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aev [14]
3 years ago
8

The picture shows the Alamillo bridge in Seville, Spain. In the picture m<1=58 and m<2=24. Find the measure of the supplem

ent of <1= and of <2. Someone please help!

Mathematics
1 answer:
Anna11 [10]3 years ago
6 0

Answer:

Supplement of <1 = 122°

Supplement of <2 = 156°

Step-by-step explanation:

Two pairs of angles are said to be supplementary, if their measures in degrees add up to give us 180°. A supplement of am angle is simply 180° - the measure of that angle.

Given that meadure of angle 1 = 58°, the supplement of angle 1 = 180° - 58° = 122°

Also, if the measure of angle 2 = 24°, the supplement of angle 2 = 180° - 24° = 156°.

Thus:

Supplement of <1 = 122°

Supplement of <2 = 156°

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1 year ago
A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In o
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Answer:

(a) Null Hypothesis, H_0 : \mu = 22 ounces  

    Alternate Hypothesis, H_A : \mu\neq 22 ounces

(b) The value of the test statistic is -2.687.

(c) The critical values are -1.96 and 1.96.

(d) We conclude that Reject H_0 since the value of the test statistic is less than the negative critical value.

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.

He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

<em>Let </em>\mu<em> = average ounces of passion fruit juice used to fill each bottle.</em>

(a) So, Null Hypothesis, H_0 : \mu = 22 ounces     {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}

Alternate Hypothesis, H_A : \mu\neq 22 ounces     {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                        T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean weight of the passion fruit juice = 21.54 ounces

            \sigma = population standard deviation = 1.38 ounce

            n = sample of bottles = 65

So, <u><em>test statistics</em></u>  =  \frac{21.54-22}{\frac{1.38}{\sqrt{65} } }  

                               =  -2.687

(b) The value of z test statistics is -2.687.

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

<em>Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that Reject H_0 since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.

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I cant do #2 without seeing the formula

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