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3 years ago

Is 10/3 irrational or rational

Mathematics

1 answer:

Masteriza [31]3 years ago

7 0

It's rational

Rational numbers are numbers that you can put into the form of a/b.

You might be interested in

Which of the following shows the length of the third side, in inches, of the triangle below? (1 point)

GrogVix [38]

Answer:

2, sqrt of 7633

Step-by-step explanation:

for now, we'll leave the 3rd side as x

y using the Pythagorean theorem, we can conclude that (1) 48^2+x^2=73^2 or (2) 48^2+73^2=x^2.

1. If we conclude the x isn't the longest, the x would be 3025 which is the sqrt of 55, now is invalid because the 2 smallest sides can't be more than the longest side when they add up 55+48>73 so that is invalid

2. Then, it leaves us with only 1 choice left! concluding that the 3rd side is the longest. Which the answer is sqrt of 7633

6 0

3 years ago

Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11

Bond [772]

Answer:

a) Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

b) $t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

c) $12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

Step-by-step explanation:

Information given

$\bar X=12.50$ represent the mean for the daily iron intake

$s=4.75$ represent the sample deviation

$n=51$ sample size

$\mu_o =14.44$ represent the reference value

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Part a

We want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:

Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

Part b

Since we don't know the population deviation the statistic would be given by:

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

Replacing the info we got

$t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

Part c

The confidence interval would be given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$