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2 years ago

A shopper buys cat food in bags of 3 lbs. Her cat eats 34 lb each week. How many weeks does one bag last?

Mathematics

1 answer:

Vika [28.1K]2 years ago

3 0

11 weeks hope this helps

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Did I do this correctly?

Rina8888 [55]

Yes!
- you applied the distributive property
- combined like terms
- and answered the rest right!

8 0

3 years ago

PLEASE HELP ME ITS DUE IN 1 HOUR

Vadim26 [7]

Answer: Answer:Median: 10.5Lower quartile: 4Upper quartile: 16Explanation:Median:First, write the observations in ascending order or descending order.The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observationsn = 12 (even)Median = Mean of (6)th and (7)th observations = 10 + 11 divided by 2 = 21/2 = 10.5Therefore the median is 10.5.After writing in ascending or descending order the first and last terms justify the lower and upper limits respectively.They are:Lower quartile: 4Upper quartile: 16Question 2.the lengths of 10 pencils, in centimeters:18, 15, 4, 9, 14, 17, 16, 6, 8, 10Type below:_________________Answer:Median: 12Lower quartile: 4Upper quartile: 18Explanation:Median:First, write the observations in ascending order or descending order.The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observationsn = 10 (even)Median = Mean of (5)th and (6)th observations = 10 + 14 divided by 2 = 24/2 = 12Therefore the median is 12.After writing in ascending or descending order the first and last terms justify the lower and upper limits respectively.They are:Lower quartile: 4Upper quartile: 18Question 3.Make a box plot to display the data set in Exercise 2.Type below:_________________Answer:The above box plot represents the lower and upper quartiles, the median.Explanation:Box plot is drawn using the number line and the rectangle which is drawn above it.The ends of the rectangles say about the lower and upper limits and the middle line indicates the median.Question 4.The numbers of students on several teams are 9, 4, 5, 10, 11, 9, 8, and 6. Make a box plot for the data.Type below:_________________Answer: Explanation:Box plot is drawn using the number line and the rectangle which is drawn above it.The ends of the rectangles say about the lower and upper limits and the middle line indicates the median.Therefore the lower and upper quartiles are 4 and 11 respectively.Median:First, write the observations in ascending order or descending order.The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observationsn = 8 (even)Median = Mean of (4)th and (5)th observations = 8 + 9 divided by 2 = 17/2 = 8.5Therefore the median is 8.5.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Which technique is most appropriate to use to solve each equation? Drag the name of the technique into the box to match each equ

NemiM [27]

For the first equation, the answer is C) completing the square.
For the second equation, the answer is B) zero product property.

For the first equation, we can easily complete the square by finding half of b and squaring it; then we can take the square root of both sides and solve the equation.

For the second equation, since it is already factored, we use the zero product property to solve it.

7 0

3 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$