The total weight of 30 bags of flour and 4 bags of sugar is 42.6 kg. If each bag of sugar weighs 0.75 kg, what is the weight of
each bag of flour?
2 answers:
The weight of each bag of flour is 1.32 kgs
<h3>Further explanation </h3>Flour is the powder made by grinding raw grains, roots, beans, nuts, or seeds. Flour is used to make many different foods. Cereal flour is the main ingredient of bread, that is a staple food for most cultures. A flour sack or flour bag is a cloth sack that usually made of cheap cotton to store flour. Flour sacks are often printed with simple designs and trademarks to indicate the millers and companies making or selling the flour. While A bag of sugar is a bag that currently contains sugar, whether it is a specially-designed bag for holding sugar or just an ordinary bag.
If each bag of sugar weighs 0.75 kg, we have 4 bags of sugar, therefore
4 bags of sugar is 42.6 kg and the total of sugar weighs is 3kg, so
There are 30 bags of flour, so
Therefore the weight of each bag of flour is 1.32 kgs
<h3>Learn more</h3>- Learn more about flour brainly.com/question/1305734
- Learn more about bag brainly.com/question/244631
- Learn more about kg brainly.com/question/9363051
.
<h3>Answer details</h3>
Grade: 5
Subject: Math
Chapter: The total weight
Keywords: flour, bag, sugar, weight, kg
You might be interested in
as you notice above, is the first-row components from A, multiplying all the columns subsequently on B, and you add the products of that row, that gives you one component on the AB matrix
in the one above, we end up with a 2x3 AB matrix
Step-by-step explanation:
3
Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].
Therefore, by mid-point formula:
4 (a)
Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:
is the equation of line AB.
Now we have to check whether C(4, 7) lie on line AB or not.
Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.
Hence, point C (4, 7) lie on the straight line AB.
4(b)
Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:
And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221