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Dennis_Churaev [7]
3 years ago
11

Solve three-fourths b minus one-sixth equals one-half for b

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
7 0
3/4b-1/6=1/2×b
1.75b=.3
b=0.2
I believe this is right. But not 100%
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Look at the figure. Name the postulate or theorem you can use to prove the train goes congruent?
Nikolay [14]

Let us recall parallelogram properties, which states that opposite angles of parallelogram are congruent.

We can see from graph that side US is parallel to TR and measure of angle U equals to measure of angle R, therefore, quadrilateral drawn in our given graph is a parallelogram.

Since we know that opposite sides of parallelogram are congruent. In our parallelogram UT=SR and US=TR.

In our triangle STU and triangle TSR side TS=TS by reflexive property of congruence.

Therefore, our triangles are congruent by SSS congruence.        

6 0
3 years ago
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What is the area of the trapezoid?
DanielleElmas [232]

Answer:

C) 100 m^2

Step-by-step explanation:

Area of a trapezoid= h*[(b1+b2)/2]

h=10

b1=8

b2=12

Therefore:

A=10*[(8+12)/2]

A=10*(20/2)

A=10*10

A=100

So the area of the trapezoid is 100 m^2

7 0
2 years ago
The scores of the first 6 competitors in a
Kipish [7]

the Answer is -1 think

8 0
2 years ago
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BRAINLIEST***
ANTONII [103]

Answer:

C, ON/MN

Step-by-step explanation:

Since tan is opposite/adjacent, and the opposite of angle M is NO, then NO must be the numerator. The only answer with this as the numerator is C.

7 0
3 years ago
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A hockey team is convinced that the coin used to determine the order of play is weighted. The team captain steals this special c
fredd [130]

Answer:

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

Step-by-step explanation:

Let p be the probability of heads in a single toss of the coin. Then our null hypothesis that the coin is fair will be formulated as

H0 :p 0.5   against   Ha: p ≠ 0.5

The significance level is approximately 0.05

The test statistic to be used is number of heads x.

Critical Region: First we compute the probabilities associated with X the number of heads using the binomial distribution

Heads (x)        Probability (X=x)                        Cumulative     Decumulative

0                        1/16384 (1)             0.000061     0.000061

1                         1/16384  (14)         0.00085             0.000911

2                       1/16384 (91)           0.00555             0.006461

3                       1/16384(364)         0.02222

4                       1/16384(1001)         0.0611

5                       1/16384(2002)       0.122188

6                        1/16384(3003)      0.1833

7                         1/16384(3432)      0.2095

8                        1/16384(3003)       0.1833

9                        1/16384(2002)       0.122188

10                       1/16384(1001)        0.0611

11                       1/16384(364)        0.02222

12                      1/16384(91)            0.00555                             0.006461

13                     1/16384(14)              0.00085                           0.000911

14                       1/16384(1)            0.000061                            0.000061

We use the cumulative and decumulative column as the critical region is composed of two portions of area ( probability) one in each tail of the distribution. If  alpha = 0.05 then alpha by 2 - 0.025 ( area in each tail).

We observe that P (X≤2) =   0.006461 > 0.025

and

P ( X≥12 ) = 0.006461 > 0.025

Therefore true significance level is

∝=  P (X≤0)+P ( X≥14 ) = 0.000061+0.000061= 0.000122

Hence critical region is (X≤0) and ( X≥14)

Computation x= 12

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

3 0
3 years ago
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