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Katena32 [7]
3 years ago
8

Find the mode for the price of movie tickets at various theaters. $5, $4, $7, $9, $5, $8, $5, $3, $6, $5, $7, $4, $5, $4

Mathematics
1 answer:
kupik [55]3 years ago
3 0

5 is the most occurring number

Step-by-step explanation:

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fgiga [73]
I hated probability in middle school. I'm not 100% sure but I would go with the 3rd answer choice, 3/100.
3 0
3 years ago
A pair of $36 jeans were on sale at Jim's Jeans for 1/3 off. The same pair of jeans were $42 at David's Denims. The jeans were o
nikklg [1K]

Complete question :

A pair of $36 jeans were on sale at Jim's Jeans for 1/3 off. The same pair of jeans were $42 at David's Denims. The jeans were on sale for 2/3 off.

A.) which store had the better buy

B.) how much will you pay at the store that has the better buy

Answer:

A.)David's denim has the better buy

B.) $14

Step-by-step explanation:

Jim's Jean store :

Cost of Jean = $36

Discount = 1/3 off

Discounted cost = $36 - (1/3 * $36)

Discounted cost = $36 - $12 = $24

David's Denim:

Cost of Jean = $42

Discount = 2/3 off

Discounted cost = $42 - (2/3 * $42)

Discounted cost = $42 - $28 = $14

David's denim has the better buy with a discounted price lower than the discounted price at Jim's store.

Cost at store with the better buy is $14

8 0
3 years ago
What are the zeros of the function y = 2x2 + 5x + 2?
Dvinal [7]

Answer: not sure

but I need points

Step-by-step explanation:

JUST KIDDING It’s A I did this in school before and got it correct :)

4 0
3 years ago
320, 305, 308, 340, 345,<br>315, 330, 315, 330, 318, 325 the mode median​
grin007 [14]

Answer:

mode= 315, 330

median=320

Step-by-step explanation:

4 0
3 years ago
According to a survey, the average American person watches TV for 3 hours per week. To test if the amount of TV in New York City
Neporo4naja [7]

Answer:

Test statistic (t-value) of this one-mean hypothesis test is -2.422.

Step-by-step explanation:

We are given that according to a survey, the average American person watches TV for 3 hours per week. She surveys 19 New Yorkers randomly and asks them about their amount of TV each week, on average. From the data, the sample mean time is 2.5 hours per week, and the sample standard deviation (s) is 0.9 hours.

We have to test if the amount of TV in New York City is less than the national average.

Let Null Hypothesis, H_0 : \mu \leq 3  {means that the amount of TV in New York City is less than the national average}

Alternate Hypothesis, H_1 : \mu > 3   {means that the amount of TV in New York City is more than the national average}

The test statistics that will be used here is One-sample t-test statistics;

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

where, \bar X = sample mean time = 2.5 hours per week

             s = sample standard deviation = 0.09 hours

             n = sample size = 19

So, <u>test statistics</u> = \frac{2.5 - 3}{\frac{0.9}{\sqrt{19} } } ~ t_1_8

                              = -2.422

Therefore, the test statistic (t-value) of this one-mean hypothesis test (with σ unknown) is -2.422.

8 0
4 years ago
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