Answer will give brainliest

1. A 5 kg box is lifted from the floor to a height of 1.5 m. How much work was done in this proces

iren [92.7K]

Answer:

Work done = 73.575 J

Step-by-step explanation:

Given,

Mass (m) = 5 kg

Height (h) = 1.5 m

Gravity (g) = 9.81 m/s²

Work done = ?

Work done

= mgh

= 5 kg × 9.81 m/s² × 1.5 m

= 73.575 joule (J)

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4 0

2 years ago

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A classic rock radio station claims to play an average of 50 minutes of music every hour. However, it seems that every time you

tia_tia [17]

Answer:

a) $47.917-1.796\frac{2.811}{\sqrt{12}}=46.460$

$47.917+1.796\frac{2.811}{\sqrt{12}}=49.374$

b) Null hypothesis: $\mu =50$

Alternative hypothesis: $\mu \neq 50$

And after analyze the interval obtained the upper limit not contains the value of 50, so then we can conclude that the claim is not satisfied for this case, and the population mean seems to be lower than 50 at 10% of significance.

Step-by-step explanation:

Previous concepts

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".

We have the following data: 44 49 45 51 49 53 49 44 47 50 46 48

And we can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And replacing we got:

$\bar X= 47.917, s= 2.811$

$\mu$ population mean (variable of interest)

n=12 represent the sample size

Part a: Calculate the confidence interval

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given:

$df= n-1= 12-1=11$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,11)".And we see that $t_{\alpha/2}=1.796$

Now we have everything in order to replace into formula (1):

$47.917-1.796\frac{2.811}{\sqrt{12}}=46.460$

$47.917+1.796\frac{2.811}{\sqrt{12}}=49.374$

Part b

For this case the system of hypothesis are:

Null hypothesis: $\mu =50$

Alternative hypothesis: $\mu \neq 50$

And after analyze the interval obtained the upper limit not contains the value of 50, so then we can conclude that the claim is not satisfied for this case, and the population mean seems to be lower than 50 at 10% of significance.

6 0

3 years ago

one half multiplyed by 4 pluse one quarter times 4 pluse three quarters times 4 pluse three 8s times. 3 pluse 5 8s times 3

elena-14-01-66 [18.8K]

.5 times four is 3 plus one quarter is 2.33 times 4 is 9.32 plus 75 is 10.07plus 24 is 24.07 then 32 and answer is 56.07

4 0

3 years ago

100 points!

GenaCL600 [577]

Answer: If i was living there usually , The climate in Ancient Greece generally has hot summers and mild winters. Because it can get so hot, most people wore lightweight clothing throughout most of the year. They would put on a cloak or wrap during the colder days of the winter months. The environment would make a HUGE impact on the Greeks by teaching them how to collect food and survive in a harsh environment. Therefore, I would have to adapt to adapt to their environment efficiently. The environment also affected them because they had to learn to fish instead of hunt on

Step-by-step explanation: This is how I would do it.

6 0

3 years ago

A movie theater charges $6.50 for matinee showings and $8.75 for evening showings. Yesterday the theater sold 378 tickets for a

harina [27]

The number of matinee tickets sold would be 168

6 0

3 years ago

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