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schepotkina [342]
4 years ago
11

Find The area of the rectangle below

Mathematics
1 answer:
Natasha_Volkova [10]4 years ago
7 0

Answer:

4r^2-14r-30

Step-by-step explanation:

Multiply the binomials

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Solve: -4x &lt; 20 <br> x &lt; 5<br> x &gt; 5<br> x &lt; -5<br> x &gt; -5
Thepotemich [5.8K]
The answer is x > -5

8 0
3 years ago
Read 2 more answers
Using traditional methods, it takes 10.1 hours to receive a basic driving license. A new license training method using Computer
Law Incorporation [45]

Answer:

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:  

Null hypothesis:\mu =10.1  

Alternative hypothesis:\mu \neq 10.1

t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990    

p_v =2*P(t_{(11)}>0.990)=0.343    

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL  reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=10.5 represent the sample mean

s=1.4 represent the sample standard deviation

n=12 sample size  

\mu_o =10.1 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

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 technique performs differently than the traditional method, the system of hypothesis would be:  

Null hypothesis:\mu =10.1  

Alternative hypothesis:\mu \neq 10.1  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=12-1=11  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(11)}>0.990)=0.343  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL  reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.  

7 0
4 years ago
1. Use the figure below to answer the questions.<br><br> (a) Solve for x.
Dmitry [639]

Let's see what to do buddy...

_________________________________

The sum of the angles of each n-sided figure is found in the following equation :

(n - 2) \times 180

The question's figure is 5-sindes figure

so the sum the angles equal :

(5 - 2) \times 180 = 3 \times 180 = 540 \\

So we have :

90 + 107 + 144 + 138 + x = 54 0 \\

479 + x = 540

Subtract the sides of the equation minus 479

x = 540 - 479

x = 61

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

5 0
3 years ago
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nydimaria [60]
Answer is 0= -2

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8 0
2 years ago
Read 2 more answers
Someone please help me with this!! i really need this to pass :(
DochEvi [55]

Answer:

Option :" Use distance formula to prove that the lengths of the diagonals are equal" is correct

Step-by-step explanation:

Option : Use distance formula to prove that the lengths of the diagonals are equal" is correct.

Because " By using the coordinate geometry to prove that the diagonals of the rectangle are congruent, first we have to find the lengths of the top and bottom of the rectangle and then solve it for the lengths of the diagonals by using the distance formula".

3 0
3 years ago
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