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omeli [17]
3 years ago
13

A food manufacturer makes 32-oz boxes of pasta. not every box weighs exactly 32 oz. The allowable difference from the ideal weig

ht is at most 0.05 oz. Write and solve an absolute value inequality to find the range of allowable weights.
I will give brainliest
Mathematics
2 answers:
Vlada [557]3 years ago
6 0

Answer:

You are An army????

Step-by-step explanation:

Because I am an Army

BARSIC [14]3 years ago
3 0

Answer:

The range of allowable weights is between 31.95 oz  and 32.05 oz  both included

Step-by-step explanation:

w= allowable weights

  I w-32 I ≤ 0.05

-0.05 ≤  w-32 ≤0.05

31.95 ≤ w ≤ 32.5

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Estimate a 20% tip on a dinner bill of $161.86 by first rounding the bill amount to the nearest ten dollars.
MAXImum [283]

Answer:

See below

Step-by-step explanation:

<u>If you just want the tip</u>

<u />0.20*161.86\approx\frac{1}{5}*160=32

<u>If you want the total bill</u>

<u />1.20*161.86\approx160+32=192

7 0
2 years ago
Correctly order the proceed for solving for x using the algebraic properties of equality.
Snezhnost [94]

Answer:

3

1

2

Step-by-step explanation:

Step 1: Subtract 9

50.24 = 3.14x²

Step 2: Divide by 3.14

x² = 16

Step 3: Square root both sides

x = ±4

6 0
3 years ago
If you like graphing this is for you :D​
Pavel [41]

Answer:

You are trying to find the x-value when the y-value is 2. On the graph, the x-value is -3 when the y-value is 2. So, the x-value when f(x)=2 is -3.

:)

3 0
2 years ago
ms.white is making book marks for her first grade class each bookmark has a gold tassel.8 book marks are in one package and she
ioda
Maybe its about 3 so yea
7 0
3 years ago
Read 2 more answers
Two ballpoint pens are selected at random from a box that contains 3 blue pens, 2 red pens, and 3 green pens. If X is the number
Flauer [41]

Answer:

a) f(x,y) =\frac{\binom{3}{x}\binom{2}{y}\binom{3}{2-x-y}}{\binom{8}{2}} ;   x = 0, 1 , 2;  y = 0, 1 , 2; 0 ≤ x+y ≥ 2

b) = \frac{9}{14}

Step-by-step explanation:

joint probability is a function that characterizes the distribution of a random variable. If X and Y be two random variables then the joint probability will be P(X = x, Y=y)

Given Data,

X = The number of blue Pens

Y = The number of red Pens

a)

possible outcomes(X, Y) are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), (2,0)

Please refer fig. also

total number ways of selecting any 2 pens = \binom{8}{2}= \frac{8!}{2! 6!} =28

f(x,y) = \frac{\binom{3}{x}\binom{2}{y}\binom{3}{2-x-y}}{\binom{8}{2}} ;   x = 0, 1 , 2;  y = 0, 1 , 2; 0 ≤ x+y ≥ 2

b)

P(X,Y)∈A = P(X + Y ≤ 1)

= P(0,0) + P(1,0) + P(0,1)

= \frac{3}{28} + \frac{3}{14} + \frac{9}{28}

= \frac{9}{14}

4 0
3 years ago
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