6= 2(y+2) i need help on this

daren has 10 sweets .hellen has 30 sweets.faith has 4 fewer sweets than the average number of sweets daren,hellenand faith have.

Sunny_sXe [5.5K]

Answer:

16

Step-by-step explanation:

Average = the sum of all numbers divided by the amount of items in the set, so:

10+30=40

40/2=20

20-4=16

Hopefully this helps :) Good luck!

8 0

3 years ago

I givee brainlilsttt

shutvik [7]

Answer:

( 2, 1 )

Step-by-step explanation:

x remains the same and y coordinate changes signs

4 0

3 years ago

Read 2 more answers

Refer to the accompanying TI-83/84 Plus calculator display of a 95% confidence interval. The sample display results from using a

nikklg [1K]

Answer:

The confidence interval is $22.805$

Step-by-step explanation:

We have given,

The Z interval (23.305,25.075)

Mean $\bar{x}=23.69$

Sample n=30

To find : The confidence interval?

Solution :

We know, The confidence interval is in the format of

$\bar{x}-E$

E denotes the margin of error,

$E=\frac{U-L}{2}$

Where, U is the upper limit U=25.075

L is the lower limit L=23.305

$E=\frac{25.075-23.305}{2}$

$E=\frac{1.77}{2}$

$E=0.885$

Substitute the value of E and $\bar{x}$ in the formula,

$23.69-0.885$

$22.805$

Therefore, The confidence interval is $22.805$

8 0

3 years ago

Plsss hurry i need help brainliest involed

dybincka [34]

45.5 is your answer, you will have to multiple the height times the base ( 5 x 9.1 )

7 0

3 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago