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3 years ago

What is the perimeter of rhombus ABCD? 10 units 20 units units units

2 answers:

3 0

Answer:
20 units
Step-by-step explanation:
We are given that
On a coordinate plane , rhombus ABCD in which point A is at (1,1) and point B is at (-2,-3), point C is at (-5,1) and D is at (-2,5).
We have to find the perimeter of rhombus ABCD.
Distance formula :
Using the distance formula
Side AB=
We know that sides of rhombus are equal
Therefore, AB=BC=CD=AD=5 units
Perimeter of rhombus=
Perimeter of rhombus=
Answer: 20 units

Read more on Brainly.com - brainly.com/question/13829807#readmore

andrey2020 [161]3 years ago

3 0

Answer: B on edg 2020

Step-by-step explanation:

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Write 10 5/12 as an equevilent improper fraction

dexar [7]

10 5/12 can be turned into a improper fraction by multiplying 10 and 12. You get 120. Then you can add 5 and get 125/12.

THE ANSWER IS 125/12!

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4 years ago

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The image shows parallel lines cut by a transversal. The expressions represent unknown angle measurements. What is the value of

Triss [41]

Answer) C. 77
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4 years ago

PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST

ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

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3 years ago

Please help with the question in the picture!

Stella [2.4K]

Answer:

Tan C = 3/4

Step-by-step explanation:

Given-

∠ A = 90°, sin C = 3 / 5

METHOD - I

Sin² C + Cos² C = 1