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3 years ago

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Lillian bought snacks for her team's practice. She bought a bag of apples for $2.36 and a 4-pack of juice bottles. The total cos

t before tax was $6.48. How much was each bottle of juice?

1 answer:

stiks02 [169]3 years ago

5 0

Answer:

$1.030 ($1.03 rounded.)

Step-by-step explanation:

First, we need to start of making an equation.

$2.36 + x = 6.48.

We will use x as the variable for the amount of each bottle of juice, because it is unknown.

Then, we must put the variable on one side of the equation. To do this, we must subtract $2.36 from $6.48; and $2.36 - $2.36.

$2.36 - $2.36 crosses out; 0.

$6.48 - $2.36 = $4.12.

We get the result that x = $4.12. However, only 1 pack is $4.12, and we must find the cost of 4 juice bottles. Therefor, we divide 4.12 by 4.

4.12 / 4 = 1.030.
Now, to round the answer, this would be $1.03 per bottle.

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In a grouped frequency distribution one interval is listed as 50-54. Assuming that the scores are measuring a continuous variabl

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Answer:

a. 49.5 and 54.5

Step-by-step explanation:

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Class limit is the minimum and maximum value the class interval may contain. The minimum value is called the lower class limit and the maximum value is called the upper class limit. For class interval 50-54, the lower class limit is 50 and the upper class limit is 54.

Class boundaries are the numbers used to separate classes. It is the real limits of a class. For non-overlapping classes, the lower class boundary of each class is calculated by subtracting 0.5 from the lower class limit. The upper class boundary of each class is calculated by adding 0.5 to the upper class limit. Example: For class interval 50-54, the lower class boundary is 49.5 and the upper class class boundary is 54.5

Considering the question given, to get the real limits of the interval 50-54, 0.5 is subtracted from the lower class limit to give 49.5. Also, 0.5 is added to the upper class limit to give 54.5.

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4 years ago

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Answer:

$S_3=39$

Step-by-step explanation:

The nth term of the sequence is

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To get the second term, substitute n=2,

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To get the third term, substitute n=3,

$a_3=3(3)^{3-1}=27$

The sum of the first three terms is

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We could also use the formula

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3 years ago

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Slav-nsk [51]

Answer:

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Step-by-step explanation:

Given

$\dfrac{4^{-3}}{4^{-1}}$

Required

Choose equivalent expressions

Choosing the first answer:

$\dfrac{4^{-3}}{4^{-1}}$

Split expressions

$4^{-3} * \frac{1}{4^{-1}}$

Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

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Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

$\frac{1}{4^3} * \frac{1}{1/4}$

$\frac{1}{4^3} * \frac{4^1}{1}$

$\frac{4^1}{4^3}$

Hence:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}$

Choosing the second:

$\dfrac{4^{-3}}{4^{-1}}$

Apply law of indices: $(\frac{a^m}{a^n} = a^{m-n})$

So,

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3-(-1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3+1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-2}$

Apply law of indices: $(a^{-b} = \frac{1}{a^b})$

So:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}$

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4 years ago