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IceJOKER [234]
3 years ago
12

Given: <2 and <3 form a linear pair, <1 and <3 are vertical angles, and m<1,m<4, and m<5 form a straight an

gle

Mathematics
1 answer:
topjm [15]3 years ago
8 0

Attached is the drawing that shows that

  • 1, 4, and 5 form a straight angle (straight line)
  • 2 and 3 form a linear pair (next to each other and create a straight line)
  • 1 and 3 are vertical angles (opposite sides of the X)

You might be interested in
Help and show your work
OLga [1]

Answer:

-3=k y=24

Step-by-step explanation:

y=kx.

-18=k(6)

k=-3.

So our value of k is -3 now, let solve for y.

y=kx

y=(-3)(-8)

y=24

3 0
3 years ago
2.3=7.9+y how do I solve
riadik2000 [5.3K]
Simplify both sides
2.3=y+7.9
flip the equation.
y+7.9=2.3
Subtract 7.9 from both sides.
y+7.9-7.9=2.3-7.9
y=-5.6
hope this helps!
5 0
4 years ago
If 2x/3+x/2=1/6, then x=?
Svetradugi [14.3K]
May I get brainlist? Answer: x= 1/7
7 0
3 years ago
Researchers obtained a sample of n = 10 children who attended daycare before starting school. The children are given a standardi
attashe74 [19]

Answer:

t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093  

p_v =2*P(t_{9}>4.093)=0.00084  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.  

We can say that at 5% of significance the true mean is significantly different from 50.

Step-by-step explanation:

Data given and notation  

53, 57, 61, 49, 52, 56, 58, 62, 51, 56

We can calculate the sample mean and deviation with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X))^2}{n-1}}

And we got:

\bar X=55.5 represent the sample mean  

s=4.249 represent the sample standard deviation  

n=10 sample size  

\mu_o =50/tex] represent the value that we want to test  [tex]\alpha=0.05 represent the significance level for the hypothesis test (ASSUMED).  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is equal or not to 50 :  

Null hypothesis:\mu = 50  

Alternative hypothesis:\mu \neq 50  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093  

P-value  

We need to calculate the degrees of freedom first given by:  

df=n-1=10-1=9  

Since is a two-sided tailed test the p value would given by:  

p_v =2*P(t_{9}>4.093)=0.00084  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.  

We can say that at 5% of significance the true mean is significantly different from 50.

6 0
4 years ago
What is 17% of 700 cm =
Gnoma [55]

17% of 700 cm is 1.19 meters.


8 0
3 years ago
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