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3 years ago

8

Kathy John Barbara Dominic and Troy want to go to a concert but they only have 3 tickets. If they decide to select randomly who

will go, what is the probability Kathy will be selected

1 answer:

Lilit [14]3 years ago

6 0

1/5 chance kate will be selected.

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The GRE is an entrance exam that most students are required to take upon entering graduate school. In 2014, the combined scores

Ede4ka [16]

$\mathbb P(286$
$\mathbb P(286$

The empirical rule says that approximately 68% of any normal distributed data set lies within one standard deviation of the mean.

$\mathbb P(-2$
$\implies\mathbb P(-2$

The same rule states that about 95% lies within two standard deviations of the mean. Using this rule, and the fact that any normal distribution is symmetric about its mean, you have

$\mathbb P(-2$
$0.95=2\mathbb P(-2$
$\implies\mathbb P(-2$

$\implies\mathbb P(-2$

8 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago

Lady bird [3.3K]

Answer:

19 1-point shots and 32 2-point shots

Step-by-step explanation:

Let x be the amount of 1-point shots they made and y be the amount of 2-point shots they made then...

x+y=51

x+2y=83

We can solve this equation using elimination. In this case, we can eliminate x be subtracting the two equations.

-y=-32

Divide both sides by -1

y=32

Plug this back into the equation to solve for x

x+32=51

Subtract 32 from both sides

x=19

They scored 19 1-point shots and 32 2-point shots.

8 0

2 years ago

Brums [2.3K]

Answer:

<u>Yes</u>

Step-by-step explanation:

<u>When there are 2 rows</u>

Number of erasers per row = No. of erasers / No. of rows
Number of erasers per row = 18/2
Number of erasers per row = 9

<u>When there are 3 rows</u>

Number of erasers per row = No. of erasers / No. of rows
Number of erasers per row = 18/3
Number of erasers per row = 6

<u>Therefore, there will be more erasers per row in 2 equal rows than in 3 equal rows.</u>

7 0

2 years ago

Math help!<br><br> Question 5

Rudiy27

Answer:

a. 4 teachers b. 58 students c. 62 people

Step-by-step explanation:

multiply 80 by 0.05, you get 4.

Multiply 1160 by 0.05, you get 58.

Add 80 and 1160 together, you get 1240. Multiply 1240 by 0.05, you get 62.

5 0

3 years ago