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nikklg [1K]
3 years ago
12

a rocket is scheduled to launch from a command center in 3.75 hours. what time is it now if launch time is 11:20am. i need to kn

ow how to write equation to get answer
Mathematics
2 answers:
gtnhenbr [62]3 years ago
5 0

The equation would be 3.75+t=11.2 so:

3.75+t=11.2

3.75+t-3.75=11.2-3.75

t=7.45, which is 7:35.


Over [174]3 years ago
4 0

Answer:

t + 3:45 = 11:20 am

Step-by-step explanation:

A rocket is scheduled to launch from a command center in 3.75 hours.

The launch time is 11:20 am.

We need to know the time before 3.75 hours of 11:20 am.

3.75 hours = 3 hours (0.75 × 60) = 45 minutes

Let the current time be 't'

Now the equation will be

t + 3:45 = 11:20 am

t = 11:20 - 3:45

t = 7 : 35 am

Before 3.75 hours of 11:20 am it will be 7:35 am.

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Pls help me solve surds , √75 ,√48, √12,√125,√28. pls help​
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√48 = √(4²×3) = √4² √3 = 4√3 = 6.93

√12 = √(2²×3) = √2² √3 = 2√3 = 3.46

√125 = √(5²×3) = √5² √3 = 5√3 = 11.18

√28 = √(2²×7) = √2² √7 = 2√7 = 5.29
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Which graph shows y=3⌈x⌉+1?
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A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su
Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }   ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0
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