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Feliz [49]
3 years ago
13

Find the lengths of all four sides : P (2,2), Q (1,-3),R (-4,2),S (-3,7)

Mathematics
1 answer:
Sveta_85 [38]3 years ago
4 0
PQ = sqrt [(-3-2)^2 + (1-2)^2] =  sqrt 26 =  5.1
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Can sides of a triangle have lengths 1, 13, and 14 ???
victus00 [196]

Answer: Yes it can!

Step-by-step explanation:

This is called a scalene triangle, where all 3 sides are different lengths.

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3 years ago
I’m not going on my own right path I
IrinaK [193]

Answer:

The answer is It is your opinion.

Step-by-step explanation:

thats your opinion .-.

7 0
3 years ago
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A glass of skim milk supplies 0.1 mg of iron, 8.5 g of protein, and 1 g of carbohydrates. A quarter pound of lean red meat provi
Tems11 [23]

Answer:

glasses of milk=4

quatar pounds servings of meat =5

two slices of whole grains bread =8

Step-by-step explanation:

Let m= milk

r=meat

b=bread

0.1m+3.4r+2.2b=35 (1)

8.5m+22r+10b=224 (2)

m+20r+12b=200 (3)

Multiply (1) by 85 and (2) by 1

8.5m+289r+187b=2,975 (4)

8.5m+22r+10b=224 (5)

Subtract (5) from (4)

267r+177b=2,751 (6)

Recall,

8.5m+22r+10b=224 (2)

m+20r+12b=200 (3)

Multiply (2) by 1 and (3) by 8.5

8.5m+22r+10b=224 (7)

8.5m+170r+102b=1700 (8)

Subtract (7) from (8)

148r+92b=1,476 (9)

Recall,

267r+177b=2,751 (6) and

148r+92b=1,476 (9)

Multiply (6) by 148 and (9) by 267

26196b=407,148 (10)

24564b=394,092 (11)

Subtract (11) from (10)

1,632b=13,056

Divide both sides by 1,632

b=13,056/1,632

b=8

Substitute the value of b in (9)

148r+92b=1,476 (9)

148r+92(8)=1,476

148r+736=1,476

148r=1,476-736

148r=740

Divide both sides by 148

r=740/148

r= 5

Substitute the value of b and r in (1)

0.1m+3.4r+2.2b=35 (1)

0.1m+3.4(5)+2.2(8)=35

0.1m+17+17.6=35

0.1m+34.6=35

0.1m=35-34.6

0.1m=0.4

Divide both sides by 0.1

m=0.4/0.1

m=4

4 glasses of milk, 5 quatar pounds servings of meat and 8 two slices of whole grains bread will supply the required nutrients.

8 0
3 years ago
17 - 19 please and thank you!
Nataly [62]
The answer and equation is 17 - 19 = -2
8 0
3 years ago
Read 2 more answers
The USDA conducted tests for salmonella in produce grown in California. In an independent sample of 252 cultures obtained from w
Marat540 [252]

Answer:

 The decision rule is  

Fail to reject the null hypothesis

  The conclusion is  

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

Step-by-step explanation:

From the question we are told that

   The first  sample size is n_1   =  252

    The number that tested positive is  k_1  =  18

     The second sample size is  n_2   =  476

     The number that  tested positive is  k_2 =  20

     The level of significance is  \alpha  = 0.01

Generally the first sample proportion is mathematically represented as

      \^ p _1 =  \frac{k_1 }{ n_1 }

=>    \^ p _1 =  \frac{18 }{ 252 }

=>    \^ p _1 = 0.071

Generally the second sample proportion is mathematically represented as

      \^ p _2 =  \frac{k_2 }{ n_2 }

=>    \^ p _2 =  \frac{20 }{ 476}

=>    \^ p _2 = 0.042

The  null hypothesis is            H_o  :  p_1 - p_2 = 0

The alternative hypothesis is  H_a :  p_1 - p_2 \ne 0

Generally the test statistics is mathematically represented

       z =  \frac{ \^ p_1 - \^ p_2  -  ( p_ 1 - p_2 )}{ \sqrt{\frac{\^ p_1 (1-\^ p_1)}{ n_1  } + \frac{\^ p_2 (1-\^ p_2)}{ n_2  }  } }

=>    z =  \frac{ 0.071 - 0.042  - 0 }{ \sqrt{\frac{0.071  (1-0.071)}{  252  } + \frac{0.042 (1-\^ 0.042)}{ 476  }  } }

=>    z =  1.56

From the z table  the area under the normal curve to the right corresponding to  1.56   is  

        P(Z >  1.56 ) =0.05938

Generally the p-value is mathematically represented as

         p-value =  2 * P(Z >  1.56 )

=>      p-value = 2 *  0.05938

=>      p-value = 0.1188

From the value obtained we see that   p-value  >  \alpha hence

  The decision rule is  

Fail to reject the null hypothesis

  The conclusion is  

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

 

5 0
3 years ago
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