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myrzilka [38]
4 years ago
9

X/5=2/3=5/y 2/3 4/9 1

Mathematics
1 answer:
Mademuasel [1]4 years ago
5 0
\frac{x}{5}= \frac{2}{3}  = \frac{5}{y}
Break the equation into parts;
\frac{x}{5} = \frac{2}{3} ;  \frac{2}{3}= \frac{5}{y}
Solving for x;
\frac{x}{5} = \frac{2}{3}
x=\frac{10}{3}
Solving for y;
\frac{2}{3} = \frac{5}{y} 
y= \frac{3*5}{2} 
y= \frac{15}{2}
\frac{x}{y} = \frac{10/3}{15/2}
\frac{x}{y} = \frac{4}{15}
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Read 2 more answers
In an experiment to determine whether there is a systematic difference between the weights obtained with two different mass bala
Maslowich

Answer:

H0: μd=0 Ha: μd≠0

t= 0.07607

On the basis of this we conclude that the mean weight differs between the two balances.

Step-by-step explanation:

The null and alternative hypotheses as

H0: μd=0 Ha: μd≠0

Significance level is set at ∝= 0.05

The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Specimen        A               B           d = a - b         d²

1                     13.76        13.74         0.02           0.004

2                    12.47        12.45          0.02         0.004

3                    10.09        10.08           0.01        0.001

4                       8.91       8.92          -0.01          0.001

5                     13.57      13.54           0.03        0.009

<u>6                     12.74      12.75          -0.01        0.001</u>

<u>∑                                                      0.06         0.0173</u>

d`= ∑d/n= 0.006/6= 0.001

sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882

sd= 0.05368

t= 0.001/ 0.05368/ √6

t= 0.18629/2.449

t= 0.07607

Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.

3 0
3 years ago
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