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2 years ago

9

Of the 50 tables in a restaurant, 40% seat 2 people. The rest of the tables seat 4 people. How many seats are there in the resta

urant

1 answer:

Kipish [7]2 years ago

8 0

There are 50 tables.

20 tables seat two people and 30 tables seat 4 people.

You might be interested in

) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa

asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

<u><em>Let </em></u> $p_1$ <u><em> = population proportion of drivers 16-24 years old who buckle up .</em></u>

<u><em /></u> $p_2$ <u><em> = population proportion of drivers 25-69 years old who buckle up .</em></u>

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no significant difference in seat belt use}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a difference in seat belt use}

The test statistics that would be used here <u>Two-sample z proportion statistics;</u>

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers 16-24 years old who buckle up = 79%

$\hat p_2$ = sample proportion of drivers 25-69 years old who buckle up = $\frac{924}{1100}$ = 84%

$n_1$ = sample of 16-24 years old drivers = 1000

$n_2$ = sample of 25-69 years old drivers = 1100

So, <u><em>test statistics</em></u> = $\frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }$

= -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u><em>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that there is a difference in seat belt use.

4 0

3 years ago

The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.

mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

0.96516 probability of a chipmunk living through the year, thus $p = 0.96516$

Item a:

Two is P(X = 2) when n = 2, thus:

$P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315$

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

Six is P(X = 6) when n = 6, then:

$P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834$

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

At least one not living is:

$P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166$

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0

2 years ago

Ye has his own business. He checks his sales receipts three times a day, his afternoon sales were 50$ more than his mourning sal

IRINA_888 [86]

Answer:

The answer is 150

Step-by-step explanation:

It's 150 because ye evening sales are three times of his afternoon sales.

SO that means you have to times 50x3=150 or you can add 50+50+50=150. there is your answer.

3 0

2 years ago

Read 2 more answers

On Friday night 225 people saw the dinosaur exhibit at the museum this amount represents 20% of the people who visited the museu

Igoryamba

20 % = 225
1% = 225 / 20
100% = 225 * 100 / 20
= 1125 answer

8 0

3 years ago

Why are tenths, fifths, and fourths easy to convert to percent's?

meriva

Because they all go into 100 easily. for example 10*10=100 5*20=100

5 0

2 years ago