If 2/5 of a tumor weighs 1 pound, how much will the whole tumor weigh?

Solve the equation when x=12 1/3 Please show your work 7x+8(x+1/4)=3(6x-9)-8

alina1380 [7]

7(12 1/3) + 8(12 1/3 + 1/4) = 3(6(12 1/3) - 9) - 8

12 1/3 = 12.33
1/4 = 0.25

7(12.33) + 8(12.33 + 0.25) = 3(6(12.33) - 9) - 8
86.31 + 8(12.58) = 3(73.98 - 9) - 8
86.31 + 100.64 = 3(64.98) - 8
186.95 = 186.94 - 8
186.95 ≠ 178.94

False. The x, when 12 1/3 does not work inside this equation

hope this helps

5 0

3 years ago

#14 quick answer pls :)

Andrej [43]

Answer:

35.56

Step-by-step explanation:

4 0

2 years ago

The population of a town was 7652 in 2016. The population grows at a rate of 1.6% annually.

Serggg [28]

Answer:

(A) The population's growth rate in equation form is y = (0.016t * 7652) + 7652

(B) y = (0.016t * 7652) + 7652 =

y = (0.016(8) * 7652) + 7652 =

y = (0.128 * 7652) + 7652 =

y = 979.456 + 7652 =

y = 8631.456 (Or About) 8631

Step-by-step explanation:

(A) Y = the total population of the town. 0.016 is 1.6% just in its original form. T = the year in which were trying to find the town's total population. 7652 is the total population of the town in 2016. With this information, the equation reads, The total population of the town (Y) is equal to 16% (0.016) of the current year's population (T) added to 2016's population of 7652. (This last sentence can also be read what is 1.6% of the towns population in the year were trying to find. Because the population is always growing, 1.6% gets multiplied as to scale with the total population in year T)

(B) We just substitute (T) for 2024, or 8 years after 2016 (2024-2016) and simplify the equation we made.

7 0

3 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of trials before q successes is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c: