PLEASE HELP

Ariel has 20 bouncy balls. He later then sold them for a price of $2 per ball to his best friend, Mike Hawk. Later, Ariel has go

ValentinkaMS [17]

Ariel has 40 balls left

6 0

2 years ago

Factories 5x+10 <br> Somebody please help

jolli1 [7]

Answer:

5(x+2)

Step-by-step explanation:

to factor it we need to find a common term(5) and then we take it out(divide) resulting in the answer abover

5 0

2 years ago

Find the range of the function f(x)=(x-2)^2+2

kari74 [83]

The function is in vertex form of a quadratic function.
Vertex form is f(x)=a(x-h)^2+k
The vertex is (h, k)

In this case, our vertex is (2, 2). The a value is positive which means that the graph points up. Since the graph points up, the vertex is the minimum, the lowest point of the graph. The range refers to y-values, so the range of the function is y/y>2. It can also be written as 2≤y<span>≤positive infinity.</span>

8 0

3 years ago

Read 2 more answers

Ms. Monroe mixes 200 cubes of 4 different colors for a project. She then uses a scoop to place 20 cubes into each of 10 paper ba

Scilla [17]

Answer:

The mean no. of red cubes is more than the median no. of red cubes.

Step-by-step explanation:

According to the question, we can prepare the following table denoting the no. of bags containing any number of red cube(s). The no. of red cubes in each bag can theoretically be any no. between 0 and 20 (both end inclusive) as in total each bag contains 20 cubes of different colors.There are as a whole 10 bags (given).

No. of red cubes No. of bags $(x_{i}f_{i})$

$(x_{i})$ $(f_{i})$

0 2 0

1 0 0

2 2 4

3 1 3

4 1 4

5 0 0

6 1 6

7 2 14

8 1 8

9 0 0

10 0 0

Total 10 39

Now, no. of bags containing no. of red cubes above 10 is 0 [since there are as a whole 10 bags only]

So, from the table we get that mean no, of red cubes in each bag

= $\frac {\sum_{i =1}^{10}x_{i}f_{i}}{\sum_{i=1}^{10}f_{i}}$

= 39/10 = 3.9

Also, from the table, we get that, no. of bags with no. of red cubes ≤ 3

= (2 + 2 + 1) = 5

and, no. of bags with no. of red cubes ≥ 4

= (1 + 1 + 2 + 1) = 5

Also there are 1 bag each for no. of red cubes = 3 and 4

So. median no. of red cubes = (3 + 4)/2 = 3.5

So, we get that the mean no. of red cubes is more than the median no. of red cubes.

5 0

3 years ago

Read 2 more answers

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >5.860)=0.32$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

4 0

3 years ago