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Anon25 [30]
3 years ago
7

The temperature outside is 15 degrees Fahrenheit . If the temperature drops 20 degrees , will the outside temperature be represe

nted by a positive integer or negative integer?Explain your reasoning
Mathematics
2 answers:
tamaranim1 [39]3 years ago
7 0

Answer:

The temperature would be -5 degrees Fahrenheit

Step-by-step explanation: It's represented by a negative integer because 15 - 20 = -5. This means the temperature outside would be -5 degrees Fahrenheit.

Hope this helps! (:

Andrei [34K]3 years ago
7 0
It’s -5 Cus you are taking away 15 by 20 instead of 20-15
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5%3×4+14-15 answers thes questions

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2 years ago
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Is 7 a solution of 5x - 3 = 12?
V125BC [204]

Answer:

no

Step-by-step explanation:

5x - 3 = 12

5x=12+3

x=15/5=3

x=3

substitute x with 7 to find if it is a solution :

5(7)-3=12

35-3 ≠12 ( so no 7 is not a solution)

3 0
3 years ago
It will cost Amber an estimated $3,723 a year in tuition to attend her preferred college. She has been putting $350 a month in a
kenny6666 [7]

Answer:

Yes; she will have $4008 for other expenses.

Step-by-step explanation:

First you find out how much she saved by multiplying 350 by 36 as she saved for 3 years which is 36 months.

350 \times 36 = 12600

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3723 \times 4 = 14892

If her parents add to her account half the amount she saved that means that they give her an additional 6300 as that's half of 12600.

This means she has 18900 and if you take away 14892 you will be left with 4008 for expenses.

4 0
3 years ago
Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
quester [9]
Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



5 0
3 years ago
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