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Lemur [1.5K]
3 years ago
14

I need help, The temperature was 2°F below zero. The temperature drops by 5°F. What is the temperature now? What is the initial

temperature written as an integer?
Mathematics
2 answers:
wariber [46]3 years ago
7 0
-7 degres is the awnser
kolezko [41]3 years ago
3 0
-7 degrees Fahrenheit
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Which graph correctly represents 5/2x-y< 3
kompoz [17]
5/2x - y < 3
Add y
5/2x < 3 + y
Subtract 3
5/2x - 3 < y
y > 5/2x - 3
Graph "5/2x - 3" and shade the area that is above it
5 0
3 years ago
A rental car company has noticed that the distribution of the number of miles customers put on rental cars per day is skewed to
kodGreya [7K]

Using the Central Limit Theorem, the correct option is:

(c) Average number of miles put on a rental car per day across 25 customers.

--------------------------

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

  • The interpretation related to this problem is that the larger the sample size, the smaller the standard deviation.
  • Thus, among the options, the largest sample is 25, thus, option c will have the smallest standard deviation.

A similar problem is given at brainly.com/question/23088374

5 0
2 years ago
Solve the question will mark branliest and with an explantion
erma4kov [3.2K]

Answer:

The car loses $5,635 every year. Multiply 24,500 by 23% (0.23)

5 0
3 years ago
The asset turnover of Ryan Company is 7.2. The total assets of Ryan are $88,000. Ryan's net sales were:
nikitadnepr [17]
Let A be asset turnover, R revenue, and S assets. You already know the equation for asset turnover:
A = R/S

Since we need to calculate revenue, we need to modify the equation a bit:
R = A*S
R = 7.2*88000
R = 633,600

This means that Ryan's net sales (revenue) were $633,600. 
5 0
3 years ago
A company with a large fleet of cars wants to study the gasoline usage. They check the gasoline usage for 50 company trips chose
Nookie1986 [14]

Answer:

The 95% confidence interval is given by (25.71536 ;28.32464)

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

Step-by-step explanation:

Notation and definitions  

n=50 represent the sample size  

\bar X= 27.2 represent the sample mean  

s=5.83 represent the sample standard deviation  

m represent the margin of error  

Confidence =88% or 0.88

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".  

Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 88% of confidence, our significance level would be given by \alpha=1-0.88=0.12 and \alpha/2 =0.06. The degrees of freedom are given by:  

df=n-1=50-1=49  

We can find the critical values in R using the following formulas:  

qt(0.06,49)

[1] -1.582366

qt(1-0.06,49)

[1] 1.582366

The critical value tc=\pm 1.582366  

Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

m=t_c \frac{s}{\sqrt{n}}  

m=1.582366 \frac{5.83}{\sqrt{50}}=14.613  

With R we can do this:

m=1.582366*(5.83/sqrt(50))

m

[1] 1.304639

Calculate the confidence interval  

The interval for the mean is given by this formula:  

\bar X \pm t_{c} \frac{s}{\sqrt{n}}  

And calculating the limits we got:  

27.02 - 1.582366 \frac{5.83}{\sqrt{50}}=25.715  

27.02 + 1.582366 \frac{5.83}{\sqrt{50}}=28.325

Using R the code is:

lower=27.02-m;lower

[1] 25.71536

upper=27.02+m;upper

[1] 28.32464

The 95% confidence interval is given by (25.71536 ;28.32464)  

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

6 0
3 years ago
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