There are 7.55 grams of sugar in 5 servings of strawberries. How many grams of sugar are in a single serving of strawberries?

Mathematics

2 answers:

balandron [24]2 years ago

7 0

Answer:

1.51

Step-by-step explanation:

Divide 7.55 by 5 and it equals 1.51

Svetradugi [14.3K]2 years ago

6 0

The answer is 1.51. If you just divide 7.55 by 5 then you get your total of 1.51. To check your answer you can just take 1.51 and multiply it by 5 and you get 7.55.

You might be interested in

Can someone help me solve these three problems please??

Tasya [4]

I'll solve 21, and you then should be able to solve the rest on your own!

Since ADC is 135, that means that that whole angle is 135 degrees. In addition, since angles ADB and BDC add up to ADC, we get ADB+BDC=ADC=135=11x+9+7x=18x+9. Subtracting 9 from both sides, we get 126=18x. Dividing both sides by 18, we get x=7. Plugging that into 11x+9=BDC, we get 11*7+9=77+9=86

4 0

3 years ago

Can y’all help me on question 20?!

denis23 [38]

Answer:

yes I can help you but ???

8 0

3 years ago

Read 2 more answers

2. Solve x - 19 < 81

vovikov84 [41]

The answer would be X =90

5 0

3 years ago

Read 2 more answers

Simplify completely: the quantity 10 times x to the 6th power times y to the third power plus 20 times x to the third power time

eimsori [14]

Correct answer is D.

$\frac{10x^6y^3 + 20x^3y^2}{5x^3y}
\\
\\ \text{common in the numerator is }
\\
\\ 10x^3y^2
\\
\\ \text{so}
\\
\\ \frac{10x^3y^2(x^3y + 2)}{5x^3y}
\\
\\ \text{cancel common factors... }
\\
\\ 2y(x^3y + 2) = 2x^3y^2 + 4y$

6 0

3 years ago

Dr. gavin is conducting a 2 x 4 independent-groups factorial design. how many interactions will dr. gavin need to examine?

Sholpan [36]

For the answer to the question above, I believe the answer is simply 8.

2 groups divided into four participants. So all in all people needed is 8.
I hope this helped you. Have a nice day!


8 0

3 years ago