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3 years ago

If it takes a leopard to run 500 yards in 1/2 a min how how many miles could it run in 2 hours

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Answer:120000

Step-by-step explanation:

bc in a minute their is 60 seconds and if the leoperd runs 500 yards every 30 seconds u would multiply 60 times 2 bc their is 60 min in a hour and u would get 120.then u would multiply that by 2 getting u 240 which is two hours then u would do 500 times 240

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satela [25.4K]

Answer:

a) H0: $\mu \leq \mu_o$

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$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

b) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

c) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 25-1=24$

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

Step-by-step explanation:

Part a

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \leq \mu_o$

H1: $\mu > \mu_o$

n = 13 represent the sample size

$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part b

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part c

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 25-1=24$

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

8 0

3 years ago

If it takes a leopard to run 500 yards in 1/2 a min how how many miles could it run in 2 hours​

If it takes a leopard to run 500 yards in 1/2 a min how how many miles could it run in 2 hours