A class 4 truck weighs between 14,000 and 16,000. What is the weight range in tons?

jocelyn is playing a game she gains 4 points, looses 8 points, and gains 3 points.what is her final score?

Lana71 [14]

Answer: Her final score is negative 1.

Step-by-step explanation:

She started off with a score of 0, then gained 4. So you would add 4 to zero. Then, she lost 8 points so you would subtract 8 from her current score of 4, making the new total negative 4. Last, she gained three more points so you would add 3 to -4, giving you the final answer of -1.

4 - 8 = -4

-4 + 3 = -1

5 0

3 years ago

A spinner numbered 1 through 5 is spun 3 times. what is the probability of spinning a number less than 3 each time? are these ev

olya-2409 [2.1K]

Prob(spinning a number < 3on 1 spin) = Prob( spinning 1 or 2) = 2/5

Prob( spinning <3 3 times) = 2/5 * 2/5 * 2/5 = 8/125

These events are independent so we multiply the probabilities.

4 0

4 years ago

Change in y/Change in x is used to calculate rate of change from a________

zepelin [54]

3 to a 8 with 5 hope this helps

6 0

4 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

Feliz [49]

Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34$

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so $z = 2.327$

$M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45$

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 1.96*\frac{0.78}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.78$

$\sqrt{n} = \frac{1.96*0.78}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}$

$n = 14.6$

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.2 = 2.575*\frac{0.78}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.78$

$\sqrt{n} = \frac{2.575*0.78}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}$

$n = 100.85$

Rounding up

A sample size of 101 is needed.

8 0

4 years ago

What is 10% of 4.25 (moneywise)

Feliz [49]

The answer will be .425. Hope it help!

4 0

3 years ago

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