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3 years ago

ALGEBRA By bundling cellular, land line, high-speed Internet

Mathematics

1 answer:

Kruka [31]3 years ago

6 0

<h2> My Answer:</h2>

1 303.3$

Okay, I'm not 100% sure it's this because I suck at Math, but I'm 80% sure. I hope you do good on your assignment or test! ^^

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In ΔRST, s = 58 inches, ∠R=115° and ∠S=19°. Find the length of r, to the nearest 10th of an inch.

Lorico [155]

Answer:

365.9 Inches

Step-by-step explanation:

To find the value of Sin A can be found using the sine rule. the rule states that the ratio of the length facing an angle in a triangle to the sine of the angle is constant. As such,

a/Sin A = b/Sin B = c/Sin C

Where is a is the side facing angle A, b is the side facing angle B and c is the side facing angle C.

Hence,

58/Sin 19 = r/Sin 115

r = 58 Sin 115/ Sin 19

= 365.9 Inches

8 0

3 years ago

Yellow Cab Taxi charges a $1.75 flat rate in addition to $.55 per mile. Katie has no more than $10 to spend on a ride. How many

katrin2010 [14]

Take $1.75 from the $10 then dived the 55cents into the sum. Thus $10 - $1.75 = $8.25. / .55 =15 miles exactly

5 0

3 years ago

Please help with this math problem ASAP! I will give brainliest!

borishaifa [10]

x = 112°
y = 68°
z = 68°

Step-by-step explanation:

x = 112°

y = z = 180° - 112° = 68°

7 0

2 years ago

What linear function can be represented by the set of ordered pairs?

Kryger [21]

Y = 3x - 7

just find the slope then use two point and point slope form to solve for slope intercept form

8 0

2 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis 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

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

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

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 $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago