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RideAnS [48]
3 years ago
5

Tell whether the lines through the given points are parallel,perpendicular or neither line 1 (2, 3) and (4, 12) line 2 (5, 10) a

nd (14, 8)
Mathematics
1 answer:
slamgirl [31]3 years ago
3 0
Perpendicular , I hope I’m right
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farmer weighed 20 dairy goats in pounds. Of the goats, 10 were males and 10 were females. Males Females First Quartile 29 21 Sec
aliina [53]

Answer:

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Step-by-step explanation:

4 0
3 years ago
NO LINKS!! Part 2<br><br>Similarity Flow Chart Proofs​
STatiana [176]

Answer:

  • See below

Step-by-step explanation:

<h3>#4</h3>

According to diagram we have

  • QR ≅ QT
  • QS ≅ QS (common side)
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Considering above we can state

  • QSR ≅ QST by HL (hypotenuse-leg)
<h3>#5</h3>

Two angles are congruent but the order of angles is not same

Triangles are not similar.

<h3>#6</h3>

Two angles and a side are congruent but the order of angles is not same.

Triangles are not similar.

4 0
3 years ago
How do you simplify the radical: square root of 28
Lilit [14]

Answer:

√

28

Rewrite

28

as

2

2

⋅

7

.

Tap for more steps...

√

2

2

⋅

7

Pull terms out from under the radical.

2

√

7

The result can be shown in both exact and decimal forms.

Exact Form:

2

√

7

Decimal Form:

5.29150262

…

Step-by-step explanation:


3 0
3 years ago
Can someone help me find y PLEASE
Maurinko [17]
pretty sure it’s 42
4 0
3 years ago
Read 2 more answers
Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. ​(a) Descri
OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

\mu = 74

and

\sigma = 6

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

\mu = 74

The standard error of the means becomes the standard deviation of the sampling distribution.

\sigma_ { \bar X }  =  \frac{ \sigma}{ \sqrt{n} }  \\ \sigma_ { \bar X }  =  \frac{ 6}{ \sqrt{36} }  = 1

Part b) We want to find

P(\bar X \:>\:75.9)

We need to convert to z-score.

P(\bar X \:>\:75.9)  = P(z \:>\: \frac{75.9 - 74}{1} )  \\  = P(z \:>\: \frac{75.9 - 74}{1} ) \\  = P(z \:>\: 1.9) \\  = 0.0287

Part c)

We want to find

P(\bar X \: < \:71.95)

We convert to z-score and use the normal distribution table to find the corresponding area.

P(\bar X \: < \:71.95)  = P(z \: < \: \frac{71.9 5- 74}{1} )  \\  = P(z \: < \: \frac{71.9 5- 74}{1} ) \\  = P(z \: < \:  - 2.05) \\  = 0.0202

Part d)

We want to find :

P(73\:

We convert to z-scores and again use the standard normal distribution table.

P( \frac{73 - 74}{1} \:< \: z

5 0
3 years ago
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