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3 years ago

What is -3 and 2/7 minus 4 and 1/3 and can it be simplified

Mathematics

2 answers:

bixtya [17]3 years ago

7 0

Answer: -7 and 13/21

Step-by-step explanation:

(-3 2/7) - (4 1/3)

Least common multiple for the denominator= 21

1. -23/7 - 13/3

2. -69/21 - 91/21 = -160/21

3. -7 13/21

Elanso [62]3 years ago

5 0

Answer:

Yes it can be simplified.

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What is the area of a garden that measures 4mx4mx6mx4mx10mx8m

topjm [15]

The area of the garden is 120 square metre

<h2>Explanation:</h2>

The picture of the garden is shown below. In this problem, we have:

One square
Two rectangles

Recall that the area of a square is given by:

$A_{s}=L^2 \\ \\ A_{s}:Area \ of \ a \ square \\ L:Side \ of \ the \ square$

On the other hand, the area of a rectangle is given by:

$A_{r}=L_{1}\times L_{2} \\ \\ \\ A_{r}:Area \ of \ a \ rectangle \\ \\ L_{1}:Side \ 1 \ of \ the \ rectangle \\ \\ L_{2}:Side \ 2 \ of \ the \ rectangle$

Therefore:

FOR THE SQUARE:

$L=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{s}=4^2 \\ \\ \boxed{A_{s}=16m^2}$

FOR THE RECTANGLE 1:

$L_{1}=6m \\ L_{2}=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{1}}=6(4) \\ \\ \boxed{A_{r_{1}}=24m^2}$

FOR THE RECTANGLE 2:

$L_{1}=10m \\ L_{2}=8m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{2}}=10(8) \\ \\ \boxed{A_{r_{2}}=80m^2}$

Finally, the area of a garden is the sum of the three areas:

$A_{total}=A_{s}+A_{r_{1}}+A_{r_{2}} \\ \\ A_{total}=16m^2+24m^2+80m^2 \\ \\ A_{total}=16+24+80 \\ \\ \boxed{A_{total}=120m^2}$

<h2>Learn more:</h2>

Area of a pizza: brainly.com/question/12878495

#LearnWithBrainly

4 0

3 years ago

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.