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3 years ago

PLEASE HELP MEEEEEEEEEEEEEEEE

Mathematics

2 answers:

trapecia [35]3 years ago

7 0

Answer: 89 metres squared

Step-by-step explanation:

large area = 13 x 8 = 104

small area = 5 x 3 = 15

104 - 15 = 89

podryga [215]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

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

Sarah has some chocolates.

kakasveta [241]

Answer:

24:8 which is also just 6:2 or 3:1

Step-by-step explanation:

There are many ways you can write this, and this is what I think hope this helps

8 0

4 years ago

Multiples of 3 between 81 and 110

Tpy6a [65]