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2 years ago

15

Ava, Bharat, and Carmen are shopping at the farmers market. The table below shows prices for various items available at the mark

et.
Fresh Produce
tomatoes $3 per pound
cucumbers $1 each
peaches
$4 per pound
watermelon
58 each
Write an expression that describes Ava's cost if she buys x watermelons.
Answer

1 answer:

Vanyuwa [196]2 years ago

7 0

Answer:

Step-by-step explanation:

Watermelon

1=58 cents

2=$1.16

3=$1.74

4=$2.32

5=$2.90

6=$3.48

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Please help me with this:

spayn [35]

Answer:

Step-by-step explanation:

one way to describe the prime version of ABCD is that the sides are all 3 times that of the non- prime version. So they are when scaled by 3 .. the same.

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2 years ago

Tell whether the given value is a solution of the inequality

kipiarov [429]

Answer:

Yes it is, because...

Step-by-step explanation:

This is an inequality, so you can treat it as an algebraic expression.

The first step is to multiple both sides by 2, to get rid of the 2 on the left side.
Your equation will now look like this:
y >= 2y - 22

The next step is to get all the y variables to one side, now that its a lot more simplified. Subtract 2y from both sides to get:
-y >= -22

Finally, cancel out the negative on both sides of the equation to get the y as a positive y, all by itself. This will get you:
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2 years ago

Read 2 more answers

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

$P(B) = \frac{77}{100}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

So, we have:

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

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3 years ago

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user100 [1]

P^2 could also have -4 as an answer. Any time a number is squared, it is positive

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3 years ago

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spayn [35]

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3 years ago