Please answer this thanks

Evaluate each expression, m= 4, x= 9, and r= 1/6

bagirrra123 [75]

4. 14
5. 6
6. 1/12
These are the answers to the questions

8 0

3 years ago

Ɪ ᴅᴏɴ'ᴛ ɢᴇᴛ ᴛʜɪs ᴄᴀɴ sᴏᴍᴇʙᴏᴅʏ ʜᴇʟᴘ ᴍᴇ ᴏɴ ᴛʜɪs ᴘʀᴏʙʟᴇᴍ ʀᴇᴀʟʟʏ ǫᴜɪᴄᴋ

Paraphin [41]

The answer would be reading

7 0

3 years ago

Read 2 more answers

The Arc Electronic Company had an income of 90 million dollars last year. Suppose the mean income of firms in the same industry

jok3333 [9.3K]

Answer:

0.0869

Step-by-step explanation:

The arc electronic company had an income of 90 million dollars last year.

Mean(μ) = 75 million dollars

Standard deviation (σ) = 11 million dollars

Probability that the randomly selected will earn more than arc did last year = Pr(x>90)

Using normal distribution,

Z = (x - μ) / σ

Z = (90 - 75) / 11

Z = 15/11

Z = 1.36

From the normal distribution table, 1.36 = 0.4131

Φ(z) = 0.4131

Recall that when Z is positive, Pr(x>a) = 0.5 - Φ(z)

= 0.5 - 0.4131

= 0.0869

8 0

3 years ago

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa

xxMikexx [17]

Answer:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

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

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

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

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{50*189}{225}=42$

$E_{2} =\frac{50*36}{225}=8$

$E_{3} =\frac{75*189}{225}=63$

$E_{4} =\frac{75*36}{225}=12$

$E_{5} =\frac{100*189}{225}=84$

$E_{6} =\frac{100*36}{225}=16$

And the expected values are given by:

Size Company/ Heal. Ins. Yes No Total

Small 42 8 50

Medium 63 12 75

Large 84 16 100

_____________________________________

Total 189 36 225

And now we can calculate the statistic:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

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

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0

3 years ago

Plz i need this asap, if you do not know all of them, plz do not answer

Lelechka [254]

1. sqrt 52 = 7.211 rounds to 7

2. irrational number

3. sqrt 441 = 21....rational

4. 12^2 + 3^2 = h^2
144 + 9 = h^2
153 = h^2
sqrt 153 = h
12.4 = h <===

5. 6^2 + b^2 = 18^2
36 + b^2 = 324
b^2 = 324 - 36
b^2 = 288
b = sqrt 288
b = 16.97 rounds to 17 <==

6. 39^2 + 52^2 = c^2
1521 + 2704 = c^2
4225 = c^2
sqrt 4225 = c
65 = c
(39 + 52) - 65 = 91 - 65 = 26 miles shorter <==

7. 4^2 + b^2 = 16^2
16 + b^2 = 256
b^2 = 256 - 16
b^2 = 240
b = sqrt 240
b = 15.49 rounds to 15.5 <==

8. ?

8 0

3 years ago