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3 years ago

The percent equivalent of 1.01 ________

Mathematics

2 answers:

RUDIKE [14]3 years ago

8 0

101% is 1.01 as a percentage

Hope this helps!

Harrizon [31]3 years ago

6 0

Would be 101% because 1.01 multiplied by 100 is 101

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1) Here is a list of numbers your teacher drew at random from a possible 49 numbers for the last 7 days:

notka56 [123]

Answer b is correct have a great day :) it has very special value

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3 years ago

Emily is entering a bicycle race for charity. Her mother pledges $0.10 for every 0.25 mile she bikes. If Emily bikes 8 miles, ho

Arlecino [84]

Answer:

0.10 for 1/4 of a mile so 8 would be 32 times more. Which would be .1 x 32 or $3.20

Step-by-step explanation:

Plz mark brainliest :3

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3 years ago

Simplify the rational expression.

Vadim26 [7]

Answer:

$\frac{x-6}{x-9}$

Step-by-step explanation:

Given

$\frac{x^2-2x-24}{x^2-5x-36}$ ← factor the numerator and denominator

= $\frac{(x-6)(x+4)}{(x-9)(x+4)}$ ← cancel (x + 4 ) on numerator/ denominator

= $\frac{x-6}{x-9}$

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3 years ago

A friend of mine is giving a dinner party. His current wine supply includes 9 bottles of zinfandel, 10 of merlot, and 12 of cabe

e-lub [12.9K]

Answer:

Step-by-step explanation:

From the given information:

The total number of wine = 9 + 10 + 12 = 31

(1)

The number of distinct sequences used for serving any five wines can be estimated by using the permutation of the number of total wines with the number of wines served.

i.e

= $^{31}P_5$

$=\dfrac{31!}{(31-5)!}$

$=\dfrac{31!}{(26)!}$

$=\dfrac{31\times 30\times 29\times 28\times 27\times 26!}{(26)!}$

= 20389320

(2)

If the first two wines served = zinfandel and the last three is either merlot or cabernet;

Then, the no of ways we can achieve this is:

= $^9P_2\times ^{22}P_3$

$= \dfrac{9!}{(9-2)!}\times \dfrac{22!}{(22-3)!}$

$= \dfrac{9!}{(7)!}\times \dfrac{22!}{(19)!}$

$= \dfrac{9*8*7!}{(7)!}\times \dfrac{22*21*20*19!}{(19)!}$

= 665280

(3)

The probability that no zinfandel is served is computed as follows:

Total wines (with zinfandel exclusion) = 31 - 9 = 22

Now;

the required probability is:

$= \dfrac{^{22}P_5 }{^{31}P_5}$

$= \dfrac{\dfrac{22!}{(22-5)!} } {\dfrac{31!}{(31-5)!} }$

$= \dfrac{\dfrac{22!}{17!} } {\dfrac{31!}{(26)! }}$

$= \dfrac{(\dfrac{22*21*20*19*18*17!}{17!})} {(\dfrac{31*30*29*28*27*26!}{(26)! })}$

= 0.1549

≅ 0.155

4 0

3 years ago

What is cos 45º?<br> A<br> B<br> C<br> D

Keith_Richards [23]

0.53

Step-by-step explanation:

8 0

3 years ago