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3 years ago

Plsss help due in 5 minutes

Mathematics

1 answer:

77julia77 [94]3 years ago

7 0

Answer:

2 sqrt41

Step-by-step explanation:

(12^2)+(sqrt20)= (3rd side)

144+20=164

sqrt 164=2sqrt41

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In mathematics, the nth harmonic number is defined to be 1 + 1/2 + 1/3 + 1/4 + ... + 1/n. so, the first harmonic number is 1, th

WITCHER [35]

(1.0 + 1.0/2.0 + 1.0/3.0 + 1.0/4.0 + 1.0/5.0 + 1.0/6.0 + 1.0/7.0 + 1.0/8.0)

3 0

4 years ago

I don't get this. I could really use help.

olga nikolaevna [1]

The y intercept is what the y value is when x is 0.

Set X to 0 and solve:

-2/9(0) + 1/3

0 + 1/3 = 1/3

The y-intercept is 1/3

6 0

3 years ago

How to to this? Can you please help? 1.A

eimsori [14]

First you need to expand them
(x^2 + 9) + (2x + 6)
then you add them together

x^2 + 2x + 15
now you can factorise it

i hope this helped you

8 0

4 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

We used Total - 1 because it is a probability without replacement

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

3 years ago

Researchers are studying rates of homeowners in a certain town. They believe that the proportion of people ages 36-50 who own ho

Ira Lisetskai [31]

Answer:

Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:

$H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0$

Step-by-step explanation:

A hypothesis test on the difference of proportions needs to be performed for this case.

We have two sample proportions and we want to test if the true population proportions differ from each other, usign the information given by the sample statistics.

The claim is that the proportion of people of ages 36-50 who own homes is significantly greater than the proportin of people age 21-35 who own homes.

The term "higher" will define the alternative hypothesis, that is the hypothesis that represents what is claimed. The null hypothesis always include the equal sign, and will state that both proportions do not differ.

Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:

$H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0$

6 0

4 years ago