All categories

3 years ago

PLS HELP!

Mathematics

1 answer:

wolverine [178]3 years ago

8 0

Given:

Alexis has a rectangular backyard that is 50 yards by 55 yards.

She wants to build a fence that stretches diagonally from one corner to the opposite corner.

To find:

The length of the fencing she needs.

Solution:

We have,

Length = 50 yards

Width = 55 yards

We know that, the diagonal of a rectangle is

$Diagonal=\sqrt{length^2+width^2}$

$Diagonal=\sqrt{(50)^2+(55)^2}$

$Diagonal=\sqrt{2500+3025}$

$Diagonal=\sqrt{5525
}$

On further simplification, we get

$Diagonal=74.3303437$

$Diagonal\approx 74.3$

Therefore, the length of the required fencing is 74.3 yards.

You might be interested in

Help me on this plz I need help !!

12345 [234]

Answer:

2/1

Step-by-step explanation:

x1=1 ,y1=3

x2=2 , y2=5

then

m= 5-3/2-1= 2/1

6 0

3 years ago

A bank balance is more than $35. Julia wants to write an inequality for the bank balance. Which inequality symbol should she use

Aleks04 [339]

X>35 meaning that it is greater than

5 0

2 years ago

Read 2 more answers

5/6-1/8 Solve by finding a common denominator A)30/48 B)34/48 C)38/48 D)42/48

GrogVix [38]

Answer:

B) 34/48

Step-by-step explanation:

The easiest common denominator to find is the product of the denominators you have.

The difference is ...

$\dfrac{5}{6}-\dfrac{1}{8}=\dfrac{5}{6}\cdot\dfrac{8}{8}-\dfrac{1}{8}\cdot\dfrac{6}{6}\\\\=\dfrac{40-6}{48}=\boxed{\dfrac{34}{48}}$

Additional comment

The difference numerator and denominator have a common factor of 2, so can be reduced to 17/24. The denominator 24 could have been used as the common denominator from the start.

5/6 -1/8 = 20/24 -3/24 = 17/24 . . . . reduced result

Using the least common denominator does not always result in a reduced sum or difference. If some other denominator is used (as here), then you can be assured that the result will need to be reduced to get to lowest terms.

8 0

2 years ago

What's the answer for 24

scZoUnD [109]

X= -2 or X= 3 either one works.

3 0

3 years ago

Half of a set of the parts are manufactured by machine A and half by machine B. Ten percent of all the parts are defective. Six

baherus [9]

Answer:

The probability that a part was manufactured on machine A is 0.3

Step-by-step explanation:

Consider the provided information.

It is given that Half of a set of parts are manufactured by machine A and half by machine B.

P(A)=0.5

Let d represents the probability that part is defective.

Ten percent of all the parts are defective.

P(d) = 0.10

Six percent of the parts manufactured on machine A are defective.

P(d|A)=0.06

Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective :

$P(A|d) =\frac{P(A \cap d)}{P(d)}$

$P(A|d) =\frac{P(d|A)\times P(A)}{P(d)}\\P(A|d)= \frac{0.06\times 0.5}{0.10}$

$P(A|d)= 0.3$

Hence, the probability that a part was manufactured on machine A is 0.3

8 0

3 years ago