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Murljashka [212]
2 years ago
7

A company bought New Year's greeting cards to send to its valued customers. Sixty of the cards they sent cost $2.30 each. Thirty

of the cards cost $2.75 each The remaining 34 cards cost $1.95 each. What was the total amount spent on these cards?​
Mathematics
1 answer:
const2013 [10]2 years ago
3 0

Answer:

$287.82

Step-by-step explanation:

60 * 2.30 = 138

30 * 2.75 = 82.5

34 * 1.98 = 67.32

138 + 82.5 + 67.32 = 287.82

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Travka [436]

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either the first or second but id go with the second one!

Step-by-step explanation:

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2 years ago
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Based on a poll 62% of internet users are more careful about personal information when using a public wifi hotspot. What is the
Setler79 [48]

Answer:

0.9451

Step-by-step explanation:

Remaining question? <em>"How is the result affected by the additional information that the survey subjects volunteered to​respond?"</em>

Probability that at least 1 user is more careful about personal information when using a public Wi-Fi hot spot is:

P(X≥1) = 1 − P(X<1)

= 1 − P(X=0)

= 1 - [(3,0) (0.62)^0 (1-0.62)^3-0

= 1 - 0.054872

= 0.945128

= 0.9451

Thus, the probability that among three randomly selected Internet users; at least one is more careful about personal information when using a public Wi-Fi hotspot is 0.9451

7 0
2 years ago
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.

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kirza4 [7]

Answer:

a) x=7

b) x=4

c) x= 1/9

I hope this helps! Do you need more help?

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On someone plz help it is timed i will give Brainliest if correct
Oksanka [162]

Answer:

D.

Step-by-step explanation:

6 0
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