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3 years ago

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A chef got 5 bags of onions. The red onions came in bags of 3 and the yellow onions came in bags of 6. If the chef got a total o

f 18 onions, how many bags of each type of onion did he get?
?=bags of red onions
?=bags of yellow onions
you have to guess and check word problems

1 answer:

Sauron [17]3 years ago

4 0

Answer: 1 bag of yellow and 4 bags of red onions

Step-by-step explanation:

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Match the mean, median, mode, and range for the data set below. 2, 7, 8, 4, 8, 2, 5, 5, 7, 2

dusya [7]

Answer:

Mean - 5

Median - 5

Mode - 2

Range - 6

Step-by-step explanation:

Mean-

If you add 2 + 7 + 8 + 4 + 8 + 2+ 5+ 5+ 7 +2= 50

50 divided by 10 (the amount of numbers in the equation) = 5 for the mean

Median-

Put the numbers in order which is

2,2,2,4,5,5,7,7,8,8 and the number is the middle is 5

Mode-

The number that repeats the most is 2

Range-

The largest number is 8 and the smallest number is 2 so 8-2=6

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3 years ago

You just found a positive trend line showing a strong relationship between homework scores and test scores for five students. Ex

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The positive trend among the homework scores and the test scores would reveal that if a student does well in homework, he is likely to do well in exams. This can be explained in the sense that the homework will serve as the practice for the student before finally taking the exam.

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2 years ago

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Brand A has a recognition value worth $47 billion more than brand B. If the total value of these two brands is $101 billion,

gladu [14]

A+B=$101 billion
A=B+$47 billion

2B+$47 billion=$101 billion.
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A=B+$47 billion
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2 years ago

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par

lyudmila [28]

Answer:

$a(\frac{1}{5e})=5e$

Step-by-step explanation:

we are given equation for position function as

$s(t)=tln(5t)$

Since, we have to find acceleration

For finding acceleration , we will find second derivative

$s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)$

$=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t$

$=1\cdot \ln \left(5t\right)+\frac{1}{t}t$

$s'(t)=\ln \left(5t\right)+1$

now, we can find derivative again

$s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)$

$=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)$

$=\frac{1}{t}+0$

$a(t)=\frac{1}{t}$

Firstly, we will set velocity =0

and then we can solve for t

$v(t)=s'(t)=\ln \left(5t\right)+1=0$

we get

$t=\frac{1}{5e}$

now, we can plug that into acceleration

and we get

$a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}$

$a(\frac{1}{5e})=5e$

5 0

3 years ago