Why is 2+2=4?????????

Factor the following polynomial: <br> 5x(a-b)-2y(a-b)

Nookie1986 [14]

Answer:

(a - b)(5x - 2y)

Step-by-step explanation:

Given

5x(a - b) - 2y(a - b) ← factor out (a - b) from each term

= (a - b)(5x - 2y) ← in factored form

5 0

3 years ago

What is d in the following arithmetic sequence? 5, 7, 9, 11, 13....

GarryVolchara [31]

The answer is C + 2

4 0

3 years ago

GUYS I need HELP PLSSSSS I'll give you brainliest.

makkiz [27]

Answer:

C

Step-by-step explanation:

Technically there is a one here -1(5). Meaning its a positive

8 0

3 years ago

Read 2 more answers

For a sample of nequals37, find the probability of a sample mean being less than 12 comma 751 or greater than 12 comma 754 when

Irina18 [472]

Answer:

50% probability of a sample mean being less than 12,751 or greater than 12,754

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 12751, \sigma = 2.1, n = 37, s = \frac{2.1}{\sqrt{37}} = 0.3456$

Find the probability of a sample mean being less than 12,751 or greater than 12,754

Less than 12,751

pvalue of Z when X = 12751.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{12751 - 12751}{0.3456}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

50% probability of the sample mean being less than 12,751.

Greater than 12,754

1 subtracted by the pvalue of Z when X = 12,754.

$Z = \frac{X - \mu}{s}$

$Z = \frac{12754 - 12751}{0.3456}$

$Z = 8.68$

$Z = 8.68$ has a pvalue of 1

1 - 1 = 0

0% probability of the sample mean being greater than 12754

Less than 12,751 or greater than 12,754

50 + 0 =50

50% probability of a sample mean being less than 12,751 or greater than 12,754

6 0

3 years ago

(a)Find all integer solutions to the equation 105x + 83y = 1.

Mashcka [7]

Answer:

(a) (34+83t,-43-105t) where t is an integer

(b) (272+83t,-344-105t) where t is an integer.

(c) 62

Step-by-step explanation:

a)

We are going to perform Euclidean's Algorithm.

Let's begin with seeing how many times 83 goes int 105.

105=83(1)+22 (eq1)

83=22(3)+17 (eq2)

22=17(1)+5 (eq3)

17=5(3)+2 (eq4)

5=2(2)+1 (eq5)

Now let's go backwards through those equations.

5-2(2)=1 (eq5 rewritten so that the remainder was by itself)

5-2[17-5(3)]=1 (replaced the 2 in ( ) with eq4 solved for the remainder)

5-2(17)+5(6)=1 (distributive property was performed)

-2(17)+5(7)=1 (combined my 5's)

-2(17)+7(5)=1 (multiplication is commutative)

-2(17)+7(22-17)=1 (used eq3)

-2(17)+7(22)-7(17)=1 (distribute property was performed)

-9(17)+7(22)=1 (combined my 17's)

-9(83-22(3))+7(22)=1 (used eq2)

-9(83)+22(27)+7(22)=1 (distributive property was performed)

83(-9)+22(34)=1 (multiplication is commutative and combined my 22's)

83(-9)+34(105-83)=1 (used eq1)

105(34)+83(-43)=1 (after distributive property and reordering)

So we have a point on the line being (x,y)=(34,-43).

We can use the slope to figure out all the other integer pairs from that initial point there.

The slope of ax+by=c is -a/b.

So the slope of 105x+83y=1 is -105/83.

So every time we go down 105 units we go right 83 units

This says we have the following integer pairs on our line:

(34+83t,-43-105t) where t is an integer.

Let's verify:

Plug it in!

105[34+83t]+83[-43-105t]

105(34)+105(83)t+83(-43)-83(105)t

105(34)+83(-43)

1

We are good!

(b)

We got from part (a) that 105(34)+83(-43)=1.

Multiply both sides we get 8 on the right hand side:

105(34*8)+83(-43*8)=8

Simplify:

105(272)+83(-344)=8

So the integer pairs is (272+83t,-344-105t) where t is an integer.

Let's verify:

105[272+83t]+83[-344-105t]

105(272)+105(83)t+83(-344)-83(105)t

105(272)+83(-344)

8

(c)

Let u=83^(-1) mod 105.

Then 83u=1 mod 105.

This implies:

83u-1=105k for some integers k.

Add 1 on both sides:

83u=105k+1

Subtract 105k on both sides:

83u-105k=1

Reorder:

105(-k)+83u=1.

We found all (x,y) integer pairs such that 105x+83y=1.

We go (34+83t,-43-105t) where t is an integer.

So k=-34-83t while u=-43-105t.

Since we want to find an integer t such that u is between 0 and 104, we could solve 0<-43-105t<104.

Add 43 on all sides:

43<-105t<147

Divide all sides by -105:

-43/105>t>-147/105

-147/105<t<-43/105

This says t is approximately between -1.4 and -0.4 . This includes only the integer -1.

When t=-1, we have u=-43-105(-1)=-43+105=62.

3 0

3 years ago

Why is 2+2=4?????????​

Why is 2+2=4?????????