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Anettt [7]
3 years ago
13

Will give brainliest! What is the range for the third size of the triangle 2,10?

Mathematics
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

8 < x < 12

Step-by-step explanation:

Given 2 sides of a triangle then the third side x is in the range

difference of sides < x < sum of sides , that is

10 - 2 < x < 10 + 2

8 < x < 12

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PLEASE HELP ASAP!!!
xeze [42]

Answer:

JM = 10

Step-by-step explanation:

Because MN║JL,

\frac{MK}{JM} = \frac{KN}{NL}

\frac{5}{2x + 4}  = \frac{6}{12}  =  \frac{1}{2}

2(5) = 1(2x + 4)

10 = 2x + 4

2x = 6

x = 3

JM = 2x + 4 = 2(3) + 4 = 6 + 4 = 10

6 0
3 years ago
Help Me With This Show Work If You Want
puteri [66]
Use the Pythagorean therm , you know the a^2+b^2=c^2 
8 0
3 years ago
Consider the following hypothesis test H0 p 20 Ha p 20 A sample of 400 provided a sample proportion p 175 a Compute the value of
Bogdan [553]

Answer:

a) -1.25

b) 0.2112

c) -1.96

Step-by-step explanation:

Data provided in the question:

Sample size, n = 400

H0 : p = 20

\bar{p} = 175

Now,

a) The test statistic is  given as:

Z = \frac{(\bar{p}-p)}{\sqrt{\frac{p(1-p)}{n}}}

on substituting the respective values, we get

Z = \frac{(0.175-0.2)}{\sqrt{\frac{0.2\times0.8}{400}}}

= -1.25

b) The p-value = 2 × P(Z <-1.25)

Now from the standard normal table

P(Z <-1.25) = 10.56% = 0.1056

Thus,

p-value = 2 × 1056 = 0.2112

c) for a = 0.05,

the critical value is Z_{\frac{a}{2}}=Z_{\frac{0.05}{2}} i.e Z_{0.025}

Now from standard normal table

Z_{0.025} = -1.96

3 0
3 years ago
2(2t+4)=34(24−8t)<br><br> solve for t
goldfiish [28.3K]
2(2t + 4) = 34(24 − 8t)
4t + 8 = 816 - 272t
4t + 272t = 816 - 8
276t = 808
t = 808/276
= 2.93 (to three significant figures)

The value of t is 2.93.
8 0
3 years ago
Read 2 more answers
Picture question middle school math
ludmilkaskok [199]

Answer:

<h2>\huge \: b) \: 59 \: i {n}^{2}</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about</h3>
  • area of triangles
  • area of rectangle
  • PEMDAS
<h3>tips and formulas:</h3>
  • rectangle opposite sides are equal
  • area of rectangle:l×w
  • area of triangle:½×h×b
<h3>let's solve:</h3>

the area of the rectangle is

4×7 in²

28 in²

the area of above triangle is

½×6×7 in²

3×7 in²

21 in²

the area of left side triangle is

½×5×4 in²

2×5 in²

10 in²

therefore,

the area of the polygon is

28+21+10 in²

<h3>59 in²</h3>

6 0
3 years ago
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