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Vladimir79 [104]
3 years ago
10

John is creating a Thanksgiving display at the store where he works, using only canned pumpkin and canned green beans. He needs

to maintain a ratio of pumpkin to green beans of 3 to 1. If he wants to use a total of 64 cans, how many cans of green beans should he use?
Mathematics
1 answer:
mezya [45]3 years ago
3 0

Answer:

The number of Pumpkin can used   = 48 cans

The number of Green Beans can used =   16 cans

Step-by-step explanation:

The ratio of Pumpkin to Green Beans =  3 :  1

So, for every 3 can of Pumpkin, 1 can of green beans is used.

Now, the total number of  John  wants to use = 64

Let us assume the common factor in the ratio is m.

⇒ The number of Pumpkin can used  = 3 m

The number of Green Beans can used = 1 m = m

So, the Total cans used = Number of (Pumpkin + Green Beans) can

⇒ 3 m + m = 64

or, 4 m =  64

⇒ m = 64/ 4 = 16

Hence, The number of Pumpkin can used  = 3 m = 3 x 16 = 48 cans

The number of Green Beans can used = m =  16 cans

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3 years ago
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Step-by-step explanation:

It seems here that they are asking us to solve for x

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7 0
3 years ago
The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the
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Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average sodium content.</em></u>

So, Null Hypothesis, H_0 : \mu = 1100 milligrams      {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, H_A : \mu \neq 1100 milligrams      {means that the sodium content is different from what the nutrition label states}

The test statistics that would be used here <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 average sodium content = 1088.64 milligrams

            s = sample standard deviation = 234.12 milligrams

            n = sample of packages of Oriental Spice Sauce = 40

So, <u><em>test statistics</em></u>  =  \frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}  ~ t_3_9

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The value of z test statistics is -0.307.

<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.</em><em> </em>

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

Therefore, we conclude that the sodium content is same as what the nutrition label states.

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