Since 111 is greater than 33, I will assume that you meant 11 votes for a candidate. Also, assuming that each student can vote only once, the answer would be

11/33 x 100=33.333333...

About 33.333 of the voters voted for candidate a.

3 0

3 years ago

PLEASE HELP!! WILL MARK BRAINlEST!!

mina [271]

Answer:

length is 9in and width is 5 in

Step-by-step explanation:

28/4= 7

7+2=9

got u fam

7-2=5

6 0

3 years ago

In a normally distributed data set a mean of 55 where 99.7% of the data fall between 47.5 and 62.5, what would be the standard d

NeX [460]

Given that normally distributed data set has a mean of 55 and 99.7% of data fall between 47.5 and 62.5.

Let s be the standard deviation of data set.

Since 99.7% data fall within 3 standard deviations of mean, z-value for 47.5 and 62.5 has an absolute value of 3.

That is |z|=3

But z= $\frac{x-mean}{standard deviation}$

Let us plugin x=47.5 and mean =55 and equate it to 3.

That is $|\frac{47.5-55}{s}| = 3$

$|\frac{-7.5}{s} | =3$

Since x is always positive ( being standard deviation), $|\frac{-7.5}{s} | = \frac{|-7.5|}{s} = \frac{7.5}{s}$

Hence $\frac{7.5}{s}= 3$

$s=\frac{7.5}{3} = 2.5$

We will get same value with 62.5 as well.

Hence standard deviation of data set is 2.5.

3 0

4 years ago