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mixas84 [53]
3 years ago
11

What is the area of the shaded segment?

Mathematics
1 answer:
juin [17]3 years ago
5 0

Answer:

Watch

Step-by-step explanation:

Home.

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45-8÷(4-2)×9+10 how do you do this
kati45 [8]
Do PEMDAS 

so do 4-2 which is 2

2 x 9 is 18 x 10 is 180

45 - 8 is 37 

180 divided by 37 is 4.9
6 0
4 years ago
Read 2 more answers
A line passes through the point (-8, 7) and has a slope of negative 3/4
yanalaym [24]
Y = mx + b
slope(m) = 3/4
(-8,7)...x = -8 and y = 7
now we sub and solve for b, the y int
7 = 3/4(-8) + b
7 = - 6 + b
7 + 6 = b
13 = b

so ur line is : y = 3/4x + 13

3 0
3 years ago
. You are interested in the amount of time teenagers spend weekly working at part-time jobs. A random sample of 15 teenagers was
agasfer [191]

Answer:

t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115    

Now we can calculate the degrees of freedom

df=n-1=15-1=14  

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got t_{critc}= 1.761

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

Step-by-step explanation:

Information given

\bar X=147.3 represent the sample mean for the amount of time spent at part time jobs

s=50 represent the sample standard deviation

n=15 sample size  

\mu_o =120 represent the value to check

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value for the test

System of hypothesis

We want to analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:  

Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

Since we don't know the population deviation the statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115    

Now we can calculate the degrees of freedom

df=n-1=15-1=14  

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got t_{critc}= 1.761

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

4 0
3 years ago
Can anyone answer this
aniked [119]

Answer:

108ft²

Step-by-step explanation:

6×6=36

12×12=144

144-36=108ft for the shaded part

5 0
3 years ago
Read 2 more answers
Solve
polet [3.4K]

Answer:

3

4

5

9

2

Step-by-step explanation:

4 0
3 years ago
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