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vladimir2022 [97]
3 years ago
7

Y = 1.2 when x = 8 y varies inversely with x

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
5 0

Answer: 9.6

Step-by-step explanation:

y=k/x

1.2=k/8

K=1.2*8

K=9.6

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2x+5y=10<br><br>x intercept=<br><br>y intercept=​
Stells [14]

Answer:

x= 0/-8

y= 0/5

Step-by-step explanation:

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45% of the class is girls. There are 90 girls in the class. How many students are<br> in the class?
Stels [109]

Answer:

189

Step-by-step explanation:

45% x2= 90%

10% of 90= 9

90+90=180

plus that 10%

180 + 9 = 189

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3 years ago
75% of blank peaches is 15 peaches?
satela [25.4K]

.75 * p = 15

divide by .75

p = 15/.75 = 20

20 peaches

.15 * d = puppies

.15 * 60 = puppies

puppies = 9


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divide by .09

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4 0
3 years ago
Two​ shooters, Rodney and​ Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark
Sedaia [141]

Complete Question

Answer:

a

  SE  = 0.66}

b

-3.29 <  \mu_1 - \mu_2 <  -0.70  

Step-by-step explanation:

From the question we are told that

  The sample size is  n  = 60

   The first sample mean is  \= x _1  =  8

    The second sample mean is   \= x _2  =  10

    The first variance is  v_1 =  0.25

    The first variance is  v_2 =  0.55

Given that  the confidence level is 95% then the level of significance is 5% =  0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the first standard deviation is  

     \sigma_1 =  \sqrt{v_1}

=>   \sigma_1 =  \sqrt{0.25}

=>   \sigma_1 =  0.5

Generally the second standard deviation is

     \sigma_2 =  \sqrt{v_2}

=>   \sigma_2 =  \sqrt{0.55}

=>   \sigma_2 =  0.742    

Generally the first standard error is

     SE_1  =  \frac{\sigma_1}{\sqrt{n} }

      SE_1  =  \frac{0.5}{\sqrt{60} }

     SE_1  =  0.06

Generally the second standard error is

     SE_2  =  \frac{\sigma_2}{\sqrt{n} }

      SE_2  =  \frac{0.742}{\sqrt{60} }

     SE_2  =  0.09

Generally the standard error of the difference between their mean scores is mathematically represented as    

      SE  =  \sqrt{SE_1^2 + SE_2^2 }

=>     SE  =  \sqrt{0.06^2 +0.09^2 }

=>     SE  = 0.66}

Generally 95% confidence interval is mathematically represented as  

      (\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } *  SE) <  \mu_1 - \mu_2 <  (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } *  SE)

=> (8 -10) -(1.96 *  0.66) <  \mu_1 - \mu_2 <  (8-10) +(Z_{\frac{\alpha }{2} } *  0.66)  

=>  -3.29 <  \mu_1 - \mu_2 <  -0.70  

 

5 0
3 years ago
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