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Aneli [31]
3 years ago
7

A coach gives her players the option of running around their field twice or around their entire stadium once. The following diag

ram shows the field and stadium dimensions on a coordinate grid.
Mathematics
1 answer:
Elodia [21]3 years ago
7 0
Get a call to me when I can do it on Saturday
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Tems11 [23]
5x - y = -4 is the standard form
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What are the rate of changes for
astraxan [27]
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if the values are y values you will subtract them from each other to get the numerical value and because they all go from a larger number to a smaller number the rate is a negative. if they were from a smaller value to a larger value teh rates of change would be positive.
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3 years ago
WILL MARK BRAINLIEST
____ [38]

Answer:

512

Step-by-step explanation:

∛n  = 8

n = 512

double check

∛512 = ∛(8)^3   = 8

3 0
3 years ago
Read 2 more answers
The solution is (7,7)
Georgia [21]
B is the answer I chose but there are the steps so you can see how to solve it

8 0
2 years ago
The American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day. Legacy
inessss [21]

Answer:

We conclude that the null hypothesis is rejected which means residents of Legacy Ranch use less water on average.

Step-by-step explanation:

We are given that the American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day.

Twenty-three owners responded, and the sample mean water use per day was 63 gallons with a standard deviation of 8.1 gallons per day.

<u><em>Let </em></u>\mu<u><em> = average water usage by residents of Legacy Ranch.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 67 gallons     {means that the residents of Legacy Ranch uses more or equal to 67 gallons water on average}

Alternate Hypothesis, H_A : \mu < 67 gallons    {means that the residents of Legacy Ranch use less water on average}

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

                                 T.S.  = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean water use per day = 63 gallons

             s = sample standard deviation = 8.1 gallons

             n = sample of responded = 23

So, <u><em>test statistics</em></u>  =  \frac{63-67}{\frac{8.1}{\sqrt{23} } }  ~ t_2_2

                               =  -2.368

The value of the test statistics is -2.368.

<em>Now at 0.025 significance level, the </em><u><em>t table gives critical value of -2.074 at 22 degree of freedom for left-tailed test</em></u><em>. Since our test statistics is less than the critical values of t as -2.368 < -2.074, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to </em><u><em>which we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the residents of Legacy Ranch use less water on average.

8 0
3 years ago
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