The larger number is 18 more than twice the smaller. If the sum of the two numbers is 93, find

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

7907÷24=? help answer and remander

cupoosta [38]

Answer:

329.45833333333

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A university administrator was interested in determining if there was a difference in the distance students travel to get from c

eduard

Answer:

1) Fail to reject the Null hypothesis

2) We do not have sufficient evidence to support the claim that the mean distance students traveled to school from their current residence was different for males and females.

Step-by-step explanation:

A university administrator wants to test if there is a difference between the distance men and women travel to class from their current residence. So, the hypothesis would be:

$H_{o}: \mu_{F}-\mu_{M}=0\\\\ H_{a}: \mu_{F}-\mu_{M}\neq 0$

The results of his tests are:

t-value = -1.05

p-value = 0.305

Degrees of freedom = df = 21

Based on this data we need to draw a conclusion about test. The significance level is not given, but the normally used levels of significance are 0.001, 0.005, 0.01 and 0.05

The rule of the thumb is:

If p-value is equal to or less than the significance level, then we reject the null hypothesis
If p-value is greater than the significance level, we fail to reject the null hypothesis.

No matter which significance level is used from the above mentioned significance levels, p-value will always be larger than it. Therefore, we fail to reject the null hypothesis.

Conclusion:

We do not have sufficient evidence to support the claim that the mean distance students traveled to school from their current residence was different for males and females.

7 0

3 years ago

Slope of (-7,-2) (9,4)

Olenka [21]

Answer:

m= $\frac{3}{8}$