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RideAnS [48]
3 years ago
7

The temperature dropped from 6 Celsius to -5 Celsius.How many degrees did the temperature drop?

Mathematics
2 answers:
kakasveta [241]3 years ago
7 0
-11 degrees. 6+5=11, and it went down, so negative 11 degrees

zlopas [31]3 years ago
6 0
Hello

To solve this do the following
x-6=-5
x=1
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boyakko [2]
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A student group claims that first-year students at a university must study 2.5 hours (150 minutes) per night during the school w
AlekseyPX

Answer:

We have strong evidence that on average, students study less than 150 minutes per night during the school week

Step-by-step explanation:

Normal distribution:

mean     μ₀ = 150

Sample:

Sample size    n = 272

Sample mean   x = 141

Sample standard deviation  s  = 66

The standard error of the sample mean  SE = σ /√n

SE = 66/√272

SE = 66 / 16,49

SE = 4

Test Hypothesis:

Null hypothesis                            H₀             x  =   μ₀

Alternative hypothesis                Hₐ            x <    μ₀

z(s)  test statistics is:

z(s)  =  ( x  -  μ₀ ) / s/√n

z(s) = - 9 /4

z(s) =  -  2,25

p-value  for that z(s)      p-value  = 0,0122

Then for α =  0,05 p-value < 0,05

We are in the rejection region we need to reject H₀

7 0
3 years ago
Do the lines y=5x+2 and y=5x+4 intersect? if so, at what point?<br><br> Thanks for your help!!
strojnjashka [21]
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Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.
Mkey [24]

Answer:

1.~e^{-2} \\2.~e^{\frac{7}{2}}\\3.~e^8\\4.~e^{\frac{-11}{2}}

Step-by-step explanation:

We have to simplify the given exponential exponents.

Exponential Properties:

e^0 =1\\e^a.e^b = e^{a+b}\\\\\displaystyle\frac{e^a}{e^b} = e^{a-b}\\\\(e^a)^b = e^{ab}\\\\e^{-a} = \frac{1}{e^a}

Simplification takes place in the following manner:

a)

(e^{-3})^\frac{2}{3}\\(e^a)^b = e^{ab}\\=e^{-3\times \frac{2}{3}}\\=e^{-2}

b)

(e^4)(e^{\frac{-1}{2}})\\e^a.e^b = e^{a+b}\\=e^{(4+\frac{-1}{2})} \\= e^{\frac{7}{2}}

c)

(e^{-2})^{-4}\\(e^a)^b = e^{ab}\\= e^{-2\times -4}\\=e^8

d)

(e^{-4})(e^{\frac{-3}{2}})\\e^a.e^b = e^{a+b}\\=e^{(-4+\frac{-3}{2})} \\= e^{\frac{-11}{2}}

4 0
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