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jeka94
2 years ago
11

27) Harry made 4 hits in 9 times at-bat. If she keeps the same success level, how many hits should she make in 18 times at bat?

Mathematics
1 answer:
9966 [12]2 years ago
6 0

Answer:

<u>8 hits</u>

Step-by-step explanation:

The ratio is :

  • hits : times-at-bat = 4 : 9

Now, we have 18 times at bat.

Multiply both sides of the ratio with 2 :

  • 4 x 2 hits : 9 x 2 times at bat
  • <u>8 hits</u> : 18 times at bat

Harry should make <u>8 hits</u> in 18 times at bat.

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Is there statistically significant evidence that the districts with smaller classes have higher average test​ scores? The t​-sta
Musya8 [376]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  95% confidence interval is [670.03  , 673.97 ]

The  test statistics is t = 7.7

The  p-value  is    p-value  =  0

The <u>p-value</u>  suggests that the null hypothesis is<u> rejected </u>with a high degree of confidence. Hence  <u>there is</u> statistically significant evidence that the districts with smaller classes have higher average test score  

Step-by-step explanation:

From the question we are told that

   The sample size is  n =  408

    The sample mean is  \= y  =  672.0

   The standard deviation is  s = 20.3

Given that the confidence level  is 95% then the level of significance is  

   \alpha = (100 -95 )\% = 0.05

From the normal distribution table  the critical value of \frac{\alpha }{2} = \frac{0.05 }{2} is  

    Z_{\frac{\alpha }{2} } =  1.96

Generally  the margin of error is mathematically represented as  

     E  =  Z_{\frac{\alpha }{2} } *  \frac{s}{\sqrt{n} }

=>   E  =  1.96 *  \frac{20.3}{\sqrt{408} }

=>     E  =  1.970

Generally the 95% confidence interval is mathematically represented as

       \= y -E  < \mu <  \= y + E

=>     672.0 -1.970  < \mu < 672.0 +1.970

=>     670.03  < \mu < 673.97

=>     [670.03  , 673.97 ]

From the question we are told that

   Class size                                  small                                      large

  Avg.score(\= y)         \= y_1 = 683.7   \= y_2 =  676.0

   S_y                          S_{y_1} =20.2    S_{y_2} = 18.6

   sample size                             n_1 = 229        n_2 =  184

The  null hypothesis is  H_o :  \mu_1 - \mu_2 = 0

The alternative hypothesis is  H_a :  \mu_1 - \mu_2 > 0

Generally the standard error for the difference in mean is mathematically represented as

       SE =  \sqrt{\frac{S_{y_1}^2 }{n_1} +\frac{S_{y_2}^2 }{n_2}   }

=>     SE =  \sqrt{20.2^2 }{229} +\frac{18.6^2 }{184_2}   }

=>     SE =  1.913

Generally the test statistics is mathematically represented as

      t = \frac{\= y _1 - \= y_2 }{SE}

=>    t = \frac{683.7 - 676.0 }{1.913}

=>   t = 7.7

Generally the p-value is mathematically represented as

    p-value  =  P(t >  7.7 )

From the  z-table

        P(t >  7.7 ) =  0

So

   p-value  =  0

From the values we obtained and calculated we can see that p-value  <  \alpha

This mean that

The p-value  suggests that the null hypothesis is rejected with a high degree of confidence. Hence  there is statistically significant evidence that the districts with smaller classes have higher average test score  

4 0
3 years ago
Point M belongs to AN , point N belongs to BM , AB = 18 in and AM : MN : NB = 1:2:3. Find MN.
yulyashka [42]

MN = 6 in

add the parts of the ratio, 1 + 2 + 3 = 6

divide AB by 6 to find 1 part of the ratio

\frac{18}{6} = 3 ← 1 part

AM = 3 in ← 1 part

MN = 2 × 3 = 6 in ← 2 parts

NB = 3 × 3 = 9 in ← 3 parts

and 3 + 6 + 9 = 18 in = AB


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3 years ago
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Answer:

Step-by-step explanation:

Let x = the age of the youngest child

Let x + 1 = the age of the middle child

Let x + 2 = the age of the oldest child

x + x + 1 + x + 2 = 39

3x + 3 = 39

3x = 36

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tatyana61 [14]
The x-intercepts are readily identified on the graph:  (2,0) and (4,0).  The midpoint of a line connecting those 2 points would be (3,0).
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3 years ago
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erma4kov [3.2K]

Answer:

yw

Step-by-step explanation:

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3 0
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