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3 years ago

You can use a common denominator and rewrite 3/5 divided by 1/10 as _ (divison sign) _

Mathematics

2 answers:

goldenfox [79]3 years ago

6 0

Answer:

(6/10) / (1/10)

Step-by-step explanation:

you just multiply 3 by 2 and 5 by 2

Ulleksa [173]3 years ago

5 0

The common denominator is 10 and 3/5 is equivalent to 6/10. So 6/10 divided by 1/10 would 60/10 which is 6/1. The answer is 6.

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Study the graph below. Describe the functional relationship between Quantity A and Quantity B. What happens as Quantity A increa

Alex

There is an inverse corollation between the values of A and B. Specifically, a 3 unit change increase in A results in a 4 unit decrease in B

4 0

3 years ago

I cannot figure this out. Any help would be appreciated.

Lesechka [4]

So you are finding M
90+x+15+3x+15=180
4x=60
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LMN = 30
hopefully this is correct :)

7 0

3 years ago

A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading

valkas [14]

Answer:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

$p_v =2*P(t_{11}$

Step-by-step explanation:

1) Data given and notation

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

The results obtained are:

$\bar X=98.375$ represent the sample mean

$s=0.6.109$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =100$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

4) P-value

First we need to find the degrees of freedom for the statistic given by:

$df=n-1=12-1=11$

Since is a two sided test the p value would given by:

$p_v =2*P(t_{11}$

5) Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.