Round 508.0219 to the nearest hundredth

Three out of seven students in the cafeteria line are chosen to answer survey questions. How many different combinations of thre

Hatshy [7]

Answer:

35 possible combinations would be there.

Step-by-step explanation:

We have total of seven students we have to choose three out of those seven

$^{n}C_r=\frac{n!}{r!\cdot (n-r)!}$

n is the total possibilities n =7 in the given case

And r is the chosen ones

Hence, r=3 in the given case.

Substituting the values of n and r in the formula we get

$^{7}C_3$

$\frac{7!}{3!(7-3)!}$

$\frac{7!}{3!\cdot 4!}$

Using n!=n(n-1)----1

$\frac{7\cdot 6\cdot 5\cdot 4!}{3\cdot 2\cdot 1\cdot 4!}$

Cancel the common factor we get

$35$ possible combinations would be there.

8 0

3 years ago

I only get one part of this problem ?!

Otrada [13]

Answer:

Light Blue

Step-by-step explanation:

The first thing to do is to take 1 point on the graph. Lets say the bottom left point. You would move down the number of spaces to the same point.

It is 8 units down. Then, move however many units to the right. In this case, it is five units right, so the answer would be light blue.

8 0

3 years ago

What is the value of 6 (2b-4) When b=5

e-lub [12.9K]

Answer:

36

Step-by-step explanation:

Replace the 2b with 10 and then do the parenthesis, so 10-4=6

now multiply it with the 6 outside the parenthesis and you get 36

7 0

3 years ago

Read 2 more answers

The operations manager of a manufacturer of television remote controls wants to determine which batteries last the longest in hi

Margarita [4]

Answer:

$t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802$

The degrees of freedom are given by:

$df=n_{1}+n_{2}-2=4+4-2=6$

The p value for this case would be given by:

$p_v =2*P(t_{(6)}$

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

Step-by-step explanation:

Information given

Battery 1 106 111 109 105

Battery 2 125 103 121 118

We can calculate the mean and the deviation with the following formulas"

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

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

$\bar X_{1}=107.75$ represent the mean for the Battery 1

$\bar X_{2}=116.75$ represent the mean for the Bettery 2

$s_{1}=2.75$ represent the sample standard deviation for the Battery 1

$s_{2}=9.604$ represent the sample standard deviation for the battery 2

$n_{1}=4$ sample size selected for the Battery 1

$n_{2}=4$ sample size selected for the Battery 2

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to check if the difference in longevity between the two batteries, the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

The statistic is given by:

$t=\frac{\bar X_{s1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

The statistic is given by:

$t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802$

The degrees of freedom are given by:

$df=n_{1}+n_{2}-2=4+4-2=6$

The p value for this case would be given by:

$p_v =2*P(t_{(6)}$

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

8 0

3 years ago

Solve and graph this system of equations x+2y=5 and 3x=15-6y

Bogdan [553]

Answer:

(x,y)=(5-2y,y), y belongs to all real numbers

Step-by-step explanation:

x=5-2y

3x=15-6y

3(5-2y)=15-6y

15-6y=15-6y

y∈All Real Numbers

(x,y)=(5-2y,y), y∈All Real Numbers

3 0

4 years ago