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3 years ago

How do i solve (ax+bx)/3=c in terms of a and b and c

Mathematics

1 answer:

elena-s [515]3 years ago

3 0

The problem is already in terms of c.
ax+bx/3=c

ax+bx=3c
x(a+b)=3c
a+b=3c/x
a=3c/x +b or b=3c/x -a

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If the length of AO is double the length of CO, the length of BO is?

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BO is double that of DO. In other words, BO is two times longer than DO. The answer is choice B

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Side note: we can prove the triangles to be similar using the AA (angle angle) similarity theorem. The first pair of angles are the vertical angles DOC and BOA which are congruent to each other.The second pair is CDB and ABD which are congruent by the alternate interior angle theorem. Check out the attached image for a visual.

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The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie

Fittoniya [83]

Answer:

The confidence interval is $25.16 < \mu < 26.85$

Step-by-step explanation:

From the question we are given a data set

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is

$\= x = \frac{\sum x_i}{n}$

where is the sample size with values n = 10

$\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

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The standard deviation is evaluated as

$\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

$= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

$\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

$\sigma = 1.625$

The confidence level is given as 90% hence the level of significance is calculated as

$\alpha = 100 -90$

$\alpha =10$ %

$\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

$Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining the critical values of $\frac{\alpha }{2}$ instead of $\alpha$ is because we are considering two tails of the area under the normal curve