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3 years ago

(12) Check all that apply. Inequalities

Mathematics

1 answer:

MrRa [10]3 years ago

6 0

The circle is open i think is the right answer i hope i got it right if not im very sorry :(

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A 15-meter by 23-meter garden is divided into two sections. Two sidewalks run along the diagonal of the square section and along

Thepotemich [5.8K]

Solution:

We have re-drawn the given diagram, as we can see the given rectangle is actually divided into two parts.

one is Large square and the other one is a smaller rectangle.

Now we will apply the Pythagoras theorem in the larger triangle which is inside the square.

So we can write

$Diagonal_1=\sqrt{15^2+15^2}= 21.213$

Now we will apply the Pythagoras theorem in the smaller triangle which is inside the smaller rectangle.

So we can write

$Diagonal_2=\sqrt{15^2+8^2}= 17$

The approximate sum of the lengths of the two sidewalks, shown as dotted lines $=21.2+17=38.2$ m

Hence the correct option is D

6 0

3 years ago

Read 2 more answers

A cheerleading squad consists of ten cheerleaders of ten different heights. How many ways are there for the cheerleaders to line

Sindrei [870]

Answer:

11340

Step-by-step explanation:

We have 10 cheerleaders of ten different heights. There are two rows each having 5 positions so that five cheerleaders occupy the first row that have height less than the five cheerleaders who occupied the second row.

Therefore, Out of ten cheerleaders, two cheerleaders will be selected out of which the tallest one will occupy the position in the upper row and the second one will occupy in the lower row, therefore these two positions will be occupied in $10C_{2}$ ways. Now, we are left with 8 cheerleaders and following the same path, the next two positions will also be occupied in $8C_{2}$ ways, similarly, following the same path, the number of ways the cheerleaders will occupy the position will be: $10C_{2}$ × $8C_{2}$ × $6C_{2}$ × $4C_{2}$ × $2C_{2}$

=11340

8 0

3 years ago

Find the product. Simplify if possible -1/3 x (8 1/7)

gladu [14]

- 1/3 × 8 1/7 = - 2 5/7

~Hope this helped!~

4 0

3 years ago

What is an angle that is complementary to 5?

Elis [28]

angle 4 is complementary

4 0

4 years ago

For the given pair of events A and B, complete parts (a) and (b) below. A: When a page is randomly selected and ripped from

Kitty [74]

Answer:

The two events are dependent.

the probability that events A and B both occur is:

$\frac{1}{2424}\times \frac{1}{2423}=1.7026052585\times 10^{-7}\ or\ 0.00000017$

Step-by-step explanation:

Consider the provided information.

Event A: When a page is randomly selected and ripped from a 2424-page document and destroyed, it is page 2020.

Event B: When a different page is randomly selected and ripped from the document, it is page 1616.

Part(A)

The occurring of one event affects the probability of the other event.

Because if we ripped one page then the probability of ripping second page will going to change as the size of sample space will decrease.

For example: The document has 2424 page, that means size of sample space is 2424. If we ripped another page, the sample size will be 2423. That means event B is depending on event A.

Thus, the two events are dependent.

Part(B)

The probability that events A and B both occur.

If we select a page randomly from 2424-page document, the probability will be: 1/2424

Now we have 2423 pages left in the document as one page is destroyed.

The probability of selecting another page is: 1/2423.

Thus, the probability that events A and B both occur is:

$\frac{1}{2424}\times \frac{1}{2423}=1.7026052585\times 10^{-7}\ or\ 0.00000017$

4 0

4 years ago