All categories

2 years ago

What is 64% out of 150?

Mathematics

2 answers:

motikmotik2 years ago

8 0

64 x

----- = ------

100 150

100x = 9600

x = 96

rosijanka [135]2 years ago

4 0

150 mean ----- 100%
x mean---------- 64%
--------------------

x = 64*150/100 = 96

hope this is understandably

You might be interested in

Kira drove 522 miles in 9 hours.<br> At the same rate, how many miles would she drive in 7 hours?

Ipatiy [6.2K]

GIven:522 miles in 9 hours.

So if we divide 522 by 9 we get that Kira drives 58 miles per hour.

Now in 7 hours Kira just drives 7 times that which means Kira drives.

58*7=406 miles

Hope this helped!

3 0

2 years ago

Read 2 more answers

3/4 times -4 2/9 <br> I have no idea what that equals please help

anygoal [31]

Answer:

-3.16666667

Step-by-step explanation:

7 0

2 years ago

True or False, the chance of flipping heads on a coin 3 times in a row is 1/2

bogdanovich [222]

true flip a coin it's fifty fifty

3 0

2 years ago

Read 2 more answers

A number that has exactly two distinct factors is prime what is the inverse of the statement?

Anastasy [175]

Answer:

Composite number

Step-by-step explanation:

A number that has more than 2 distinct factors is a composite number

plzzzz mark this as brainliest... and I hope this helped :)

8 0

2 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

2 years ago