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3 years ago

5

Andrew and his father are scuba diving at -38 feet and tackle box Canyon has an elevation of -83 feet right and inequality to co

mpare the depths explain the meaning of the inequality

2 answers:

zepelin [54]3 years ago

7 0

-38 > -83
Where Andrew and his father are scuba diving has a higher elevation than tackle box Canyon.

xz_007 [3.2K]3 years ago

4 0

Answer:

-38 > -83

Step-by-step explanation:

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An active boy between the ages of 9 and 13 should eat about

ella [17]

Answer:

In addition, a need to lose, maintain, or gain weight and other factors affect how many ... These estimates are based on the Estimated Energy Requirements (EER) ... Estimates range from 1,600 to 2,400 calories per day for adult women and ... is for sedentary individuals; the high end of the range is for active individuals.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Suppose you select a sample of 12 individuals and find that 10 of them do not exercise regularly. Assuming that the Surveillance

Leto [7]

Answer:

a. mean = 6.96 , standard deviation = 1.71 b. 0.0641

Step-by-step explanation:

Here is the complete question

According to the Behavioural Risk Factor Surveillance System, 58% of all Americans adhere to a sedentary lifestyle (sedentary means does not exercise).

a. If you selected repeated samples of 12 from the U.S. population, what would be the mean number of individuals per sample who do not exercise regularly? What would be the standard deviation?

b. Suppose you select a sample of 12 individuals and find that 10 of them do not exercise regularly. Assuming that the Surveillance System is correct, what is the probability that you would have obtained results as bad or worse than expected?

<em>Solution</em>

a. Since we can only have two types of outcomes, that is, those who exercise and those who do not exercise, the problem follows a binomial distribution.

Since the probability of those who do not exercise, p = 58 % = 0.58,

the mean of the binomial distribution μ = np where n = sample number = 12

So, μ = 12 × 0.58 = 6.96

The standard deviation of a binomial distribution σ = √(npq) where q = probability that people exercise = 1 - p = 1 - 0.58 = 0.42

So, σ = √(12 × 0.58 × 0.42) = √2.9232 = 1.71

b. To find the worst case, we consider the probability that at best, two exercise. Since two exercise, the probability that at best two exercise is ¹²C₂q²p¹⁰ + ¹²C₁qp¹¹ + ¹²C₀q⁰p¹²

= (12 × 11/2)(0.42)²(0.58)¹⁰ + 12(0.42)(0.58)¹¹ + 1 × (0.42)⁰(0.58)¹²

= 0.05 + 0.0126 + 0.00145

= 0.06405

≅ 0.0641

6 0

3 years ago

The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The

Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(c) The necessary assumption is that the population is normally distributed.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let <em>X</em> = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

(a)

The general form of a (1 - <em>α</em>)% confidence interval is:

$CI=SS\pm CV\times SD$

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a <em>z</em>-confidence interval.

The critical value of <em>z</em> for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the confidence interval as follows:

$CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)$

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

$37\pm 0.1=SS\pm CV\times SD$

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

$P(-1$

The percentage of <em>z</em>-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the <em>z</em>-interval or <em>t</em>-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a <em>t</em>-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For <em>n</em> = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - <em>α</em>)% <em>t</em>-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The critical value of <em>t</em> is:

$t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262$

*Use a <em>t</em>-table for the critical value.

The 95% confidence interval is:

$CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)$

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

3 0

2 years ago

the sale tax rate in Jans town is 7.5% if she buys 3 lamps for $23.59 each and a sofa for $769.99, how much sales tax does she o

masha68 [24]

Cost of Lamps =
23.59*3= $70.77

Sofa = $769.99
Add them
769.99+70.77
=840.76
Percentage
7.5/100 *840.76
= 63.057

She owes $63.057 to sales tax

4 0

3 years ago

Please help! PLSSS! Its for a math test.

Maslowich

Answer:

330(D)

Step-by-step explanation:

So lets go over what we know:

#1: 44 out of 80 people think they have a average fitness level.

#2: 80 out of the 600 people in the school were surveyed.

Using what we know, lets try and find the answer.

So lets divide the total amount of people, by the amount of people that took the survey:

600/80

=

7.5

Ok, so now we know that we need to take 7.5 surveys to find out what all the students think of their fitness level.

Now lets take teh amount of people in a single survey, with average fitness(44), and multpliy it by how many surveys are needed for the whole school(7.5):

44*7.5

=

330

This is the amount of people in the whole school who think they have an average fitness level.

I hope this helps! :)

<u>Answer:</u>

<u>330</u>

8 0

3 years ago