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Alex
3 years ago
11

A certain brand of coffee comes in two sizes. A 10.15-ounce package costs 2.98. A 27.8 -ounce package costs 8.99. Find the unit

price for each size. Then state which size is the better buy based on the unit price. Round your answers to the nearest cent.
Mathematics
1 answer:
never [62]3 years ago
4 0

Answer:

Size 2 with 3.1 units price

Step-by-step explanation:

Size 1:

A 10.15-ounce package costs 2.98.

Units price of size 1 = 10.15 / 2.98

= 3.4060

Approximately 3.4

Size 2:

A 27.8 -ounce package costs 8.99.

Units price of size 2 = 27.8 / 8.99

= 3.0923

Approximately 3.1

The best size which is better buy based on unit price is size 2 with 3.1 units price

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Yes you are correct

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2 years ago
Solve the quadratic equation<br> x2-10x+49=4x+1
CaHeK987 [17]

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x=8 & x=6

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
What is 6(7-2p) =12p+42
mafiozo [28]

answer:

p = 0

step-by-step explanation:

6 (7-2p) = 12p + 42

42 - 12p = 12p + 42

     +12p   +12p

_______________

42 = 24p + 42

-42            -42

_____________

0 = 24p

_    ___

24    24

0 = p


<em>hope this helps! ❤ from peachimin (aka kayla)</em>


4 0
3 years ago
Line b passes through points (9, 1) and (1, 6). Line c passes through points (10, 7) and (2, 12). Are line b and line c parallel
andre [41]
<h2>Answer:  Lines b & c are parallel.⇄</h2>

Step-by-step explanation:

  • When lines are perpendicular, the product of their slope is -1 (the slopes are negative reciprocal).
  • When lines are parallel, they have the same slope.

Slope = ( y₂ - y₁) / (x₂ - x₁)

Slope of Line b = ( 6 - 1 ) ÷ ( 1 - 9 )

                         =  5  ÷  (- 8)

                         =   - ⁵/₈

Slope of Line c =  ( 12 - 7 )  ÷  ( 2 - 10 )

                         =  - ⁵/₈

⇒ Line b ║ Line c

8 0
2 years ago
A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall
Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : \mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7

Standard deviation : \sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}

=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}

=\dfrac{30.265}{10}=3.0265

The confidence interval for the population mean (for sample size <30) is given by :-

\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given significance level : \alpha=1-0.95=0.05

Critical value : t_{n-1,\alpha/2}=t_{9,0.025}=2.262

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0
3 years ago
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