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2 years ago

Dominique paid $15.50 to join an online music service. If

Mathematics

2 answers:

frozen [14]2 years ago

6 0

Answer:

94.7

Step-by-step explanation:

mote1985 [20]2 years ago

5 0

First you will have to calculate how much she spends on the songs per month so you will have to times how many songs she purchases (10) by how much they cost ($0.99) so

10x$0.99=$9.90

then you will need to times the amount she spends per month ($9.90) by how many months (8)

$9.90x8=$79.20

last you will need to add the amount she paid to join ($15.50) and the total amount she payed on songs in 8 months ($79.20)

$15.50+$79.20=$94.70

so the total amount she spent after 8 months is $94.70

You might be interested in

Ramina found the length of two pieces of ribbon to be 47.6 inches and 39.75 inches. Which is an effective estimation strategy fo

lukranit [14]

Answer:

<h3>Add 47.6 and 39.75, then round the answer</h3>

Step-by-step explanation:

If Ramina found the length of two pieces of ribbon to be 47.6 inches and 39.75 inches, the effective strategy of finding the sum of the two lengths is to:

1) First is to add the two values together

47.6 + 39.75

= (47+0.6)+(39+0.75)

= (47+39)+(0.6+0.75)

= 86 + 1.35

= 87.35

2) Round up the answer to nearest whole number.

87.35 ≈ 87 (note that we couldn't round up to 88 because the values after the decimal point wasn't up to 5)

Option C is correct

7 0

2 years ago

Read 3 more answers

What is a3 in an arithmetic sequence in which a10=41 and a15=61

USPshnik [31]

$\bf \begin{array}{llll}
term&value\\
-----&-----\\
a_{10}&41\\
a_{11}&41+d\\
a_{12}&(41+d)+d\\
&41+2d\\
a_{13}&(41+2d)+d\\
&41+3d\\
a_{14}&(41+3d)+d\\
&41+4d\\
a_{15}&(41+4d)+d\\
&41+5d=61
\end{array}
\\\\\\
41+5d=61\implies 5d=20\implies d=\cfrac{20}{5}\implies \boxed{d=4}\\\\
-------------------------------\\\\$

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=10\\
a_{10}=41
\end{cases}
\\\\\\
41=a_1+(10-1)4\implies 41=a_1+36\implies \boxed{5=a_1}$

thus

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=3\\
a_{1}=5
\end{cases}
\\\\\\
a_3=a_1+(3-1)4\implies a_3=5+(3-1)4$

and surely you know how much that is.

8 0

2 years ago

Read 2 more answers

Create a set of 5 positive numbers (repeats allowed) that have median 7 and mean 10.

damaskus [11]

Answer:

Step-by-step explanation:

Two cases arise to obtain the median.

Case (1): Sample size (n) is odd

First, arrange the data in ascending order. Then, the median is the middle value and lies at the ${\left( {\frac{{n + 1}}{2}} \right)^{{\rm{th}}}}$ position.

Case (2): Sample size (n) is even

First, arrange the data in ascending order. Then, the median is the mean of the two data values that lie on and positions.

Formula for mean is, where is the sum of all observations and n is the total number of observations.

(1)

Use the below process to create a set of 5 positive numbers (repeats allowed) that have median 7 and mean 10.

1.Select the 5 ordered numbers with middle value 7.

2.Adjust the remaining 4 numbers other than 7 to get mean 10 without changing the order of the numbers.

Based on the process to select the 5 ordered numbers with middle value 7 and change the other ordered numbers to get the mean value 10.

(2)

The product of the mean and the total number of observations is obtained. Then, the sum of four observations is subtracted from the product.

Here, the first four numbers are (5, 6, 7, 8).

The last number by using mean 10 is,

$\begin{array}{c}\\\bar x = \frac{{5 + 6 + 7 + 8 + a}}{5}\\\\10 = \frac{{5 + 6 + 7 + 8 + a}}{5}\\\\50 = 26 + a\\\\a = 50 - 26\\=24\end{array}$

The last number is 24.

6 0

2 years ago

Cassie made a picture graph to record the points that third grade students scored on test.Mrs Wilson's group scored 40 points in

lara31 [8.8K]

Answer:

4 students scored 3 points each

7 students scored 4 points each

Step-by-step explanation:

<u>Undetermined equations </u>

There are problems where there are not enough data to provide a unique (determined) solution. We form as many equations as possible and try to find the combination of values of the variables to fulfill all the conditions

For this question, we are assuming the grades go from 1 to 5. We know 11 students in Mrs. Wilson's group scored 40 points in all. We must find a combination of points to produce a sum of 40. We're also assuming only integer points can be scored.

To make things easier, let's find the mean value of the total scores: 40/11=3.6

If we stick to the closest integer near 3.6 we can find a quick solution. Let's think all the students scored 4. It would give a total of 44 points (4 more than required). To adjust the excess, we simply set the score of 4 students to 3 points. Our solution will be

4 students scored 3 points each: 12 points

7 students scored 4 points each: 28 points

Total: 40 points

Note: You can find more solutions apart from this one. For example, set the score of one of the 4-points students to 3. To compensate, set another one from that group to 5 points. Our new solution will be

5 students scored 3 points each: 15 points

5 students scored 4 points each: 20 points

1 student scored 5 points : 5 points

It also has a total of 40 points from 11 students

6 0

2 years ago

Find the quotient 4^10/4^2

Tanya [424]

Answer:65536

4^10/4^2

4*4=16

4*4*4*4*4*4*4*4*4*4=1048576

1048576/16=65536

3 0

2 years ago