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tatuchka [14]
2 years ago
7

1. a) if 60%% students passed in Physics, 55% pass in Math and 40% fail in both subjects, what percent of students are pass in b

oth subjects?​
Mathematics
1 answer:
hammer [34]2 years ago
5 0

Answer:

75%

Step-by-step explanation:

Percent.<em> </em><em>of students passed who in physics</em> =<em>60</em><em>%</em><em> </em><em>Percent</em><em>.</em><em> </em><em>of</em><em> </em><em>students</em><em> </em><em>who</em><em> </em><em>passed</em><em> </em><em>in</em><em> </em><em>maths</em><em> </em><em>=</em><em> </em><em>5</em><em>5</em><em>%</em>

<em>Total</em><em> </em><em>percent</em><em>.</em><em> </em><em>who</em><em> </em><em>passed</em><em> </em><em>=</em><em> </em><em>(</em><em>6</em><em>0</em><em>+</em><em>5</em><em>5</em><em>)</em><em>%</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>1</em><em>5</em><em>%</em>

<em>Percent</em><em>.</em><em> </em><em>of</em><em> </em><em>students</em><em> </em><em>who</em><em> </em><em>failed</em><em>=</em><em> </em><em>4</em><em>0</em><em>%</em>

<em>Percent</em><em>.</em><em> </em><em>of</em><em> </em><em>students</em><em> </em><em>who</em><em> </em><em>passed</em><em> </em><em>=</em><em> </em><em>(</em><em>1</em><em>1</em><em>5</em><em> </em><em>-</em><em> </em><em>4</em><em>0</em><em>)</em><em>%</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>7</em><em>5</em><em>%</em><em>.</em><em> </em>

<em>please</em><em> </em><em>mark</em><em> </em><em>me</em><em> </em><em>as</em><em> </em><em>brainl</em><em>iest</em><em> </em><em>,</em><em> </em><em>pleaseeeeeeeeee</em><em /><em>please</em><em> </em>

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In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders o
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A. We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B. Null hypothesis:p=0.1  

Alternative hypothesis:p \neq 0.1  

C. z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558  

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E. Fail to the reject the null hypothesis

F. So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.  

Step-by-step explanation:

Data given and notation

n=362 represent the random sample taken

X=33 represent the number of orders not accurate

\hat p=\frac{33}{363}=0.0912 estimated proportion of orders not accurate

p_o=0.10 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

A: Write the claim as a mathematical statement involving the population proportion p

We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B: State the null (H0) and alternative (H1) hypotheses

Null hypothesis:p=0.1  

Alternative hypothesis:p \neq 0.1  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

C: Find the test statistic

Since we have all the info required we can replace in formula (1) like this:  

z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558  

D: Find the critical value(s)

Since is a bilateral test we have two critical values. We need to look on the normal standard distribution a quantile that accumulates 0.025 of the area on each tail. And for this case we have:

z_{\alpha/2}=-1.96  z_{1-\alpha/2}=1.96

P value

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

E: Would you Reject or Fail to Reject the null (H0) hypothesis.

Fail to the reject the null hypothesis

F: Write the conclusion of the test.

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.  

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Step-by-step explanation:

We have,

Rectangle ABCD with co-ordinates A(-4,2), B(-4,1), C(-1,1) and D(-1,2).

It is transformed to a new rectangle A'B'C'D' with co-ordinates A'(-4,-2), B'(-4,-1), C'(-1,-1) and D'(-1,-2).

The graph of both the triangles is shown below.

we see that,

The rectangle ABCD is reflected about y-axis and then rotated 180° to obtain A'B'C'D'.

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A= (-4,2)                     (4,2)                                              A'= (-4,-2)

B= (-4,1)                      (4,1)                                                B'= (-4,-1)

C= (-1,1)                       (1,1)                                                 C'= (-1,-1)

D= (-1,2)                      (1,2)                                                 D'= (-1,-2)

Hence, the second rectangle is formed by: Reflection over the y-axis and rotation of 180°.

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