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Aliun [14]
3 years ago
8

Find the area of the circle with the given radius or diameter. Use pi = 3.14. r = 6

Mathematics
2 answers:
Step2247 [10]3 years ago
6 0

Answer:

113.04

Step-by-step explanation:


svetoff [14.1K]3 years ago
3 0
The answer is 113.1. Hope this helps
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I need to find n and m
Cloud [144]

Answer:

n=6, m=6√3

Step-by-step explanation:

n = Opposite leg = sin(30°)*hypotenuse = (1/2)*12 =6

m= adjacent leg = cos(30°)*hypotenuse =((√3)/2)*12=6√3

3 0
3 years ago
In triangle XYZ, shown above, what is the value of cos(X)?
adell [148]

Answer:

A

Step-by-step explanation:

Sine we can figure out side XY by using Pythagorean Theorem (A^2+B^2=C^2) and find that XY= 5, We can use Soh Cah Toa (The ways I remember to use value charts) and we know that Cosine uses Adjacent over Hypotenuse in which therefore the answer would be 5/13

8 0
3 years ago
What are the domain and range of the function 2 y x x    6 8 shown in the graph below?
Nat2105 [25]

Answer:

Missing information.

Step-by-step explanation:

However, the domain is all the x- values. The range is all the y-values (that do not repeat).

So when you look at your graph, find your x values and y-values and write them as set. That is your answer.

7 0
3 years ago
Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies ac
Shtirlitz [24]

Answer:

a) \mu -3\sigma = 336-3*6=318

\mu+-3\sigma = 336+3*6=354

b) For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%

And then the % above would be 100-84= 16%

Step-by-step explanation:

Assuming this question : "Bigger animals tend to carry their young longer before birth. The  length of horse pregnancies from conception to birth varies according to a roughly normal distribution with  mean 336 days and standard deviation 6 days. Use the 68-95-99.7 rule to answer the following questions. "

(a) Almost all (99.7%) horse pregnancies fall in what range of lengths?

First we need to remember the concept of empirical rule.

From this case we assume that X\sim N(\mu = 336. \sigma =6) where X represent the random variable "length of horse pregnancies from conception to birth"

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

From the empirical rule we know that we have 99.7% of the data within 3 deviations from the mean so then we can find the limits for this case with this:

\mu -3\sigma = 336-3*6=318

\mu+-3\sigma = 336+3*6=354

(b) What percent of horse pregnancies are longer than  342 days?

For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%

And then the % above would be 100-84= 16%

7 0
3 years ago
X^2- 4x- 12 = 0 <br>Solve by factoring and solve by completing the square​
DiKsa [7]

Use the formula (b/2)^2 in order to create a new term. Solve for by using this term to complete the square.

Cannot solve by completing the square.

8 0
3 years ago
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