Which expression is not equivalent to a · c · 5 · 5? 25 ca 10 ac 25 ac

EASY BRAINLIEST PLEASE HELP!!

iogann1982 [59]

Answer:D..the diagonals bisect each other. Since bisectors divide the diagonals equally in two parts.

6 0

3 years ago

Read 2 more answers

What is the value of |PS|

VMariaS [17]

Answer:

if i draw a line from s ro hg then its value should be 12 cm. and should form a right angle triangle. it is not fulfilling pytgogerous property..

3 0

3 years ago

Compute the derivative of the following function f(x)=3x^2 at x=2

Oduvanchick [21]

Answer:

12

Step-by-step explanation:

Differentiate using the power rule

$\frac{d}{dx}$ (a $x^{n}$ ) = na $x^{n-1}$

Given

f(x) = 3x², then

f'(x) = 2 × 3 × $x^{2-1}$ = 6x, thus

f'(2) = 6 × 2 = 12

4 0

3 years ago

Suppose that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked

IgorC [24]

Answer:

(a) 658,008 different samples can be chosen.

(b) 222,111 samples will contain at least one defective board.

(c) The probability that a randomly chosen sample of five contains at least one defective board is 0.34.

Step-by-step explanation:

We are given that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked for defects.

(a) To find how many different samples can be chosen, we will use a combination formula here because the order of selecting a sample of 5 from the production run of 40 doesn't matter.

Here, n = total sample = 40 and r = selected sample = 5

So, the combination formula is; $^{n}C_r= \frac{n!}{r! \times (n-r)!}$

$^{40}C_5= \frac{40!}{5! \times (40-5)!}$

$^{40}C_5= \frac{40!}{5! \times 35!}$

$^{40}C_5$ = 658,008 ways

So, 658,008 different samples can be chosen.

(b) To find how many samples will contain at least one defective board, we will first find how many samples will contain no or 0 defective board.

For this also, we will use a combination where n = 40 - 3 = 37 non-defective computer board and a sample of r = 5 computer boards.

So, $^{n}C_r= \frac{n!}{r! \times (n-r)!}$

$^{37}C_5= \frac{37!}{5! \times (37-5)!}$

$^{37}C_5= \frac{37!}{5! \times 32!}$

$^{37}C_5$ = 435,897 ways

This means that 435,897 of the 658,008 samples will contain no defective board.

Now, the samples that will contain at least one defective board = Total samples - Samples that contain no defective board

= 658,008- 435, 897

= 222,111

(c) The probability that a randomly chosen sample of five contains at least one defective board is given by;

Required Probability = $\frac{222,111}{658,008}$

= 0.34 or 34%

3 0

4 years ago

Do the measures of center make sense? A. Only the mode makes sense since the data is nominal. B. All the measures of center mak

Ray Of Light [21]

Answer:

A. Only the mode makes sense since the data is nominal.

Step-by-step explanation:

Hello!

The objective of the study was to determine if deficiency of carbon dioxide in the soil affects the phenotype of peas.

The variable of study is X: Phenotype of a pea grown in soil with carbon dioxide deficiency.

Possible values of Phenotype codes:

1= smooth-yellow

2= smooth-green

3= wrinkled-yellow

4= wrinkled-green

The absolute frequencies for each phenotype are:

f(1)= 3

f(2)= 4

f(3)= 6

f(4)= 1

n= 14

a) Mean:

X[bar]= (∑xifi)/n= [(1*3)+(2*4)+(3*6)+(4*1)]/14= 33/14= 2.357= 2.36

The average value is always within range of definition of the variable but it does not necessarily correspond to an observation.

b) Median:

To determine the value that corresponds to the median you have to calculate its position:

For even samples the position is:

PosMe= n/2= 14/2= 7

Then you have to arrange the data from least to greatest, in this case, starting from the first category, you have to determine where the seventh observation is within the observed absolute frequencies. The phenotype that corresponds to the 7th observation is 2= smooth-green.

Me= 2= smooth-green.

c) Mode:

The mode corresponds to the most observed category/ value of the variable, i.e. the category with the most observations is 3= wrinkled-yellow

Md= 3= wrinkled-yellow

d) Midrange: (1 + 4)/2= 2.5

e)

As you can see the variable is qualitative and categorical. Even if all central tendency measurements can be calculated, truth is that the only one that shows any valuable information regarding the data set is the mode.

I hope this helps!

3 0

3 years ago