PLEASE HELP 10 POINTS

What is 1/10 of 12.5 miles?

vazorg [7]

Answer:

1.25

Step-by-step explanation:

(1/10) x 12.5

4 0

3 years ago

Can u guys make a story with this graph good luck this is velocity

enot [183]

Yeahh I could do it

8 0

3 years ago

I'm really confused on 6-26. How do you find the equation?

exis [7]

I know the B is 2 Sorry I don't know the A

6 0

3 years ago

In a game, you toss a fair coin and a fair six-sided die. If you toss a heads on the coin and roll either a 3 or a 6 on the die,

pochemuha

Using probabilities, it is found that the expected profit of one round of this game is of $0.

A probability is the number of desired outcomes divided by the number of total outcomes.

One of the two sides of the coin are heads.
2 of the 6 sides of the dice are 3 or 6.

Hence, since the coin and the dice are independent, the probability of winning is:

$p = \frac{1}{2} \times \frac{2}{6} = \frac{1}{6}$

The expected value is the sum of each outcome multiplied by its respective probability.

In this problem:

$\frac{1}{6}$ probability of earning $30.
$\frac{5}{6}$ probability of losing $6.

Then:

$E(X) = 30\frac{1}{6} - 6\frac{5}{6} = 5 - 5 = 0$

The expected profit of one round of this game is of $0.

A similar problem is given at brainly.com/question/24855677

5 0

2 years ago

Teen obesity:

Law Incorporation [45]

Answer:

95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

Step-by-step explanation:

We are given that 15% of a random sample of 300 U.S. public high school students were obese.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample % of U.S. public high school students who were obese = 15%

n = sample of U.S. public high school students = 300

p = population percentage of all U.S. public high school students

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.15-1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ , $0.15+1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ ]

= [0.110 , 0.190]

Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

6 0

3 years ago