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4 years ago

Can anyone help me do #15

Mathematics

1 answer:

makkiz [27]4 years ago

4 0

$She \: tilled \: over \: 2 \: days: 7\frac{2}{12} + 4 \frac{11}{12} = 12 \frac{1}{2} < 15 \frac{7}{12} \\ \Rightarrow She \: didn't \: till \: enough \: soils, \: she \: need \: 15 \frac{7}{12} - 12 \frac{1}{12} = 3 \frac{1}{2} \: yards \: of \: soils \: more$

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andrew-mc [135]

Answer:

Step-by-step explanation:

yes

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3 years ago

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3 years ago

At East Middle School, there are 58 left-handed students and 609 right-handed students. The numbers of left- and right-handed st

EastWind [94]

<u>Answer:</u>

C. Left-handed: 48, Right-handed: 504

D. Left-handed: 30, Right-handed: 315

<u>Step-by-step explanation:</u>

We are given that there are 58 left-handed students and 609 right-handed students at East Middle School and these numbers of students are proportional to the number of left and right handed students at East Middle School.

Given the above information, we are are to determine which two options could be the the numbers of left-handed and right-handed students at West Junior High.

Ratio of right handed to left handed students at East Middle School = $\frac{609}{58} = 10.5$

Checking for ratios of the given options:

A. $\frac{483}{42} =11.5$

B. $\frac{378}{28} =13.5$

C. $\frac{504}{48} =10.5$

D. $\frac{560}{56} =10$

E. $\frac{315}{30} =10.5$

Therefore, the possible numbers of left-handed and right-handed students at West Junior High could be C. Left-handed: 48, Right-handed: 504 and D. Left-handed: 30, Right-handed: 315.

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3 years ago

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On Monday, Deborah ran 3 2− 5miles and on Tuesday she ran 4 1− 5miles. How many miles did she run on these two days together?

notsponge [240]

Mixed fraction:

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3 years ago

How do I solve this out??

GuDViN [60]

<h3>Answer:</h3>

(x, y) = (7, -5)

<h3>Step-by-step explanation:</h3>

It generally works well to follow directions.

The matrix of coefficients is ...

$\left[\begin{array}{cc}2&4\\-5&3\end{array}\right]$

Its inverse is the transpose of the cofactor matrix, divided by the determinant. That is ...

$\dfrac{1}{26}\left[\begin{array}{ccc}3&-4\\5&2\end{array}\right]$

So the solution is the product of this and the vector of constants [-6, -50]. That product is ...

... x = (3·(-6) +(-4)(-50))/26 = 7

... y = (5·(-6) +2·(-50))/26 = -5

The solution using inverse matrices is ...

... (x, y) = (7, -5)

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3 years ago

Can anyone help me do #15 ​

Can anyone help me do #15