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3 years ago

Please help solve for the area

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

3 0

Answer:

The area of the shape is 126 cm.

Step-by-step explanation:

6 + 3 + 3 = 12 cm, all sides

12 * 12 = 144

3 * 3 = 9 / 2 = 4.5

4.5 * 4 = 18

144 - 18 = 126cm

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Our class has 50 students. Assuming that no students are born on leap days (February29), what is the probability that no two stu

madreJ [45]

Answer: a) $\dfrac{799}{31250}$ b) $\dfrac{30451}{31250}$

Step-by-step explanation:

Since we have given that

Number of days in a year = 365

Number of students = 50

Probability that no two students share the same birthday. It means each student has different birthday.

Probability(no two students share the same birthday) $=\dfrac{365-1}{365}\times \dfrac{365-2}{365}........\times \dfrac{365-50}{365}$

Probability(no two students share the same birthday) $=\dfrac{^{365}P_{50}}{365^{50}}$

Probability(no two students share the same birthday) $=\dfrac{799}{31250}\approx 0.025568$

Probability that at least one of the students has the same birthday as another student in the class is given by

$1-P(\text{no student share the same birthday})\\\\=1-\dfrac{799}{31250}\\\\=\dfrac{30451}{31250}=0.974432$

5 0

4 years ago

A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc

pogonyaev

Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

(a) What are the mean and the standard deviation of the number X of students who accept?

$n = 1200, p = 0.75$ . So

$E(X) = np = 1200*0.75 = 900$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15$

The mean is 900 and the standard deviation is 15.

(b) Use the Normal approximation to find the probability that at least 800 students accept.

Using continuity corrections, this is $P(X \geq 800 - 0.5) = P(X \geq 799.5)$ , which is 1 subtracted by the pvalue of Z when X = 799.5. So

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

$Z = \frac{799.5 - 900}{15}$

$Z = -6.7$

$Z = -6.7$ has a pvalue of 0.

1 - 0 = 1

100% probability that at least 800 students accept.

(c) The college does not want more than 950 students. What is the probability that more than 950 will accept?

Using continuity corrections, this is $P(X \geq 950 - 0.5) = P(X \geq 949.5)$ , which is 1 subtracted by the pvalue of Z when X = 949.5. So

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

$Z = \frac{949.5 - 900}{15}$

$Z = 3.3$

$Z = 3.3$ has a pvalue of 0.9995

1 - 0.9995 = 0.0005

0.05% probability that more than 950 will accept.

(d) If the college decides to increase the number of admission offers to 1300, what is the probability that more than 950 will accept?

Now n = 1300. So

$E(X) = np = 1300*0.75 = 975$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15.6$

Same logic as c.

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

$Z = \frac{949.5 - 975}{15.6}$

$Z = -1.63$

$Z = -1.63$ has a pvalue of 0.0516

1 - 0.0516 = 0.9484

94.84% probability that more than 950 will accept

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3 years ago

A group of randomly selected Apple Valley High School students were asked to pick their favorite gym class. The table below show

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bixtya [17]

Using an exponential function, it is found that f(5.5) = 19.8.

<h3>What is an exponential function?</h3>

An exponential function is a function in which the growth rate is a percentage, modeled by:

$y = ab^x$

In which:

a is the initial value.

b is the rate of change.

f(3.5) = 25 means that:

$ab^{3.5} = 25$

$a = \frac{25}{b^{3.5}}$

f(8.5) = 14 means that:

$ab^{8.5} = 14$

Hence:

$\frac{25}{b^{3.5}} \times b^{8.5} = 14$

$25b^5 = 14$

$b = \sqrt[5]{\frac{14}{25}}$

$b = 0.89$

$a = \frac{25}{0.89^{3.5}} = 37.59$

Hence, the function is given by:

$y = 37.59(0.89)^x$

Then, when x = 5.5:

$f(5.5) = 37.59(0.89)^{5.5} = 19.8$

More can be learned about exponential functions at brainly.com/question/25537936

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2 years ago

Please help solve for the area​

Please help solve for the area