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ehidna [41]
3 years ago
9

5 times a number "x" is not less than 15 as an inequality

Mathematics
2 answers:
elena-s [515]3 years ago
6 0

Answer:

Step-by-step explanation:

It could be as long as you know what "x" is.

jeka943 years ago
3 0

Answer:

x > 3

Step-by-step explanation:

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Isabella averages 152 points per bowling game with a standard deviation of 14.5 points. Suppose Isabella's points per bowling ga
Serga [27]

Answer:

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

\mu = 152, \sigma = 14.5

The z-score when x=187 is ...

Z = \frac{X - \mu}{\sigma}

Z = \frac{187 - 152}{14.5}

Z = 2.41

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.

3 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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