MATH HELP ASAP!!! MAKING BRAINLIST

salantis [7]

If 1/4 of x is 16, we multiply 16 x 4 which will give us x=64. 3/4 of 64 = 48

6 0

3 years ago

A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c

olchik [2.2K]

The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in $E^{c}$ . Find $P(E^{c})$ .

Solution:

Part 1:

$E^{c}$ means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

$E^{c}=S-E \\ \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\ \\ 
E^{c}=(4,5,6)$

Thus the set compliment of E will contain the elements {4,5,6}.So

$E^{c}$ = {4,5,6}

Part 2)

$P(E^{c})$ means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

$P(E^{c})$ = (Number of Elements in $E^{c}$ )/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set $E^{c}$ = $n(E^{c})$ =3

So,

$P(E^{c})= \frac{n(E^{c}) }{n(S)} \\ \\ 
P(E^{c})= \frac{3}{12} \\ \\ 
P(E^{c})= \frac{1}{4}$

7 0

3 years ago

In the ASA Congruence Postulate, the side between the angles is called the ___ side.

rusak2 [61]

No English plees enpanio

6 0

2 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

21.

jasenka [17]

Answer:

b = y intercept

Step-by-step explanation:

y = mx +b

This is the slope intercept form of a line

The m is the slope and the b is the y intercept

8 0

2 years ago