let u = <5, 6>, v = <-2, -6>. Find -2u + 5v.

babunello [8412]

-2(5,6)-40 That should cover it.

7 0

28 days ago

Given matrix A below, and that A = B, find the value of the elements in B. A = 9 −2 3 2 17 0 3 22 8 b11 = b12 = b13 = b21 = b22

Svet_ta [9500]

Answer:

$b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8$

Step-by-step explanation:

Review the provided matrix

$A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]$

Let matrix B be defined as

$B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$

It is stated that

$A=B$

$\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$