30 yd wide and 50 yd long find the length of the diagonal

1 answer:

8 0

It's the same as finding the hypothenuse of a right triangle. Let c be the hypothenuse.

30 squared+ 50 squared= c squared.

900+ 2500= c squared
3400= c squared

Then find the square root of 3400 and that will equal c.

So, depending in what form the teacher was the answer. The answer will be square root of 3400 ft squared or 58.3 ft squared.

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Tasya [4]

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4 0

2 years ago

Read 2 more answers

How many 1/3 inch cubes does it take to fill a box with width 2 213 inches, length 3 1/3

MAVERICK [17]

Answer:

560 cubes

Step-by-step explanation:

Is 2 213 supposed to read 2 1/3? First calculate the volume of the box

V_box = 8/3 * 10/3 * 7/3 = 560/27 cubic inches

Now calculate the volume of your 1/3 inch cube:

V_cube = 1/3 * 1/3 * 1/3 = 1/27 cubic inches

Now divide the V_box by V_cube:

Number of boxes = (560/27) / (1/27) = (560/27) * (27/1) = 560 cubes

8 0

3 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$