Which expression is equivalent to 4{12 + 5} + 3?

Rosters Chicken advertises "lite" chicken with 30% fewer calories than standard chicken. When the process for "lite" chicken

PilotLPTM [1.2K]

Answer:

UCL (x bar) = 424

LCL (x bar) = 376

Step-by-step explanation:

Given:

Average calories contained, μ = 400

Standard deviation, σ = 30 calories

Sample size, n = 25

a) UCL (x bar) = $\mu+\frac{4\sigma}{\sqrt{n}}$

On substituting the respective values, we get

UCL (x bar) = $400+\frac{4\times30}{\sqrt{25}}$

or

UCL (x bar) = 424

Similarly,

LCL (x bar) = $\mu-\frac{4\sigma}{\sqrt{n}}$

On substituting the respective values, we get

LCL (x bar) = $400-\frac{4\times30}{\sqrt{25}}$

or

LCL (x bar) = 376

4 0

3 years ago

Solve using elimination. <br> 10x + y = -19 <br> 10x + 3y = -17

RideAnS [48]

Subtract 10x

10x + y = -19

10x + 3y = -17

You would get:

-2y = -2

Two negatives make a positive number! Divide

y = 1

Plug in

10x + (1) = -19

10x = -20

Divide

x = -2

Your solutions are

x = -2

y = 1

4 0

3 years ago

Read 2 more answers

NEED HELP ASAP PLEASE WITH #7!!

AnnZ [28]

Answer:

The answer to your question is: The second option

Step-by-step explanation:

Points A (3 , 7) B (5, 11)

slope = m = (y2 - y1) / (x2 - x1)

m = (11 - 7) / (5 - 3)

m = 4 / 2 = 2

equation if the line

(y - y1) = m(x - x1)

(y - 7) = 2(x - 3) Answer

6 0

3 years ago

A village fete has a children’s running race each year, run in heats of up to ten children. For each heat the first three contes

bekas [8.4K]

Answer: 1) 1/3,654 2) 3/406 3) 72,684,900,288,000 4) 120

<u>Step-by-step explanation:</u>

1) First and Second and Third

$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}\times \dfrac{1\ remaining\ prize}{27\ remaining\ people}=\dfrac{6}{21,924}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\large\boxed{\dfrac{1}{3,654}}$

2) First and Second

$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}=\dfrac{6}{812}=\large\boxed{\dfrac{13}{406}}$

$3)\quad \dfrac{29!}{(29-10)!}=\large\boxed{72,684,900,288,000}$

$4)\quad _{10}C_3=\dfrac{10!}{3!(10-3)!}=\large\boxed{120}$

8 0

3 years ago

A tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24. The players were still hungry so the coa

g100num [7]

The cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

<h3><u>Solution:</u></h3>

Let "f" be the cost of 1 fries

Let "h" be the cost of 1 hamburger

<em><u>Given that, tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24</u></em>

8 x cost of 1 hamburger + 5 x cost of 1 fries = 24

$8 \times h + 5 \times f = 24$

8h + 5f = 24 -------- eqn 1

<em><u>The players were still hungry so the coach bought six more hamburgers and two more fries for $16.60</u></em>

6 x cost of 1 hamburger + 2 x cost of 1 fries = 16.60

$6 \times h + 2 \times f = 16.60$

6h + 2f = 16.60 ------ eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

16h + 10f = 48 ------ eqn 3

Multiply eqn 2 by 5

30h + 10f = 83 -------- eqn 4

<em><u>Subtract eqn 3 from eqn 4</u></em>

30h + 10f = 83

16h + 10f = 48

( - ) ----------------------

14h = 35

Substitute h = 2.5 in eqn 1

8(2.5) + 5f = 24

20 + 5f = 24

5f = 4

Thus cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

5 0

3 years ago