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kompoz [17]
3 years ago
14

Tina works 15 hours a week (Monday to Friday). Last week she worked 3 ½ hours on Monday, 4 hours on Tuesday, 2 ⅙ hours on Wednes

day, and 1 ½ on Thursday. How many hours did she work on Friday?
Mathematics
1 answer:
algol133 years ago
6 0
She worked four hours and thirty minutes on friday
You might be interested in
Find an equation of the line that passes through the points (-0.5,0.75) and (0.75,-0.5)
Keith_Richards [23]

The equation of the line can be shown using the slope intercept form that is

y= mx+b

where m is the slope and b is the y intercept

So firstly we have to find the value of slope m

To find the value of slope m we apply the formula

m= \frac{( y_{2} - y_{1})}{( x_{2} - x_{1})}

So we can plug the values using the two given points

(-0.5,0.75) and (0.75,-0.5)

m= \frac{(-0.5-0.75) }{(0.75-(-0.5))} = \frac{-1.25}{1.25} = -1

So m = -1

Plug m=-1 in the equation we get

y= -1x +b

Now we have to find the value of b

to find the value of b , we plug any one pint in the equation

0.75 = -1(-0.5) +b

0.75 = 0.5 +b

Subtract 0.5 from both sides

0.25 = b

or

b= 0.25

Now plug the value of b , in the equation

we get

y = -1x +0.25

hence the equation of the line is y = -1x +0.25

8 0
4 years ago
{(-4,8),(-2,4),(0,1),(2,4),(4,8)}
astraxan [27]

Answer:

Since there is one value of y for every value of x in (−4,8),(−2,4),(0,1),(2,4), ( 4,8), this relation is a function. The relation is a function.

5 0
3 years ago
(06.04)
otez555 [7]

Answer:

5x/3y2

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
URGENT: WILL MARK BRAINLIEST
saveliy_v [14]

Answer:

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5 0
4 years ago
Read 2 more answers
A random sample of 49 statistics examinations was taken. the average score, in the sample, was 84 with a variance of 12.25. the
tester [92]

Since in this case we are only using the variance of the sample and not the variance of the real population, therefore we use the t statistic. The formula for the confidence interval is:

<span>CI = X ± t * s / sqrt(n)                      ---> 1</span>

Where,

X = the sample mean = 84

t = the t score which is obtained in the standard distribution tables at 95% confidence level

s = sample variance = 12.25

n = number of samples = 49

From the table at 95% confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t is around:

t = 1.68

 

Therefore substituting the given values to equation 1:

CI = 84 ± 1.68 * 12.25 / sqrt(49)

CI = 84 ± 2.94

CI = 81.06, 86.94

 

<span>Therefore at 95% confidence level, the scores is from 81 to 87.</span>

6 0
4 years ago
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