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svetlana [45]
3 years ago
12

A recipe calls for 3 cups of sugar and 9 cups of water. How many cups of water should be used with 12 cups of sugar?

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
8 0

Answer:

36

Step-by-step explanation:

you get 36 becayse you have to multiply the amoubt of how much you got for the other one

Yuki888 [10]3 years ago
3 0

Answer:

36

Step-by-step explanation:

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A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writin
Morgarella [4.7K]

Answer:

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

<u><em>Let </em></u>\mu<u><em> = true average writing lifetime under controlled conditions</em></u>

So, Null Hypothesis, H_0 : \mu \geq 10 hr   {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, H_A : \mu < 10 hr    {means that the true average writing lifetime under controlled conditions is less than 10 hr}

<u>The test statistics that is used here is one-sample t test statistics;</u>

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

where, \bar X = sample mean

             s = sample standard deviation

             n = sample size of pens = 18

          n - 1 = degree of freedom = 18 -1 = 17

<u>Now, the decision rule based on the critical value of t is given by;</u>

  • If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then <u>we will not reject our null hypothesis</u> as it will not fall in the rejection region.
  • If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then <u>we will reject our null hypothesis</u> as it will fall in the rejection region.

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

<em>Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.</em>

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>

Here, clearly the value of test statistics is more than the critical value of t as  -1.74 < 0.57, so we fail to reject our null hypothesis.

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