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shtirl [24]
3 years ago
8

What is 2a+50=5a-4? Thank you for your time.​

Mathematics
2 answers:
Savatey [412]3 years ago
7 0

Answer:

2a+50=5a-4

5a-2a=50+4

3a=54

a=18

kotykmax [81]3 years ago
7 0

Answer:

a=18

Step-by-step explanation:

2a+50=5a-4\\\mathrm{Subtract\:}50\mathrm{\:from\:both\:sides}\\2a+50-50=5a-4-50\\Simplify\\2a=5a-54\\\mathrm{Subtract\:}5a\mathrm{\:from\:both\:sides}\\2a-5a=5a-54-5a\\\mathrm{Simplify}\\-3a=-54\\\mathrm{Divide\:both\:sides\:by\:}-3\\\frac{-3a}{-3}=\frac{-54}{-3}\\Simplify\\\frac{-3a}{-3}=\frac{-54}{-3}\\\mathrm{Simplify\:}\frac{-3a}{-3}:\quad a\\\frac{-3a}{-3}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{3a}{3}\\\mathrm{Divide\:the\:numbers:}\:\frac{3}{3}=1

=a\\\mathrm{Simplify\:}\frac{-54}{-3}:\quad 18\\\frac{-54}{-3}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{54}{3}\\\mathrm{Divide\:the\:numbers:}\:\frac{54}{3}=18\\=18\\a=18

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Will all slices created by a slicing plane that passes through a sphere be the same size
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One special case we could consider is when x=0. If h=r, then we have a hemisphere. You may have learnt that the surface area of a sphere is 4πr2, so the answer in this case is 2πr2. Likewise, we can consider the most extreme cases: when h=2r, we get an area of 4πr2 (the entire sphere), and when h=0, we get zero area.

So if we do manage to get a general formula, we will be able to check it in these special cases.

Another possible idea when x=0 is to consider what happens when h is very small. In this case, the surface area looks a lot like a circle, but the radius of the circle does not seem particularly easy to work out, so maybe we’ll leave this one for a moment. (There are things we can do to approximate the radius using more advanced techniques, but they are currently beyond us.)

Another thing we could consider is the situation where h is very small. Then the part of the sphere between the planes looks very much like the frustum of a cone, and we know how to find the (surface) area of such a shape – see Cones.

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Step-by-step explanation:

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The graph suggests that the two lines meet at (3,6)

If this is true, that point must belong to both lines.

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3 years ago
The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%
vagabundo [1.1K]

Answer:

1) \chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313

2) p_v =P(\chi^2_{3}>16.313)=0.000978

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}

Where O rpresent the observed values and E the expected values.  

State the null and alternative hypothesis

Null hypothesis: The distribution  is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution  is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

E = \% Grand total

E_{Catfish}=500*0.3=150

E_{Bass}=500*0.15=75

E_{Bluegill}=500*0.4=200

E_{Pike}=500*0.15=75

Part 1: Calculate the statistic

\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313

\chi^2 =16.313

Calculate the critical value

First we need to calculate the degrees of freedom given by:

df= (categories-1)=(4-1)= 3

Since the confidence provided is 95% the significance would be \alpha=1-0.95=0.05 and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be \chi^2_{crit}=7.815

We can calculate also the p value:

p_v =P(\chi^2_{3}>16.313)=0.000978

And we got the same decision reject the null hypothesis at 5% of significance.

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