All categories

3 years ago

Number 18 factor linear expression

Mathematics

1 answer:

notsponge [240]3 years ago

5 0

I don’t have enough information to answer this..... what expression needs to be factored?

You might be interested in

Consider the following binomial experiment: A newspaper publisher claims that 52% of the people in a certain community read his

AURORKA [14]

Answer:

$SD = 8.4932$

Step-by-step explanation:

Given

$Sample\ Size (n) = 289$

$proportion, p= 52\%$

Required

Determine the standard deviation (SD)

The standard deviation of a binomial distribution is calculated as thus:

$SD = \sqrt{np(1-p)}$

Substitute values for n and p

$SD = \sqrt{289 * 52\% * (1-52\%)}$

Convert to decimals

$SD = \sqrt{289 * 0.52 * (1-0.52)}$

$SD = \sqrt{289 * 0.52 * 0.48}$

$SD = \sqrt{72.1344}$

$SD = 8.49319727782$

$SD = 8.4932$ --- approximated

Hence, the calculated standard deviation is approximately 8.4932

4 0

3 years ago

Easy question! 2x - 3 = 8 solve plz

statuscvo [17]

2x - 3 = 8
add 3 to both sides
2x = 11
divide both sides by 2
x = 5.5

7 0

3 years ago

Question 1b.

Ainat [17]

Answer: i) Reasonably symmetric

=============================================================

Explanation:

The data set is

35,40,45,49,51,54,57,58,58,64,68,68,75,77

Erase the outer pair of values (35 and 77) to end up with this smaller set

40,45,49,51,54,57,58,58,64,68,68,75

Repeat and erase off the outer values (40 and 75) to end up with

45,49,51,54,57,58,58,64,68,68

Keep going until we have either one or two elements left

49,51,54,57,58,58,64,68
51,54,57,58,58,64
54,57,58,58
57,58

At this point, we see that the center is between these two last values. So the median is (57+58)/2 = 57.5

--------------------------------

The center is at 57.5

Let L = {35,40,45,49,51,54,57} consist of the set of numbers lower than the median

Let U = {58,58,64,68,68,75,77} be the upper set of values larger than the median.

The middle of set L is 49 and the middle of set U is 68. So Q1 and Q3 are 49 and 68 respectively.

From that, we see the IQR = Q3 - Q1 = 68-49 = 19

Let A and B be the lower and upper fence

We would then say

A = Q1 - 1.5*IQR = 49-1.5*19 = 20.5

B = Q3 + 1.5*IQR = 68 + 1.5*19 = 96.5

Notice how all of the given data values are between A = 20.5 and B = 96.5, which means we don't have any outliers.

Furthermore, if we plotted a box-and-whisker plot (see below), we can see that the two whiskers are fairly the same length. It's not perfect but they're reasonably symmetric. Similarly, the median is somewhat in the middle of the box itself. So this gives a rough idea that the data set we're dealing with is fairly symmetric.

More confirmation can be in the form of a histogram which is also part of the image below. This histogram is fairly symmetric as the left half is nearly (not perfectly) a mirror image of the right half, and vice versa. No one side pulls too far out to create a highly skewed distribution.