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Hunter-Best [27]
2 years ago
10

A car is traveling with a velocity of 40 m/s and has a mass of 1120kg. The car has __________ energy. Calculate it.

Mathematics
1 answer:
prohojiy [21]2 years ago
6 0

Answer:

Kinetic Energy

=

1

2

m

v

2

=

1

2

1120

⋅

40

2

=

896000

J

=

8.96

⋅

10

5

J

Step-by-step explanation:

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2 years ago
On Friday, a fruit stand sold a total of 92 apples and oranges. On Saturday, the fruit stand sold four times the number of apple
andreev551 [17]

Option A : a+o = 92 , 4a+2o = 284 is the right answer

Step-by-step explanation:

Let o be the number of oranges sold on Friday and

a be the number of apples sold on Friday

So,

A total of 92 was sold on Friday will yield the equation

a+o = 92

On Saturday,

4 times the number of apples = 4a

2 times the number of oranges = 2o

So,

4a+2o = 284

Hence,

The system of equations is:

a+o = 92\\4a+2o = 284

Option A : a+o = 92 , 4a+2o = 284 is the right answer

Keywords: Linear equations, variables

Learn more about linear equations at:

  • brainly.com/question/669100
  • brainly.com/question/663692

#LearnwithBrainly

5 0
3 years ago
At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul
vampirchik [111]

Answer:

a) Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b) The 95% confidence interval would be given by (910.05;959.95)    

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b. What is the 95% confidence interval estimate of the population mean examination  score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

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".

\bar X=935 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=180 represent the population standard deviation

n=200 represent the sample size  

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=3278.222

The sample deviation calculated s=97.054

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

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

935-1.96\frac{180}{\sqrt{200}}=910.05    

935+1.96\frac{180}{\sqrt{200}}=959.95    

So on this case the 95% confidence interval would be given by (910.05;959.95)    

c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your  conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we replace we got:

z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

So then since the p value is less than the significance we can reject the null hypothesis at 5% of significance.

8 0
3 years ago
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