What are the solutions to the equation -2t+10t=0

PLZ HELP ME

hoa [83]

Answer: false

Step-by-step explanation:

false because 19 - 6 =13 and 13 +12 = 25 and 25 - 18 = 7

8 0

3 years ago

A dealer paid $10,000 for a boat for 30% more than the auction. At the dealership, a salesperson sold the boat for 30% more than

Levart [38]

50,000 because they rounded it down to half price, simple math

Please Mark Brainliest

8 0

4 years ago

Read 2 more answers

What is the question 15%of258

Readme [11.4K]

The answer would be 219.3 (:
Please mark me the brainliest ty

4 0

3 years ago

Can someone please help me with 1 to 2 please . With explanation. And how you got the answers

Neko [114]

So, the rate per 7 minutes is 7:196

To find the unit rate (how many spins per minute) we need to divide the 196 by 7.

196/7 = 28

The unit rate is 1:28

At this rate, the unicycle makes 420 revolutions per 15 minutes. We find this by 28 * 15 = 420.

5 0

4 years ago

A professor knows that her statistics students' final exam scores have a mean of 79 and a standard deviation of 11.3. In his cla

OverLord2011 [107]

For each student, there are only two possible outcomes. Either they score an A, or they do not. The probability of a student scoring an A is independent of any other student, which means that the binomial probability distribution is used to solve this question.

Additionally, to find the proportion of students who scored an A, the normal distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Proportion of students that scored an A:

Scores have a mean of 79 and a standard deviation of 11.3, which means that $\mu = 79, \sigma = 11.3$

Scores of 90 or higher are graded an A, which means that the proportion is 1 subtracted by the p-value of Z when X = 90, so:

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

$Z = \frac{90 - 79}{11.3}$

$Z = 0.97$

$Z = 0.97$ has a p-value of 0.8340.

1 - 0.8340 = 0.166

The proportion of students that scored an A is 0.166.

Probability that 6 students or more will score an "A" on the final exam:

Binomial distribution.

22 students, which means that $n = 22$

The proportion of students that scored an A is 0.166, which means that $p = 0.166$

The probability is:

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Then

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

$P(X = 0) = C_{22,0}.(0.166)^{0}.(0.834)^{22} = 0.0184$

$P(X = 1) = C_{22,1}.(0.166)^{1}.(0.834)^{21} = 0.0807$

$P(X = 2) = C_{22,2}.(0.166)^{2}.(0.834)^{20} = 0.1687$

$P(X = 3) = C_{22,3}.(0.166)^{3}.(0.834)^{19} = 0.2239$

$P(X = 4) = C_{22,4}.(0.166)^{4}.(0.834)^{18} = 0.2117$

$P(X = 5) = C_{22,5}.(0.166)^{5}.(0.834)^{17} = 0.1517$

Then

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0184 + 0.0807 + 0.1687 + 0.2239 + 0.2117 + 0.1517 = 0.8551$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8551 = 0.1449$

Thus

0.1449 = 14.49% probability that 6 students or more will score an "A" on the final exam.

For a problem that used the normal distribution, you can check brainly.com/question/15181104, and for a problem that used the binomial distribution, you can check brainly.com/question/15557838

3 0

3 years ago