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lawyer [7]
1 year ago
10

in one version of a trail mix, there are 3 cups of peanuts mixed with 2 cups of raisins in another version of trail mix, there a

re 4.5 cups of peanuts mixed with 3 cups of raisins. are the ratios equivalent for the two mixes?
Mathematics
1 answer:
vlada-n [284]1 year ago
8 0

Based on the ratios of the peanuts mixed with raisins in the versions of the trail mix, the ratios are equivalent for both mixes.

<h3>What are the ratios?</h3>

The first ratio is:

3 : 2

When taken to the simplest terms it is:

3/2 : 2/2

1.5 : 1

The second ratio is:

4.5 : 3

4.5/3 : 3/3

1.5 : 1

The ratios of the peanut and raisins mix are therefore equivalent.

Find out more on equivalent ratios at brainly.com/question/2328454

#SPJ1

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The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales
mihalych1998 [28]

Answer:

We conclude that the mean number of calls per salesperson per week is more than 37.

Step-by-step explanation:

We are given that the Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 37 sales calls per week on professors.

To investigate, a random sample of 41 sales representatives reveals that the mean number of calls made last week was 40. The standard deviation of the sample is 5.6 calls.

<em><u>Let </u></em>\mu<em><u> = true mean number of calls per salesperson per week.</u></em>

SO, <u>Null Hypothesis</u>, H_0 : \mu \leq 37   {means that the mean number of calls per salesperson per week is less than or equal to 37}

<u>Alternate Hypothesis,</u> H_A : \mu > 37   {means that the mean number of calls per salesperson per week is more than 37}

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

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

where, \bar X = sample mean number of calls made last week = 40

             s = sample standard deviation = 5.6 calls

             n = sample of sale representatives = 41

So, <em><u>test statistics</u></em>  =   \frac{40-37}{\frac{5.6}{\sqrt{41} } }  ~ t_4_0

                               =  3.43

Hence, the value of test statistics is 3.43.

<em>Now at 0.025 significance level, the t table gives critical value of 2.021 at 40 degree of freedom for right-tailed test. Since our test statistics is more than the critical value of t as 3.43 > 2.021, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the mean number of calls per salesperson per week is more than 37.

4 0
3 years ago
Juan brought 4 Tshirts and leather jacket. The T shirts were all the same price, and the price of the leather jacket was 6 times
Alex

Let us say that:

x = the number of T shirts bought = 4

y = the number of leather jackets bought = 1

a = the price of the T shirts

b = the price of the leather jackets

 

With the given variables, we can say that the total cost is:

total cost = a x + b y

 

Since, b = 6 a, therefore:

total cost = a x + 6 a y

total cost = a (x + 6 y)

 

Since total cost is 209, therefore we can calculate for a since x and y are also given:

209 = a (4 + 6 * 1)

209 = a (10)

a = $20.9

 

<span>Therefore each T shirt cost $20.9</span>

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3 years ago
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In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a
Softa [21]

Answer:

95% Confidence interval = (23.4,26.2)

Step-by-step explanation:

In this problem we have to develop a 95% CI for the mean.

The sample size is n=49, the mean of the sample is M=24.8 and the standard deviation of the population is σ=5.

We know that for a 95% CI, the z-value is 1.96.

The CI is

M-z*\sigma/\sqrt{n}\leq\mu\leq M+z*\sigma/\sqrt{n}\\\\24.8-1.96*5/\sqrt{49}\leq\mu\leq 24.8+1.96*5/\sqrt{49}\\\\ 24.8-1.4\leq\mu\leq 24.8+1.4\\\\23.4\leq\mu\leq 26.2

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