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JulsSmile [24]
1 year ago
15

Hi would you be able to help me ignore my work

Mathematics
1 answer:
Ulleksa [173]1 year ago
5 0

Solution:

Given the system;

\begin{gathered} 4x+6y=32.............equation1 \\  \\ 3x-6y=3..............equation2 \end{gathered}

Add equation1 to equation2, we have;

\begin{gathered} 4x+3x+6y-6y=32+3 \\  \\ 7x=35 \\  \\ \frac{7x}{7}=\frac{35}{7} \\  \\ x=5 \end{gathered}

Substitute x = 5 in equation2;

\begin{gathered} 3x-6y=3 \\  \\ 3(5)-6y=3 \\  \\ 15-6y=3 \\  \\ 6y=15-3 \\  \\ 6y=12 \\  \\ y=2 \end{gathered}

ANSWER:

x=5,y=2

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lianna [129]
The answer to your question is D
5 0
2 years ago
A painter bought 14 gallons of paint just enough to cover 2 room
OverLord2011 [107]
<h3>Answer:</h3>

7

<h3>Step-by-step explanation:</h3>

Based\ on\ the\ given\ conditions,\ formulate:

14\div 2

\frac{14}{2}

Creoss out the common factor

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6 0
1 year ago
6/(x+1)-5/2=6/(3x+3)
jolli1 [7]
If you would like to calculate 6/(x+1)-5/2=6/(3x+3), you can do this using the following steps:

6/(x+1)-5/2=6/(3x+3)
6/(x+1)-5/2=6/(3(x+1))      /*(x+1)
6 - 5/2 * (x+1) = 6/3
6 - 2 = 5/2 * (x+1)
4 = 5/2 * (x+1)     /*2/5
4 * 2/5 = x + 1
8/5 - 1 = x
x = 8/5 - 5/5 = 3/5

The correct result would be 3/5.
6 0
3 years ago
Calculate the ratio to the nearest hundredth (Please do them all for me)<br>PLEASE!
earnstyle [38]

All of them are 3.14

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8 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
2 years ago
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