All categories

4 years ago

china receives a shipment of 125 dishes in 5 boxes, at this rate how many dishes will it receive in 15 boxes

2 answers:

Elena-2011 [213]4 years ago

6 0

Simple....

in 5 boxes you have 125 dishes...how many per box?

125/5=25

So, how many dishes in 15 boxes?

25*15=375

This means that in 15 boxes there are 375 dishes.

Thus, your answer.

sasho [114]4 years ago

3 0

125 / 5 boxes = 25 dishes a box, so 15 x 25 = 375 dishes in 15 boxes

You might be interested in

Use the graph to find the x-intercept for the equation ý +2 Graph: Y + 2 = 2(x+3)

alex41 [277]

Answer:

(-2, 0)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Write the quadratic function f(x) = 3(x + 1)(x – 7) in standard form.

Sveta_85 [38]

Answer:

Step-by-step explanation:

Given

f(x) = 3(x + 1)(x - 7) ← expand the factors using FOIL

= 3(x² - 6x - 7) ← distribute the parenthesis

= 3x² - 18x - 21 → B

4 0

3 years ago

HELP PLEASE HELP WILL GIVE POINTS PLEASE!!!!!!!!!

sdas [7]

Answer: positive association and linear association

Step-by-step explanation:

4 0

3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago

Please help this is so easy and show work!!!!

omeli [17]

the answer is yellow

6 0

3 years ago