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vichka [17]
3 years ago
11

° ~ Calculate the percentage ~ ° Help please !! With T_T operations

Mathematics
1 answer:
natulia [17]3 years ago
4 0

Step-by-step explanation:

please re ask question. its unvalid and please simplfy it

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A package of 25 wristbands cost $5.25.At this rate,what is the cost of 1 wristband in dollars and cents?
Pani-rosa [81]
$5.25 divided by 25 = .21 cents each
6 0
3 years ago
Help me with this urgent due in 5 min​
solniwko [45]

Answer:

The last option

Step-by-step explanation:

4 0
2 years ago
What is the value of x in the diagram?
dalvyx [7]

Answer:

x = 30

Step-by-step explanation:

here 50 is hypotenuse as it is opposite of 90 degree.

x and x + 10 are the two other smaller sides of a right angled triangle respectively.

using pythagoras theorem,

a^2 + b^2 = c^2

x^2 + (x + 10)^2 = 50^2

x^2 + x^2 + 20x + 100 = 2500

2x^2 + 20x + 100 = 2500

2x^2 + 20x + 100 - 2500 = 0

2x^2 + 20x - 2400 = 0

2(x^2 + 10x - 1200) = 0

x^2 + 10x - 1200 = 0

x^2 + (40 - 30) - 1200 = 0

x^2 + 40x - 30x - 1200 = 0

x(x + 40) - 30(x + 40x) = 0

(x + 40)(x - 30) = 0

either x + 40 = 0          OR x - 30 = 0

x = 0 - 40

x = -40

x - 30 = 0

x = 30

x = -40,30

since the length and distance is not measured in negative ur answer will be 30

credit goes to sreedevi102

thank u very much . At first i was wrong and giannathecookie i m really sorry

3 0
3 years ago
Read 2 more answers
Use a number line to name two numbers that are the same distance apart as 3.2 and 4.4.
Nataliya [291]

Answer:

3.3 and 4.5

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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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