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3 years ago

What is 8/9 closer to? A. 1/2 B. 1 or C. 0

Mathematics

2 answers:

nikitadnepr [17]3 years ago

5 0

9/9 = 1 and 8/9 is only 1/9 less than 9/9 which is 1.

Answer: B) 1 8/9 is closer to 1

Arisa [49]3 years ago

5 0

In order to find this out, you could convert all the numbers to have the same common denominator. If you made the denominator 18, then
8/9 = 16/18
1/2 = 9/18
1 = 18/18
0 = 0/18.

This allows a clear comparison which shows that 8/9 is closest to 1 out of those options as it is only 2/18 (or 1/9) different. Therefore, the answer is B.

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Plzzzz help!!! quick one

ludmilkaskok [199]

D= 65 degrees.

Because straight line adds up to 180
So 180-115= 65

5 0

2 years ago

What is the slope of a line parallel to the line whose equation is 2x + 4y = 56.<br><br> HELP ASAP

Anton [14]

Answer:

Let's put this equation in standard form...

$2x+4y=56\\4y=56-2x\\y=14-0.5x$

If it is parallel, it would have the SAME slope but ANY OTHER y-intercept!!

For example...

$y=-0.5x+304\\y=-0.5x-67$

Hope this helps!!

3 0

1 year ago

Read 2 more answers

Neighbor Fred built a retaining wall that was 2 feet outside his property line. What has Fred created?

yulyashka [42]

Answer:

An encroachment.

Step-by-step explanation:

In real estate,encroachment is building a structure between the boundaries of your property or land and your neighbor's property or land.

The retaining wall Fred build that was 2ft outside his property line is called encroachment

4 0

3 years ago

The first term of a geometric sequence is 15, and the 5th term of the sequence is <img src="https://tex.z-dn.net/?f=%5Cfrac%7B24

sladkih [1.3K]

The geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

Explanation:

Given that the first term of the geometric sequence is 15

The fifth term of the sequence is $\frac{243}{125}$

We need to find the 2nd, 3rd and 4th term of the geometric sequence.

To find these terms, we need to know the common difference.

The common difference can be determined using the formula,

$a_n=a_1(r)^{n-1}$

where $a_1=15$ and $a_5=\frac{243}{125}$

For $n=5$ , we have,

$\frac{243}{125}=15(r)^4$

Simplifying, we have,

$r=\frac{3}{5}$

Thus, the common difference is $r=\frac{3}{5}$

Now, we shall find the 2nd, 3rd and 4th terms by substituting $n=2,3,4$ in the formula $a_n=a_1(r)^{n-1}$

For $n=2$

$a_2=15(\frac{3}{5} )^{1}$

$=9$

Thus, the 2nd term of the sequence is 9

For $n=3$ , we have,

$a_3=15(\frac{3}{5} )^{2}$

$=15(\frac{9}{25} )$

$=\frac{27}{5}$

Thus, the 3rd term of the sequence is $\frac{27}{5}$

For $n=4$ , we have,

$a_4=15(\frac{3}{5} )^{3}$

$=15(\frac{27}{25} )$

$=\frac{81}{25}$

Thus, the 4th term of the sequence is $\frac{81}{25}$

Therefore, the geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

5 0

3 years ago

In a clothing store, 65% of the customers buy a shirt, 30% of the customers

diamong [38]

Answer:

75%

Step-by-step explanation:

List out known probabilities

$\Pr(shirt) = 0.65 = \frac{65}{100} = \frac{13}{20}\\ \Pr(pants) = 0.40 = \frac{30}{100} = \frac{6}{20}\\\Pr(shirt \cap pants) = 0.20 = \frac{20}{100}=\frac{1}{5}$

recall equation

$\Pr(A \cup B)=\Pr(A)+\Pr(B)-\Pr(A \cap B)$

Plug in values

$\Pr(shirt \cup pants)=\frac{13}{20}+\frac{6}{20}-\frac{1}{5}\\\\therefore \Pr(shirt \cup pants) = \frac{15}{20} =\frac{3}{4} = 75\%$

4 0

2 years ago