Write the prime factorization of 20 using exponents

Answers

Answer 1

Answer:

Prime factorization expresses an integer in terms of multiples of prime numbers. The prime factorization of 20 using exponents can be written as 2²×5¹.

What is prime factorization?

Factorization is expressing a mathematical quantity in terms of multiples of smaller units of similar quantities.

Prime factorization is when all those factors are prime numbers.

Thus, prime factorization expresses an integer in terms of multiples of prime numbers.

The prime factorization of 20 using exponents can be written as,

20 = 1 × 2 × 2 × 5

= 2² × 5¹

Hence, the prime factorization of 20 using exponents can be written as 2²×5¹.

Learn more about prime factorization here:

brainly.com/question/10454590

#SPJ2

Answer 2

Answer: Here you go

2 to the second power times 5

Related Questions

Which relation is a function?
In the figure below, angle X is 35 degrees. What is the measure of angle Z?
We know that if the probability of an event happening is 100%, then the event is a certainty. Can it be concluded that if there is a 50% chance of contracting a communicable disease through contact with an infected person, there would be a 100% chance of contracting the disease if 2 contacts were made with the infected person
Just cThank you sm!!!WILL GIVE BRAINLIEST!!
Mrs.Montana has five students namely:rodley,cristina,christian,elsa and made .Count all the letters in each name.In which of the following sets the order of the names least to greatest.ARodley,cristina,Christian,mae,elsa Bcristina,Christian elsa mae rodleyc Mae elsa rodley cristina Christian D Christian Cristina rodley mae elsa

(help) Write an expression for the quotient of 371 and y .

Answers

Answer: 371 / y

Step-by-step explanation: 371 is the dividend and y is the divisor

The owner of a small machine shop has just lost one of his largest customers. The solution to his problem,he says, is to fire three machinists to balance his workforce with his current level of business. The owner says that it is a simple problem with a simple solution. The three machinists disagree. Why

Answers

Answer:

It may look simple to the owner because he is not the one losing a job. For the three machinists it represents a major event with major consequences

Two points located on jk are j (-1,-9) and k (5,3). What is the slope of jk?

Answers

Answer:

Slope = 2

Step-by-step explanation:

Slope = $(rise)/(run)$

Slope = $(3+9)/(5+1)$

Slope = $(12)/(6)$

Slope = 2

In the given case, we can conclude that The slope of the line JK is 2.

To find the slope of the line that passes through the points J(-1,-9) and K(5,3), we can use the formula: slope = $(y2 - y1) / (x2 - x1).$

The slope of a line is a measure of how steep the line is. It describes the rate at which the dependent variable (usually denoted as 'y') changes with respect to a change in the independent variable (usually denoted as 'x').

Plugging in the coordinates, we get:

slope = (3 - (-9)) / (5 - (-1)) = 12 / 6 = 2.

Learn more about Slope here:

brainly.com/question/19131126

#SPJ3

A.3(f) The line graphed on the grid represents the first of two equations in a system of linear equations.20
-16
12
-8
-20 -16 -12
-
-4
48
12
16 2024
-8
-12
16
If the graph of the second equation in the system passes through (-12, 20) and (4,12), which statement is true?

Answers

I don’t any idea for this

Jimmy is a flag person and earned $321.10 last week for 32.5 hours work. What is his hourly wage? *Your answer

Answers

The answer should be 9.88

In tilapia, an important freshwater food fish from Africa, the males actively court females. They have more incentive to court a female who has already laid all of her eggs, but can they tell the difference? an experiment was done to measure the male tilapia's response to the smell of female fish. Water containing feces from females that were either pre-ovulatory (they still had eggs) or post-ovulatory (they had already laid their eggs) was washed over the gills of males hooked up to an electro-olfactogram machine which measured when the senses of the males were excited. The amplitude of the electro-olfactogram was used as a measure of the excitability of the males in the two different circumstances. Six males were exposed to the scent of pre-ovulatory females; their readings average 1.51 with a standard deviation of .25. Six different males were exposed to post-ovulatory females; their average readings of 0.87 with standard deviation is .31. Assume that the electro-olfactogram readings were approximately normally distributed within the groups.(A) test for a difference in the excitability of the males with exposure to these two types of females
(B) what is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidnece limit for the difference between population means?

Answers

Answer:

a) $t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

b) $(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

Step-by-step explanation:

Part a

Data given and notation

$\bar X_(1)=1.51$ represent the mean for scent of pre ovulatory

$\bar X_(2)=0.87$ represent the mean for post ovolatory

$s_(1)=0.25$ represent the sample standard deviation for preovulatory

$s_(2)=0.31$ represent the sample standard deviation for postovulatory

$n_(1)=6$ sample size for the group preovulatory

$n_(2)=6$ sample size for the group postovulatory

z would represent the statistic (variable of interest)

$p_v$ represent the p value

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean's are different, the system of hypothesis would be:

H0: $\mu_(1) = \mu_(2)$

H1: $\mu_(1) \neq \mu_(2)$

If we analyze the size for the samples both are lower than 30, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We have all in order to replace in formula (1) like this:

$t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

Find the critical value

We find the degrees of freedom:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking for thwo values on the t distribution with df =10 that accumulates 0.025 of the area on each tail. We can us excel or a table to find it, for example the code in Excel is:

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

Other Questions

B is the midpoint of AC, D is the midpoint of CE , and AE = 23. Find BD.12.54611.523

Find the area of the composite figure below.

A disease has hit a city. The percentage of the population infected t days after the disease arrives is approximated by p(t)equals7 t e Superscript negative t divided by 13 for 0less than or equalstless than or equals39. After how many days is the percentage of infected people a maximum? What is the maximum percent of the population infected?

Y=3x+4. Find an equation of a line that is:parallel to the given liney=3x+4. Find an equation of a line that is:perpendicular to the given linePLZ ANSWER FAST 50 POINTS