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3 years ago

Someone please solve this from 1 to 10

Mathematics

1 answer:

gladu [14]3 years ago

8 0

4x - 6 = x + 9
4x - x = 9 + 6
3x = 15
x = 15/3
x = 5
==============
4 - 7x = 1 - 6x
4 - 1 = -6x + 7x
3 = x
==============
-4x - 3 = -6x + 9
-4x + 6x = 9 + 3
2x = 12
x = 12/2
x = 6
==============
41 - 2x = 2 + x
41 - 2 = x + 2x
39 = 3x
39/3 = x
13 = x
===============
6(2 + y) = 3(3-y)
12 + 6y = 9 - 3y
6y + 3y = 9 - 12
9y = - 3
y = -3/9
y = -1/3
==============
4x = 2(x - 5) - 2
4x = 2x - 10 - 2
4x - 2x = -10 - 2
2x = -12
x = -12/2
x = -6
===============
6x - 9x - 4 = -2x - 2
-3x - 4 = -2x - 2
-4 + 2 = -2x + 3x
-2 = x
==============
-2(2x + 3) = -4(x + 1) - 2
-4x - 6 = -4x - 4 - 2
-4x - 6 = -4x - 6
infinite solutions
================
3 + 6x = 9 - 6x
3 - 9 = -6x - 6x
-6 = -12x
-6/-12 = x
1/2 = x
==============
-9x + 6 = -x + 4
6 - 4 = -x + 9x
2 = 8x
2/8 = x
1/4 = x

I hope I read these correctly....not that easy...u got weird looking x's and ur 4's can almost pass for 6's

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$z=\frac{0.0156-0.0114}{\sqrt{0.01554(1-0.01554)(\frac{1}{32601}+\frac{1}{88})}}=0.318$

$p_v =P(Z>0.318)=0.375$

So the p value is a very high value and using the significance level provided $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona is not significantly higher than the national incident rate at 1% of significance.

Step-by-step explanation:

Data given and notation

$X_{1}=507$ represent the number of children diagnosed with Autism Spectrum Disorder (ASD) in Arizona

$n_{1}=32601$ sample elected from arizona

$\hat p_{1}=\frac{507}{32601} =0.0156$ represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona

$p_{2}=\frac{1}{88}= 0.0114$ represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in the nation

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if that the incident of ASD is more in Arizona than nationally, the system of hypothesis are:

Null hypothesis: $p_{1} \leq p_{2}$

Alternative hypothesis: $p_{1} > p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{507+1}{32601 +88}=0.01554$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.0156-0.0114}{\sqrt{0.01554(1-0.01554)(\frac{1}{32601}+\frac{1}{88})}}=0.318$

Statistical decision

The significance level provided is $\alpha=0.01$ , and we can calculate the p value for this test.

Since is a one right sided test the p value would be:

$p_v =P(Z>0.318)=0.375$

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