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3 years ago

Given the two expressions shown below:

1 answer:

3 0

$\sqrt3\ is\ irrational\\\\\sqrt2\ is\ irrational\\\\\sqrt6\ is\ irrational\\\\therefore\\\\A.\ \sqrt3+\sqrt2\ is\ irrational\ and\ B.\ \sqrt3+\sqrt6\ is\ irrational$
Answer: Both are irrational.

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7×{(185-36÷3)-[(28÷4)×(28÷4)-7×2])

Paladinen [302]

7x((185-36/3)-
(28/4)x (28/4)-(7x2)
7x173-(49-14)
=1211-63
=1148

8 0

3 years ago

What is the midpoint of the longest side of the triangle with vertices (1, 4), (3, 4),

JulijaS [17]

Answer:

Step-by-step explanation:

midpoint x= 2

midpoint y=5

Midpoint= (2, 5)

6 0

3 years ago

In a class there are 75 students if there are 27 male students then the ratio of the number of female students to the total numb

seropon [69]

Answer:

48:75

Step-by-step explanation:

the 48 is the number of females

5 0

3 years ago

A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the

Rasek [7]

Answer:

$\mu_{p} = p = 0.06$

And the standard error is given by:

$SE_{p} = \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123$

And we want to find this probability:

$P(\hat p < 0.03)$

We can calculate the z score for this case and we got:

$z = \frac{\hat p -\mu_p}{\Se_p} = \frac{0.03-0.06}{0.0123}= -2.440$

And using the normal distribution table or excel we got:

$P(\hat p < 0.03)= P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

For this case we can find the mean and standard error for the sample proportion with these formulas:

$\mu_{p} = p = 0.06$

And the standard error is given by:

$SE_{p} = \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123$

And we want to find this probability:

$P(\hat p < 0.03)$

We can calculate the z score for this case and we got:

$z = \frac{\hat p -\mu_p}{\SE_p} = \frac{0.03-0.06}{0.0123}= -2.440$

And using the normal distribution table or excel we got:

$P(\hat p < 0.03)= P(Z$

8 0

4 years ago

This worth 100 point

wlad13 [49]

Answer:

cap

Step-by-step explanation:

5 0

3 years ago