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3 years ago

With a fair die, the probability of rolling any number 1 through 6 is the same.

Mathematics

2 answers:

Stella [2.4K]3 years ago

8 0

P(any number from 1 to 6) = 1/6

so the sum of P(1) + P(2) etc is 6*1/6 = 1

yaroslaw [1]3 years ago

7 0

The probability of all possible outcomes is always 1, meaning 100%.

In this case the probability of rolling a 1,2,3,4,5, or 6 is:

1/6+1/6+1/6+1/6+1/6+1/6

6(1/6)

6/6

1

You might be interested in

What is the product of 16x48?

Elis [28]

768 because 6x8 = 48 the 8 drop Down to the answer and the 4 on top of 1 and multiply 1x8= 8+1 =9 so the 9 drop down to the answer. Now that is fake because you need to multiply 4x6 and 4x1 so 4x6 is 24 below the 8 add a 0 and go 1 move to the left and put 4 and 2 on top of 1 now 4x1 is 4+1 is 5 and put 5 on the left of 4 and the final step is add 98+540=768 so 768 is your real answer.

4 0

3 years ago

A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women are randomly sample

BabaBlast [244]

Answer:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

$t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73$

$df=5+5-2=8$

$t_{crit}=\pm 2.306$

Reject H0: stress affects the menstrual cycle

Step-by-step explanation:

The statistic is given by the following formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =2$ represent the sample size for group 2

Tha data given is:

Group 1 (High stress) : 20,23,18,19,22

Group 2 (Relatively stress free): 26,31,25,26,30

We can calculate the sample mean and the sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X_1 =20.4$ represent the sample mean for the group 1

$\bar X_2 =27.6$ represent the sample mean for the group 2

$s_1=2.074$ represent the sample standard deviation for group 1

$s_2=2.702$ represent the sample standard deviation for group 2

And now we can calculate the statistic:

$t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73$

Now we can calculate the degrees of freedom given by:

$df=5+5-2=8$

Now we can calculate the critical value since the confidence is 95% the value for the significance would be $\alpha=1-0.95=0.05$ and the value for $\alpha/2 =0.025$ and if we find a critical value on th t distribution with 8 degrees of freedom that accumulates 0.025 of the area on each tail we got $t_{crit}=\pm 2.306$

Since our calculated value is lower than the critical value, we have enough evidence to reject the null hypothesis. And makes sense say that the difference between the two means are different at 5% of significance.

Reject H0: stress affects the menstrual cycle

5 0

3 years ago

ANY HELP PLEASE THANK YOU

Natalka [10]

Answer:

2:4 because there is two circles and 4 shapes?

7 0

4 years ago

If the cost of plastering 4 walls of a room at Rs 45 per sq. m is Rs 8100, find the area of the 4 walls of the room

olga55 [171]

Step-by-step explanation:

Rs8,100 / (Rs45 per square meter)

= 180 square meters.

The area of the 4 walls of the room is 180 sq.m.

3 0

3 years ago

All questions to be solved using linear combination.

monitta

1)
I:x-y=-7
II:x+y=7

add both equations together to eliminate y:
x-y+(x+y)=-7+7
2x=0
x=0

insert x=0 into II:
0+y=7
y=7

the solution is (0,7)

2)
I: 3x+y=4
II: 2x+y=5

add I+(-1*II) together to eliminate y:
3x+y+(-2x-y)=4+(-5)
x=-1

insert x=-1 into I:
3*-1+y=4
y=7

the solution is (-1,7)

3)

I: 2e-3f=-9
II: e+3f=18

add both equations together to eliminate f:
2e-3f+(e+3f)=-9+18
3e=9
e=3

insert e=3 into I:
2*3-3f=-9
-3f=-9-6
-3f=-15
3f=15
f=5

the solution is (3,5)

4)
I: 3d-e=7
II: d+e=5

add both equations together to eliminate e:
3d-e+(d+e)=7+5
4d=12
d=3

insert d=3 into II:
3+e=5
e=2

the solution is (3,2)

5)
I: 8x+y=14
II: 3x+y=4

add I+(-1*II) together to eliminate y
8x+y+(-3x-y)=14-4
5x=10
x=2

insert x=2 into II:
3*2+y=4
y=4-6
y=-2

the solution is (2,-2)

8 0

3 years ago