Draw a 150 degree angle

I hate this pain i got through i just cant anymore... y+50=700

DiKsa [7]

Answer:

y=650

Step-by-step explanation:

Just subtract 50 from each side

And Yes, please

5 0

3 years ago

Read 2 more answers

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly sel

garri49 [273]

Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation $\sigma$ = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

$S.E = \frac{\sigma}{\sqrt{n}}$

$= \frac{2}{\sqrt{45}}$

$= \frac{2}{6.7082}$

$SE = 0.298$

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

$t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802$

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

5 0

3 years ago

What is 5 7/10 - 5/18

hoa [83]

The answer is 5 19/45.

3 0

3 years ago

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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

Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati

Brrunno [24]

Remember that an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.

Lets solve our equation to find out what is the extraneous solution:
 $\sqrt{x-3} =x-5$
$(\sqrt{x-3})^2 =(x-5)^2$
$x-3=x^2-10x+25$
$x^2-11x+28=0$
$(x-4)(x-7)=0$
$x-4=0$ and $x-7=0$
$x=4$ and $x=7$

So, the solutions of our equation are $x=4$ and $x=7$ . Lets replace each solution in our original equation to check if they are valid solutions:
- For $x=7$
$\sqrt{x-3} =x-5$
$\sqrt{7-3} =7-5$
$\sqrt{4} =2$
$2=2$
We can conclude that 7 is a valid solution of the equation.

- For $x=4$
$\sqrt{x-3} =x-5$
$\sqrt{4-3} =4-5$
$\sqrt{1} =1$
$1 \neq 1$
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: D. the extraneous solution is x = 4

7 0

2 years ago