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valkas [14]
2 years ago
7

The least amount of wrapping paper needed to wrap a cube-shaped gift is 150 square inches. How long is one side of the gift?

Mathematics
1 answer:
devlian [24]2 years ago
5 0


Isn't the only thing you have to do is since a cube has 6 sides than you would have to do 150 divided by 6. If I am correct.


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Use the values on the number line to find the sampling error.
pantera1 [17]
Part 1:

Given a number line with the point \bar{x}=3.8 and the point \mu=4.27

The sampling error is given by:

\bar{x}-\mu=3.8-4.27=-0.47



Part 2:

Given a number line with the point \bar{x}=26.43 and the point \mu=24.67

The sampling error is given by:

\bar{x}-\mu=26.43-24.67=1.76
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2 years ago
If 25% of 150 is equal to 700 percent of n, then n is equal to what number?
adelina 88 [10]
25% of 150 = 37.5
700% of x = 37.5
x = 5.35714285
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2 years ago
The functions f(x) and g(x) are described using the following equation and table: f(x) = −2(1.07)x x g(x) −4 −10 −2 −7 0 −4 2 1
Natali [406]

Answer:

The y-intercept of f(x) is equal to 2 times the y-intercept of g(x).

Step-by-step explanation:

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2 years ago
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What is the median of the following 7 scores?
dybincka [34]
First, you want to put all the numbers in order from least to greatest.

21, 33, 33, 42, 67, 79, 89

Now, you can use this song to help remember how to do these problems.

Cross off the sides till you get to the center. 1 is good, 2 is bad. (If you get two numbers in the middle) add then divide by two.

Cross off 21, then 89, then 33, then 79, then 33, then 67. Now, you're left with 42 in the middle. That is your median. Hope this helps! :)
8 0
2 years ago
The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
Doss [256]

Answer:

First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0
1 year ago
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