Ask question

Ask question

All categories

4 years ago

6

Every student in the senior class is taking history or science and 85 of them are taking both. If there are 106 seniors taking h

istory and 109 seniors taking science, how many students are in the senior class?

1 answer:

MissTica4 years ago

5 0

Number of students in senior class is 130

<em><u>Solution:</u></em>

Given that Every student in the senior class is taking history or science

85 of them taking both history and science

106 seniors takes history

109 seniors takes science

To find: number of students in senior class

Let A be the set of the students of the senior class that take history

Let B be the set of students of senior class that they science

The number of students in senior class is given by:

$|A \cup B|=|A|+|B|-|A \cap B|$

Where,

A = 106 and B = 109 and |A n B| = 85

$|A \cup B|=106+109-85=130$

Thus number of students in senior class is 130

You might be interested in

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

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

In a set with mean $\mu$ and standard deviation $\sigma$ , 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

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

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

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

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0

3 years ago

Use Newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 − x −

drek231 [11]

Answer:

$x_{2} = 0.0000$

Step-by-step explanation:

The formula for the Newton's method is:

$x_{i+1} = x_{i} + \frac{f(x_{i})}{f'(x_{i})}$

Where $f' (x_{i})$ is the first derivative of the function evaluated in $x_{i}$ .

$x_{i+1} = x_{i} + \frac{x_{i}^{4}-x_{i}-3}{4\cdot x_{i}^{3}-1}$

Lastly, the value of $x_{2}$ is determined by replacing $x_{1}$ with its numerical value:

$x_{2} = x_{1} + \frac{x_{1}^{4}-x_{1}-3}{4\cdot x_{1}^{3}-1}$

$x_{2} = 1.0000 + \frac{1.0000^{4}-1.0000-3}{4\cdot (1.0000)^{3}-1}$

$x_{2} = 0.0000$

8 0

3 years ago

A=1/2•e•m <br> solve the following formula for e

Vedmedyk [2.9K]

(1 / 2) x e x meters =

1.35914091 meters

8 0

3 years ago

The radioactive substance cesium-137 has a half-life of

statuscvo [17]

A(t)=A0*(1/2)^(t/30)
where t=elapsed years
A0=initial mass at t=0.

7 0

3 years ago

Stan's cookie recipe makes 24 cookies and calls for exactly 384 sprinkles. He is wondering how many sprinkles (p ) he will need

Arisa [49]

Answer:

960

Step-by-step explanation:

<u>Given:</u>

Stan's cookie recipe makes 24 cookies.
It needs exactly 384 sprinkles.

<u>To Find:</u>

Number of sprinkles needed for 60 cookies

<h2><u>Solution:</u></h2>

<u>Find the number of sprinkles needed for 1 cookie:</u>

24 cookies = 384 sprinkles
1 cookie = 384 ÷ 24
1 cookie = 16 sprinkles

<u>Find the number of sprinkles needed for 60 cookies:</u>

1 cookie = 16 sprinkles
60 cookies = 16 x 60
60 cookies = 960 sprinkles

Answer: Stan will need 960 sprinkles for 60 cookies.

3 0

2 years ago