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LenKa [72]
3 years ago
15

How would i find y i need helppp

Mathematics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

C

Step-by-step explanation:

sin30° = y/6

sin30° = 1/2

y/6 =  1/2

y = 3

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Math homework plz help​
uysha [10]

Answer:

1/20

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Translate the phrase into a variable expression: three less than the product of a number and six
CaHeK987 [17]
I think it will be C. 

hope its correct ! 
8 0
3 years ago
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What is 8+[4+(-9)] ?
fgiga [73]

Answer:

Positive 3

Step-by-step explanation:

you have to add 4 and -9 first and you get -5 then you add that to positive 8 and get 3

7 0
3 years ago
(05.08A)Triangle ABC is transformed to similar triangle A′B′C′ below:
Romashka-Z-Leto [24]

Answer:

1st option is correct i.e., 1 over 2

Step-by-step explanation:

Given that Triangle ABC transformed to Triangle A'B'C'.

Coordinated of Triangle ABC are A(2,6) , B(2,4) and C(4,4)

Coordinated of Triangle A'B'C' are A'(1,3), B'(1,2) and C'(2,2)

Scale factor of Dilation is the no. of times coordinates of vertices of a figure increase or decrease.i.e.,

if A(a,b) ⇒ A'(c,d)

where, a.k=c & b.k=d then k is scale factor of dilation.

Value of k is same for all vertices.

For coordinates of A & A' ,we have 2.k=1\implies k=\frac{1}{2} \:\: and \:\: 6.k=3\implies k=\frac{1}{2}

For coordinates of B & B' ,we have 2.k=1\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}

For coordinates of C & C' ,we have 4.k=2\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}

Therefore, the scale factor of dilation is \:\frac{1}{2}\:  since in all cases k=\frac{1}{2}

1st option is correct i.e., 1 over 2

3 0
3 years ago
Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin
Artemon [7]
"<span>Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find \bar{d}, s_{d}, the t-test statistic, and the critical values to test the claim that \mu_{d} = 0"

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find </span><span>\bar{d}
You are asked to find the mean difference between the two variables, which is given by the formula:
\bar{d} =  \frac{\sum (x - y)}{n}

These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example: 
</span><span>\bar{d} =  \frac{6}{8} = 0.75</span>

B) Find <span>s_{d}
</span><span>You are asked to find the standard deviation, which is given by the formula:
</span>s_{d} =  \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }

These are the steps to follow:
1) Subtract the mean difference from each pair's difference 
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:
s_{d} = \sqrt{ \frac{101.5}{8-1} }
= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:
t =  \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}

where SE = standard error is given by the formula:
SE =  \frac{ s_{d} }{ \sqrt{n} }

These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract  \mu_{d}
6) divide by the standard error

In our example:
SE = 3.81 / √8
      = 1.346

The problem gives us <span>\mu_{d} = 0, therefore:
t = [(9.75 - 9) - 0] / 1.346</span>
  = 0.56

D) Find t_{\alpha / 2}
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find <span>t_{\alpha / 2} = 1.895</span>

Since our t-value is less than <span>t_{\alpha / 2}</span> we can reject our null hypothesis!!

7 0
3 years ago
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