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3 years ago

Help What is the point-slope form of the equation of the line with a slope of 5/3 that goes through point (0,3)?

Mathematics

1 answer:

AnnZ [28]3 years ago

3 0

Y - 3 = 5/3 (x - 0)
this is point slope

y - 3 = 5/3x
y = 5/3x + 3

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10^6 in standard form

nalin [4]

1000000 is the answer

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3 years ago

Read 2 more answers

Triangle 2 and triangle 3 were created by drawing an altitude in triangle 1. What is the relationship between the green highligh

Naily [24]

Answer:

Triangle 1 line segment AB and triangle 2 line segment BA correspond to each other

Step-by-step explanation:

Point b and point a don't correspond to anything else

triangle 1 and 2 are congruent my answer would be a 50/50 guess between C and D If it were me id risk it and go D but it is up to you.

4 0

3 years ago

Pls help me I don’t know I just need 5 and 6 to be answered

Vlada [557]

Answer:

5. Mark had 6 laps and John has 42 laps. So you could do a sentence like: John has 42 laps and Mark has 6 laps. What is the diffrence between the two laps?

6. Im not sure about this one :/. Im truly sorry. :(. Good luck tho!

Step-by-step explanation:

6 0

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.

5 0

3 years ago

Y'all please help me! Basically just match up the points with an equation.

Len [333]

1. Y=-1/2x+1/2
2.Y=-2x-8
3.Y=-3/2x+2
4. Y=-4x-16
5.Y=-4/7x+2

4 0

3 years ago