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3 years ago

Choose the ratio equivalent to 10/9

Mathematics

1 answer:

satela [25.4K]3 years ago

3 0

<h3>Answer:</h3>

30/27, 20/18...

<h3>Solution:</h3>

There are <em>infinitely many ratios equivalent to 10/9.</em>
Here are <em>some </em>of them:
30/27
20/18
100/90
60/54

Hope it helps.

Do comment if you have any query.

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A Statistics exam is created by choosing for each question on the exam one possible version at random from a bank of possible ve

Oduvanchick [21]

Answer: E (c *) = 13.77766

Step-by-step explanation:

the significance level of 0.05 is known

seeks to calculate the sample for the population medis using the hypothesis

H0: µ = 15

H0: µ 15

sample size n = 35

Degrees of freedom df = n-1 = 35-1 = 34

The critical value for z is -1.645

The critical value for t is -1.691

For \ sigma = 3,

Standard error = 3 /√ 35 = 0.5070926

When obtaining the results of the population standard deviation, we will use the z-score to estimate the critical value.

ex = 15-1645 * 0.5070926 = 14.16583

For s = 4.2,

Standard error = 4.2 / √ 35 = 0.7099296

When we do not know what the population standard deviation is, we can use the statistical t to obtain the critical value

c * = 15-1.691 * 0.7099296 = 13.79951

for s = 5.7,

Standard error = 5.7 / √ 3.5 = 0.9634759

c * = 15-1.691 * 0.9634759 = 13.37076

It is obtained that c * is the critical value for the rejection region in this question

P (c * = 14.16583) = 7/20

P (c * = 13.79951) = 6/20

P (c * = 13.37076) = 1-7 / 20-6 / 20

= 7/20

E (c *) = (7/20) * 14,16583 + (6/20) * 13.79951 + (7/20) * 13.37076

Outcome:

E (c *) = 13.77766

5 0

3 years ago

A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. T

CaHeK987 [17]

Answer:

We need to conduct a hypothesis in order to check if the mean is higher than 75, the system of hypothesis are :

Null hypothesis: $\mu \leq 75$

Alternative hypothesis: $\mu > 75$

A. One-Tailed

$z=\frac{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189$

$p_v =P(z>2.189)=0.0143$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

We have the following data: 84 94 80 88 77 68 90 74 96 71

We can calculate the sample mean with the following formula:

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

And replacing we got:

$\bar X=82.2$ represent the sample mean

$\sigma=10.4$ represent the population standard deviation

$n=10$ sample size

$\mu_o =75$ represent the value that we want to test

$\alpha=0.01$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 75, the system of hypothesis are :

Null hypothesis: $\mu \leq 75$

Alternative hypothesis: $\mu > 75$

A. One-Tailed

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

Calculate the statistic

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

$z=\frac{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189$

P-value

Since is a ONE-TAILED test the p value would given by:

$p_v =P(z>2.189)=0.0143$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

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3 years ago

How many times will 6 go into 1155 and what is left over?

GarryVolchara [31]

192 times with .5 left over

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4 years ago

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