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2 years ago

Need help! Can someone help, thanks

Mathematics

2 answers:

ki77a [65]2 years ago

6 0

I believe the length is 18 and the width is 3.

Shalnov [3]2 years ago

5 0

The length is 18ft and the width is 3ft.

How to get this answer?

The length = 6*W
Perimeter = 42 ft

Formulas:
Perimeter = 2Length + 2Width

42ft = 2L + 2W
We know the length is 6W so replace that with L.

42ft = 2(6W) + 2W
42ft = 12W + 2W
42ft = 14W
W = 3ft
L= W*6 = 3*6 = 18ft

Check it with formula:

P = 2L + 2W
2(18) + 2(6) = 42ft, so it is correct.

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The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o

frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 36, \sigma = 5$

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40 - 36}{5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

8 0

2 years ago

When sampling from a population, the sample mean will: Group of answer choices always be closer to the population mean as the sa

Radda [10]

Answer:

When sampling from a population, the sample mean will: be closer to the population mean as the sample size increases.

Step-by-step explanation:

The sample mean is not always equal to the population mean but if we increase the number of samples then the mean of the sample would become more and more closer to the population mean.

Usually the population size is very huge that is why we select a random sample from the population, care must be taken to ensure randomized sampling otherwise results would not be accurate. After that we have to make sure that the number of samples are enough for the given population size. The number of samples depends upon the shape of the population. If the population is normal than according to central limit theorem, a less number of samples would be enough to ensure normal distribution of sampling mean, otherwise a greater sample size will be required.

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2 years ago

If a sprinkler waters 1 over 20 of a lawn in 1 over 5 of an hour, how much time will it take to water the entire lawn?

Genrish500 [490]

5 hours bc my math add ups

7 0

2 years ago

Read 2 more answers

A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5

gladu [14]

Answer:

A) It will get to a temperature of 125°F at 9:19 PM

B) It will get to a temperature of 150°F at 9:16 PM

C) as time passes temperature approaches the initial temperature of 450°F

Step-by-step explanation:

We are given;

Initial temperature; T_i = 450°F

Room temperature; T_r = 70°F

From Newton's law of cooling, temperature after time (t) is given as;

T(t) = T_r + (T_i - T_r)e^(-kt)

Where k is cooling rate and t is time after the initial temperature.

Now, we are told that After 5 minutes, the temperature is 300°F.

Thus;

300 = 70 + (450 - 70)e^(-5k)

300 - 70 = 380e^(-5k)

230/380 = e^(-5k)

e^(-5k) = 0.6053

-5k = In 0.6053

-5k = -0.502

k = 0.502/5

k = 0.1004 /min

A) Thus, at temperature of 125°F, we can find the time from;

125 = 70 + (450 - 70)e^(-0.1004t)

125 - 70 = 380e^(-0.1004t)

55/380 = e^(-0.1004t)

In (55/380) = -0.1004t

-0.1004t = -1.9328

t = 1.9328/0.1018

t ≈ 19 minutes

Thus, it will get to a temperature of 125°F at 9:19 PM

B) Thus, at temperature of 150°F, we can find the time from;

150 = 70 + (450 - 70)e^(-0.1004t)

150 - 70 = 380e^(-0.1004t)

80/380 = e^(-0.1004t)

In (80/380) = -0.1004t

-0.1004t = -1.5581

t = 1.5581/0.1004

t ≈ 16 minutes.

Thus, it will get to a temperature of 150°F at 9:16 PM

C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.

Thus,we have;

T(t) = T_r + (T_i - T_r)

T_r will cancel out to give;

T(t) = T_i

Thus, as time passes temperature approaches the initial temperature of 450°F

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2 years ago

Solve for k<br> 1/4K=3(-1/4K +3)

Gre4nikov [31]

Answer:

k= 9

Step-by-step explanation:

hope this helps you

7 0

2 years ago