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bogdanovich [222]
4 years ago
9

Is 2 yd greater than 200 cm

Mathematics
2 answers:
SVEN [57.7K]4 years ago
7 0
No, 200 centimeters is greater than 2 yards
damaskus [11]4 years ago
6 0

Answer:

no

Step-by-step explanation:

1 foot is 30.48 cm

So 1 yard is 3(30.48 cm) = 91.44 cm (less than 100 cm = 1 meter)

So 2 yards is about 183 cm, less than 200 cm

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In Major League Baseball, the American League (AL) allows a designated hitter (DH) to bat in place of the pitcher, but in the Na
IRINA_888 [86]

Answer:

Step-by-step explanation:

The objective is to compute that more runs are scored in games for which DH is used

Let \mu_1 denote the population mean number of runs scored for DH group

Let \mu_2 denote the population mean number of runs scored for DH group

Let n_1  and  n_2  denote the sample sizes for DH and no DH

From the available information

n_1=n_2=20

The population standard deviation of runs score is 2.54 for both the groups.

That is \sigma _1^2=\sigma_2^2=2.54

The null hypothesis is H_0:\mu_1\leq \mu_2

The alternative hypothesis is H_1:\mu_1>\mu_2

Let the level of significance \alpha =0.10

Since, the population standard deviation for both the group is known , even though the sample size is less than 30 , use z test

The test statistic is

z=\frac{\bar x_1 - \bar x_2}{\sqrt{\frac{\mu_1^2}{n_1} +\frac{\mu_2^2}{n_2} } }

The sample mean for the DH group is computed as

\bar x_1=\frac{1}{n_1} \sum_{i=1}^{n_1}\\\\=\frac{1}{20} (0+6+8+2+2+4+7+7+6+5+1+1+5+4+4+5+7+11+10+0)\\\\=\frac{92}{20}\\\\=4.6

The sample mean of no DH is computed as

\bar x_2=\frac{1}{n_2} \sum_{i=1}^{n_1}\\\\=\frac{1}{20} (3+6+2+4+0+5+7+6+1+8+12+4+6+3+4+0+5+2+1+4)\\\\=\frac{83}{20}\\\\=4.15

The test statistic is

z=\frac{\bar x_1 - \bar x_2}{\sqrt{\frac{\mu_1^2}{n_1} +\frac{\mu_2^2}{n_2} } }

=\frac{4.60-4.15}{\sqrt{\frac{2.54^2}{20} +\frac{2.54^2}{20} } } \\\\=\frac{0.45}{\sqrt{0.32258+0.32258} } \\\\=\frac{0.45}{0.803219}\\\\=0.5602

P-Value

From the alternative hypothesis, it is clear that the test is one tailed test

P-value = P(Z > z)

1-P(Z\leq z)\\\\1-P(Z\leq 0.5602)

1 - normsdist (0.5602)

(using excel function "normsdist(z)")

= 1 - 0.7123

= 0.2877

Therefore, the P-Value is 0.2877

Decision Rule

Reject the null hypothesis , if the p-value is less than the level of significance . that is p-value is < 0.10

Here, the p-value is 0.2877 which is greater than the level of significance 0.10

so , fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that more runs scored in games for which DH is used.

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