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1 month ago

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casey bought sandwiches and a bag of chips. Each sandwich cost three times as much as bag of chips. She bought 5 sandwiches for

6 dollars each and spent 42 dollars. How many bags of chips b did she buy?

2 answers:

Inessa [12.5K]1 month ago

4 0

Answer:

She acquired 3 bags of chips.

Step-by-step explanation:

The price of 1 sandwich is $6.

Each sandwich costs three times as much as a single bag of chips.

Thus, the cost of a bag of chips equals $\frac{6}{3}=2$ .

Let x signify the number of chip bags she obtained.

We know that she bought 5 sandwiches priced at 6 dollars each, resulting in a total of 42 dollars.

The expense of 6 sandwiches is represented as $6 \times 6 = 36$ .

The cost of x chip bags amounts to 2x.

According to the information provided

$36+2x=42$ .

$2x=6$ .

$x=3$ .

Therefore, she purchased 3 bags of chips.

AnnZ [12.3K]1 month ago

3 0

Answer:

4

Step-by-step explanation:

Each sandwich costs 6 dollars, while chips sell for 3 dollars. She purchased 5 sandwiches at 6 dollars each, amounting to 30 dollars (5*6=30). This leaves her with 12 dollars to use since her total expenditure was 42 dollars. Given that chips are 3 dollars each, with 12 dollars left, 4 bags could be bought (4 * 3 = 12).

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a. Kalkulahin ang posibilidad na itigil ang paggawa kapag ang sample mean ay 16 inches.

$\mu = 16$ Narito ang

Mas mataas sa 16.25:

Ito ay 1 na ibinawas mula sa p-value ng Z kapag X = 16.25. Kaya

$Z = \frac{X - \mu}{\sigma}$

Sa pamamagitan ng Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{16.25 - 16}{0.1}$

$Z = 2.5$

$Z = 2.5$ may p-value na 0.9938

1 - 0.9938 = 0.0062

Mas mababa sa 15.75:

Ito ay p-value ng Z kapag X = 15.75. Kaya

$Z = \frac{X - \mu}{s}$

$Z = \frac{15.75 - 16}{0.1}$

$Z = -2.5$

$Z = -2.5$ may p-value na 0.0062

Probabilidad ng pagtigil:

2*0.0062 = 0.0124

0.0124 = 1.24% ang posibilidad na itigil ang paggawa kapag ang sample mean ay 16 inches.

b. Kalkulahin ang posibilidad na itigil ang paggawa kung ang mean ay lumipat sa 16.05 inches.

$\mu = 16.05$

Narito ang

Mas mataas sa 16.25:

$Z = \frac{X - \mu}{s}$

$Z = \frac{16.25 - 16.05}{0.1}$

$Z = 2$

may p-value na 0.9772

1 - 0.9772 = 0.0228

Mas mababa sa 15.75:

Ito ang p-value ng Z kapag X = 15.75. Kumbaga

$Z = \frac{X - \mu}{s}$ may p-value na 0.0013

$Z = \frac{15.75 - 16.05}{0.1}$

$Z = -3$

$Z = -3$ Probabilidad ng pagtigil:

0.0228 + 0.0013 = 0.0241

0.0241 = 2.41% ang posibilidad na itigil ang paggawa kung ang mean ay lumipat sa 16.05 inches.

c. Kalkulahin ang posibilidad na hindi makagambala sa produksyon kung ang mean ay lumipat sa 16.25 inches.

Sa pagitan ng 16.25 at 15.75 na may $\mu = 16.25$ . Ito ang p-value ng Z kapag X = 16.25 na ibinawas mula sa p-value ng Z kapag X = 15.75.

X = 16.25

$Z = \frac{X - \mu}{s}$

$Z = \frac{16.25 - 16.25}{0.1}$

$Z = 0$

$Z = 0$ may p-value na 0.5

X = 15.75

$Z = \frac{X - \mu}{s}$

$Z = \frac{15.75 - 16.25}{0.1}$

$Z = -5$

$Z = -5$ may p-value na 0

0.5 - 0 = 0.5

0.5 = 50% ang posibilidad na hindi makagambala sa produksyon kung ang mean ay lumipat sa 16.25 inches.

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