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2 years ago

7

Xian and his cousin Kai both collect stamps. Xian has 56 stamps, and Kai has 80 stamps. The boys recently joined different stamp

collecting clubs. Xians club will send him 12 new stamps per month. Kai’s club will send him 8 new stamps per month. After how many months will Xian and Kai have the same number of stamps? How many stamps will they each have?

1 answer:

kondaur [170]2 years ago

4 0

Answer:

all work is pictured/shown

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If a hurricane was headed your way, would you evacuate? The headline of a press release states, "Thirty-four Percent of People o

IgorC [24]

Answer:

The 98% confidence interval would be given by (0.326;0.354)

Step-by-step explanation:

1) Notation and definitions

$X=63$ number of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$n=6138$ random sample taken

$\hat p=0.34$ estimated proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$p$ true population proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by $\alpha=1-0.98=0.02$ and $\alpha/2 =0.01$ . And the critical value would be given by:

$z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.34 - 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.326$

$0.34 + 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.354$

The 98% confidence interval would be given by (0.326;0.354)

8 0

2 years ago

For a school drama performance, student tickets cost $5 each and adult tickets cost $10 each. The sellers collected $3,570 from

olchik [2.2K]

I think the right equation is 10(512 - c) plus 5c equals 3,570

4 0

1 year ago

Read 2 more answers

Mixed numbers between 0 and 2 with an interval of 1/3 between each pair of mixed numbers

Marysya12 [62]

A mixed number is a whole number plus a fraction.The smallest whole number is 1.So no mixed number can be less than 1.
Mixed numbers that are 1/3 apart and are between 0 and 2
could be (1-1/3) and (1-2/3).
They could also be (1-1/6), (1-1/2), and (1-5/6) .

3 0

2 years ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago

Consider the lengths of stay at a hospital’s emergency department. Hours Count Percent 1 18 3.44 2 55 10.50 3 81 15.46 4 109 20.

Vladimir [108]

Answer:

Probability = 0.502

Step-by-step explanation:

We are given the following data :

Hours Count Percent

1 18 3.44

2 55 10.50

3 81 15.46

4 109 20.80

5 88 16.79

6 66 12.60

7 39 7.44

8 17 3.24

9 17 3.24

10 19 3.63

15 15 2.86

We need to calculate the probability

P(Length of stay of exactly 1 is less than or equal to 4)

P( $Y \leq 4$ ) = P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)

$P(Y \leq 4) = 0.0344 + 0.1050 + 0.1546 + 0.2080 = 0.502$

We convert the percent into probabilities by dividing them with 100. This gave us the required probabilities.

8 0

2 years ago