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Luda [366]
1 year ago
13

What is the reflection across the y-axis?

Mathematics
2 answers:
Illusion [34]1 year ago
6 0
Answer : (c). W (-3,3) V (-2,2) X (-2,1)

The rule for a reflection over the y -axis is (x,y)→(−x,y) .

So

W (3,3) V (2,2) X (2,1)

Becomes

W (-3,3) V (-2,2) X (-2,1)

Allisa [31]1 year ago
3 0

Answer:

c

Step-by-step explanation:

what ever it is you do the same on the other side

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PLEASE SOMEONE HELP ME WILL MAKE YOU BRAINLEST
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Answer:

To work out how much it is for a pound of pistachio or almonds:

You subtract 8 from 48 which leaves you with 40 then you divide 40 with 5 which leaves you with 8 (meaning $8 for a pound for almonds) and you add 4 to 8 which gives you 12 ($12 for a pound of pistachios)

then you substitute the answers into a, b, c,d and e.

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3 years ago
By law, an industrial plant can discharge not more than 500 gallons of waste water per hour, on the average, into a neighboring
Alisiya [41]

Answer:

We conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

Step-by-step explanation:

We are given that an industrial plant can discharge not more than 500 gallons of wastewater per hour, on average, into a neighboring lake.

Four one-hour periods are selected randomly over a period of one week. The following are observed:

1384, 683, 1534, 405

Let \mu = <u><em>population average gallons of wastewater discharged per hour</em></u>

So, Null Hypothesis, H_0 : \mu \leq 500 gallons      {means that not more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

Alternate Hypothesis, H_A : \mu > 500 gallons     {means that more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because 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 = \frac{\sum X}{n} = 1001.5 gallons

             s = sample standard deviation = \sqrt{\frac{\sum (X - \bar X)^{2} }{n-1} } = 543.79

             n = sample of periods = 4

So, <u><em>the test statistics</em></u> =  \frac{1001.5-500}{\frac{543.79}{\sqrt{4} } }  ~ t_3

                                    =  1.844

The value of t-test statistics is 1.844.

Since in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 level of significance, the t table gives a critical value of 2.353 at 3 degrees of freedom for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.844 < 2.353, so we have <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

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