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nevsk [136]
2 years ago
10

PLEASE HELP!!!!!! URGENT!!!!!!!!! ALGEBRA! Hannah, Yu, and Becky went to buy groceries together. All three of them bought pasta,

noodles, and split peas. Hannah bought 3 packs of pasta, 10 packs of noodles, and 2 packs of split peas. Yu bought 5 packs of pasta, 12 packs of noodles, and 1 pack of split peas. Becky bought 3 packs of pasta, 15 packs of noodles, and 2 packs of split peas. Hannah, Yu, and Becky spent $8.41, $9.42, and $9.66 respectively on these items. What are the costs of a pack of pasta, a pack of noodles, and a pack of split peas? A) Pasta: $0.69 per pack, Noodles: $0.25 per pack, Split peas: $0.99 per pack B) Pasta: $1.09 per pack, Noodles: $0.20 per pack, Split peas: $1.97 per pack C) Pasta: $0.99 per pack, Noodles: $0.25 per pack, Split peas: $1.47 per pack D) Pasta: $0.69 per pack, Noodles: $0.20 per pack, Split peas: $1.97 per pack
Mathematics
1 answer:
Kaylis [27]2 years ago
7 0

Answer:

option d

sorry it took so long.

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PLEASE ANSWER< WILL GIVE BRAINLY TO BEST ANSWER!?
Mice21 [21]
A as let’s say we have an isosceles triangle two of the sides are the same but one of the sides usually the base us different . Therefore , for this question the sides will both be equal so both 6 and the base 12-different HOPE THIS HELPS (Brainly best answer pleasee )
7 0
2 years ago
1) The population of a town was7652in 2016. The population grows at a rate of 1.6%annually.(a)Use the exponential growth model t
marshall27 [118]

Answer:

1. a) population of the town in 2024 = 8697

2.) the earthquake with a magnitude of 8 is a <u>100 times stronger</u> than an earthquake with a magnitude of 6.

Step-by-step explanation:

1) the population of a town was, N_{0}   = 7652 in 2016.

a) Exponential growth model is given by N(t) = N_{0} e^{kt} where t is the time in years after 2016.

 Year 2016 is when t = 0.

  k = growth rate constant = 1.6% = 0.016

 i) therefore the population of the town in 2024

    = N = N_{0} e^{kt}  = 7652e^{0.016\times8}  = 8697   ( to the nearest whole number)

2.) log (I \times M \times S) = magnitude of earthquake  where I is the Intensity of the earthquake and S is the intensity of a standard earthquake

therefore  I \times M \times S = 10^{magnitude\hspace{0.1cm}of\hspace{0.1cm}earthquake}

therefore \frac{strength\hspace{0.1cm}of\hspace{0.1cm}earthquake\hspace{0.1cm}1 }{strength\hspace{0.1cm}of\hspace{0.1cm}earthquake\hspace{0.1cm}2}  = \frac{10^{8} }{10^{6} }  = 100

 

3 0
3 years ago
Need help (brainliest will be given)
ruslelena [56]

Answer:

-16

Step-by-step explanation:

−2(32)−

8

2

+6

=(−2)(9)−

8

2

+6

=−18−

8

2

+6

=−18−4+6

=−22+6

=−16

5 0
3 years ago
Read 2 more answers
Use the AA similarity Postulate to prove the diagram has two similar
Allisa [31]

Answer:

Triangle  PRT is similar to triangle SRQ

Step-by-step explanation:

AA similarity Postulate :If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

one  angle is already given   : angle    RPT congruent to  angle   RSQ

The other is   :   angle    TRP congruent to  angle  SRQ (both triangles share the  same angle)

Triangle PRT is similar to triangle SRQ  (AA postulate)

3 0
3 years ago
The list below shows the Highlight in inches of the player on a man college basketball team
boyakko [2]

We have the data set:

{79, 83, 82, 78, 79, 77, 80, 73, 82, 72}

This data set has 10 items.

To complete the data summary is helpful to sort the data:

{72, 73, 77, 78, 79, 79, 80, 82, 82, 83}

The minimum value is 72.

<em>The quartile divides the data set in 4 parts. </em>

<em>The first quartile is the value for which 25% of the data is below.</em>

<em>The second quartile is the value for which 50% of the data is below. The second quartile is equivalent to the median.</em>

<em>The third quartile is the value for which 75% of the data is below.</em>

In this data set, as its size is a even number, the quartile fall between two positions, so we will calculate the average of this numbers.

The first quartile will fall in the position 0.25*10=2.5. Then we can calculate the 1st quartile as the average between the 2nd and 3rd number of the data set.

This values are 73 (2nd position) and 77 (3rd position), so the average is:

Q_1=\frac{73+77}{2}=75

We can repeat this with the other quartiles:

\begin{gathered} Q_2=M=\frac{79+79}{2}=79 \\ Q_3=\frac{82+82}{2}=82 \end{gathered}

Then:

The first quartile is 75.

The median (second quartile) is 79.

The third quartile is 82.

The maximum value is 83.

7 0
1 year ago
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