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atroni [7]
3 years ago
7

The speed of object A is 1,500 mph faster than that of object B. If

Mathematics
2 answers:
Elena L [17]3 years ago
5 0

Answer:

2.42857142857

Step-by-step explanation:

11Alexandr11 [23.1K]3 years ago
4 0

Answer:

1564.42 mph

Step-by-step explanation:

Let a = Object A's speed

     b = Object B's speed

a = b + 1500

a/85000 = b/3500

Let's make subtract 1500 from both sides of the first equation

a = b + 1500

- 1500  - 1500

b = a - 1500

Cross multiply the second equation

3500a = 85000b

Plug in the new b into this new equation

3500a = 85000(a - 1500)

3500a = 85000a - 127500000

Subtract 3500a and 127500000 from both sides

3500a = 85000a - 127500000

- 3500a  - 3500a

0 = 81500a - 127500000

+ 127500000    + 127500000

127500000 = 81500a

Divide both sides by 81500

127500000/81500 = 81500a/81500

a = 1564.4171779141104

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Answer:

b) independent samples t-test

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: \mu_1 = \mu_2  

Alternative hypothesis: \mu_1 \neq \mu_2  

Or equivalently:  

Null hypothesis: \mu_1 - \mu_2 = 0  

Alternative hypothesis: \mu_1 -\mu_2\neq 0  

Our notation on this case :  

n_1 =5 represent the sample size for group 1  

n_2 =5 represent the sample size for group 2  

We can calculate the mean and deviation for eaach group with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_1 =5.92 represent the sample mean for the group 1  

\bar X_2 =5.74 represent the sample mean for the group 2  

s_1=1.88 represent the sample standard deviation for group 1  

s_2=1.81 represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:  

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}  

And now we can calculate the statistic:  

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

The degrees of freedom are given by:  

df=5+5-2=8

And now we can calculate the p value using the altenative hypothesis:  

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

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