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rjkz [21]
3 years ago
8

How much carbon dioxide is emitted from using a 1500 watt dryer for 8 hours every

Mathematics
2 answers:
nalin [4]3 years ago
6 0

9514 1404 393

Answer:

  854.88 lb/yr

Step-by-step explanation:

It is effectively a units conversion problem. The conversion factors of interest are ...

  1000 W × 1 h = 1 kWh

  52 wk = 1 yr

__

  (1500 W)×(8 h/wk) × (52wk/yr) ÷ ((1000 W)(1 h))/(1 kWh) × (1.37 lb/kWh)

  = (1500×8×52×1.37)/(1000) lb/yr

  = 854.88 lb/yr

slavikrds [6]3 years ago
5 0

ERROR!ERROR!ERROR!ERROR!

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A swimming pool is being drained so it can be cleaned. The amount of water in the pool is changing according to the función f(t)
TiliK225 [7]

Answer:

0 ≤ t ≤ 5.

Step-by-step explanation:

In the function f(t), t is the independent variable. The domain of f is the set of all values of t that this function can accept.

In this case, f(t) is defined in a real-life context. Hence, consider the real-life constraints on the two variables. Both time and volume should be non-negative. In other words,

  • t \ge 0.
  • f(t) \ge 0.

The first condition is an inequality about t, which is indeed the independent variable.

However, the second condition is about f, the dependent variable of this function. It has to be rewritten as a condition about t.

\begin{aligned} f(t) &\ge 0 &&\text{Assumption} \cr 80000 - 16000\, t& \ge 0 && \text{Definition of} ~ f \cr 80000 & \ge 16000\, t && \begin{aligned}&\text{Add $16000\, t$} \\[-0.5em] & \text{to both sides of the inequality}\end{aligned} \cr 5 &\ge t &&\begin{aligned}&\text{Divide both sides of} \\[-0.5em] & \text{the inequality by $16000$}\end{aligned} \cr t &\le 5 && \text{Flip the inequality}\end{aligned}.

Hence, t ≤ 5.

Combine the two inequalities to obtain the domain:

0 ≤ t ≤ 5.

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2 years ago
HELP WITH ONE MATH QUESTION!!! 20 POINTS AND WILL MARK BRAINLIEST
Jobisdone [24]

Answer:

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Step-by-step explanation:

6/h = 10/40

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3 years ago
The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili
Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{78 - 74}{2.5}

Z = 1.6

Z = 1.6 has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

Z = \frac{X - \mu}{s}

Z = \frac{77 - 74}{1.2274}

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

X = 71

Z = \frac{X - \mu}{s}

Z = \frac{71 - 74}{1.2274}

Z = -2.45

Z = -2.45 has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that n = 219, s = \frac{15}{\sqrt{219}} = 1.0136

This probability is the pvalue of Z when X = 74.2. So

Z = \frac{X - \mu}{s}

Z = \frac{74.2 - 74}{1.0136}

Z = 0.2

Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0
3 years ago
The price of a technology stock was $9.63 yesterday. Today, the price rose to $9.74. Find the percentage increase. Round your an
aev [14]
0.10 is the answer pretty sure
4 0
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