(50 POINTS FOR THE FIRST ANSWER)

20(x + 1) = 200 solve for x and explain in number form

omeli [17]

Answer:

X = 9

Step-by-step explanation:

Write Equation: 20(x+1) = 200

Distribute 20: 20x +20 =200

Subtract 20: 20x = 180

Divide by 20x to leave x by itself: X=9

<em>Brainliest?</em>

6 0

3 years ago

Read 2 more answers

6c^3 +30c / 3c^2 + 15

mario62 [17]

Hello there,

$6c^{3} +30c/ 3c^{2} +15 = 16c^{3}+15$

Hope this answer has helped you. If it has, Please help me out by marking it as brainliest. Thanks

Have a BRAINLY day.

7 0

3 years ago

What is the probability that a random sample of 1414 second grade students from the city results in a mean reading rate of more

Sonbull [250]

Answer with explanation:

Number of second grade students who are considered for sample =1414

Number of words that is being read in a Minute = 9292 words per minute

If number of students will be more than half of 1414, that is if, they can read or their reading rate is more than 9292 words per minute, then we can say that the sample chosen is Appropriate.

⇒Required Probability,that sample of 1414 students can read 9292 words per minute

$=\frac{707+x}{1414} \text{Where }, [1 \leq x\leq 707]$

5 0

3 years ago

What is 8(9h - 6) simplifyed

wariber [46]

Answer:

72h - 48

Step-by-step explanation:

Use distributive property

8(9h - 6)

Multiply 8 out of 9h - 6

72h - 48

7 0

3 years ago

In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and de

Naily [24]

Answer:

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

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

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:

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

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.

5 0

3 years ago