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Tema [17]
3 years ago
6

(10 points!!!) AND I’LL MAKE YOU BRAINLEST

Mathematics
1 answer:
Ket [755]3 years ago
7 0

Answer:Jose will make 901.31 more

Step-by-step explana

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Hat is the interquartile range of the data represented by the box plot shown below?
marissa [1.9K]

Answer:

Where is the box plot?

Step-by-step explanation:

3 0
4 years ago
Factor this epression 3x^2 - 15x
nika2105 [10]

Answer:

x = 0 and x = 5

Step-by-step explanation:

3x^2 - 15x. the first thing that you do is find a common factor for both of the numbers. The common factor is 3x. so the equation would now be 3x(x-5) = 0. So now all you have to do is plug in for x and make it equal 0.

6 0
3 years ago
A yodeler ascends a mountain to its peak to do some sweet yodeling. The peak is 5,800 ft above sea level. The yodeler then desce
Vesna [10]
If the yodeler is at 5800 ft above sea level.....and he descends 200 ft....he will then be at (5800 - 200) = 5600 ft above sea level where he meets his friend
6 0
4 years ago
Anyone know what the answer is???(sorry if it’s unclear)
Sergio [31]

The laws of exponents I will be using are:

(x^y)^z = x^{yz}, and x^y \cdot x^z = x^{y+z}.

For the first part, (15^3)^3 \cdot 15^{-6} = 15^9 \cdot 15^{-6} = 15^3.

For the second part, 15 \cdot (15^3)^2 \cdot (15^{-5})^2 = 15 \cdot (15^6) \cdot (15^{-10}) = 15^{1+6-10} = 15^{-3} = \frac{1}{15^3}.

5 0
3 years ago
The authors of a paper presented a correlation analysis to investigate the relationship between maximal lactate level x and musc
ArbitrLikvidat [17]

Answer:

a) Sample correlation coefficient, r = 0.7411

bi) test statistic, t = 4.102

bii) P-value = 0.000736

Step-by-step explanation:

a) The formula for the sample correlation coefficient is given by the formula:

r = \frac{S_{xy} }{\sqrt{S_{xx} S_{yy} }} }

S_{xx} = 2,648,130.357\\S_{yy} = 36.7376,\\S_{xy} = 7408.225

r = \frac{7408.225}{\sqrt{2648130.357*36.7376} }

r = 0.7511

b)

i) formula for the test statistic is given by the formula:

t = \frac{r\sqrt{n-1} }{\sqrt{1 - r^{2} } }

sample size, n = 4

t = \frac{0.7511\sqrt{14-1} }{\sqrt{1 - 0.7511^{2} } }

t = 4.102

ii) Degree of freedom, df = n -2

df = 14 -2

df = 12

The P-value is calculate from the degree of freedom and the test statistic using excel

P-value =(=TDIST(t,df,tail))

P-value = (=TDIST(4.1,12,1)

P-value = 0.000736

4 0
3 years ago
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