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Lynna [10]
4 years ago
5

Courtney had 15 hits in 60 at bats. Katie's average was .300 and she had 12 hits. Sara got a hit 32.5% of the time and she knows

she had 40 at bats. Determine each players batting average
Mathematics
1 answer:
S_A_V [24]4 years ago
3 0

Answer:

\text{Courtney's batting average}=0.25

\text{Katie's batting average}=0.300

\text{Sara's batting average}=0.325

Step-by-step explanation:

We have been given that Courtney had 15 hits in 60 at bats.

We know that batting average can be determined by dividing hits by number of at-bats.

\text{Courtney's batting average}=\frac{15}{60}

\text{Courtney's batting average}=\frac{1}{4}

\text{Courtney's batting average}=0.25

Therefore, Courtney's batting average is 0.25.

We are already told that Katie's average was 0.300. Therefore, Katie's batting average is 0.300.

We have been given that Sara got a hit 32.5% of the time. To find Sara's batting average, we need to convert 32.5% into decimal as:

32.5\%=\frac{32.5}{100}=0.325

Therefore, Sara's batting average is 0.325.

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6 0
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Read 2 more answers
An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 83 type K batteries and a sample of 77
agasfer [191]

Answer:

1. Null Hypothesis, H_0 : \mu_1 = \mu_2  {mean voltage for these two types of

                                                         batteries is same}

Alternate Hypothesis, H_1 : \mu_1\neq \mu_2 {mean voltage for these two types of

                                                            batteries is different]

2. Test Statistics value = -5.06

4. Decision for the hypothesis test is that we will reject null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, \mu_1 = true mean voltage for type K batteries.

            \mu_2 = true mean voltage for type Q batteries.

So, <em>Null Hypothesis, </em>H_0<em> : </em>\mu_1 = \mu_2<em>  {mean voltage for these two types of </em>

<em>                                                          batteries is same}</em>

<em>Alternate Hypothesis, </em>H_1<em> : </em>\mu_1\neq \mu_2<em> {mean voltage for these two types of </em>

<em>                                                             batteries is different]</em>

The test statistics we use here will be :

                          \frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }   follows t_n__1+n_2-2

where, X_1bar = 9.29         and      X_2bar =  9.65

                s_1    = 0.374       and             s_2 =  0.518

                 n_1   = 83            and             n_2  =  77

                  s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} } =   \sqrt{\frac{(83-1)0.374^{2}+(77-1)0.518^{2}  }{83+77-2} } = 0.45                                                   Here, we use t test statistics because we know nothing about population standard deviations.

      Test statistics = \frac{(9.29-9.65) - 0 }{0.45\sqrt{\frac{1}{83}+\frac{1}{77}  } }  follows t_1_5_8

                              = -5.06

<em>At 0.05 or 5% level of significance t table gives a critical value between (-1.98,-1.96) to (1.98,1.96) at 158 degree of freedom. Since our test statistics is less than the critical table value of t as -5.06 < (-1.98,-1.96) so we have sufficient evidence to reject null hypothesis.</em>

Therefore, we conclude that mean voltage for these two types of batteries is different.

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