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Stella [2.4K]
2 years ago
7

sean sold 4 more boxes of candy for the school fundraiser than Chris. The sum of the boxes that they sold was 22

Mathematics
2 answers:
sweet-ann [11.9K]2 years ago
5 0

If it was asking for the amount that Chris made, the answer would be 18.

Hope this helps!<3

Sveta_85 [38]2 years ago
3 0

Set up an equation.

x+x+4=22

the varibles x are the unknown number of boxes. the first x is Chris amount and the second is Seans amount.

Isolate the variable:

x+x+4=22

2x+4=22

2x=18

x=9

Chris has 9 boxes.

Add 4 to the variable to get Seans amount:

9+4=13

Sean has 13 boxes

To check your answer add the results together to get the correct sum.

9+13=22

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3 years ago
Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m &lt; 20
Vsevolod [243]

Answer:

<em>18</em> values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of n are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

Ax^{2} +Bx+C

and the roots are: \alpha and \beta

Then sum of roots, \alpha+\beta = -\frac{B}{A}

Product of roots, \alpha \beta = \frac{C}{A}

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots,  \alpha+\beta = -\frac{-m}{1} = m

Product of roots, \alpha \beta = \frac{n}{1} = n

We are given that m  

\alpha and \beta are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of n (= \alpha \times \beta) are:

1.\ 2,  2\Rightarrow  n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow  n = 6\\3.\ 2, 5 \Rightarrow  n = 10\\4.\ 2,  7\Rightarrow  n = 14\\5.\ 2, 11 \Rightarrow  n = 22\\6.\ 2, 13 \Rightarrow  n = 26\\7.\ 2, 17 \Rightarrow  n = 34\\8.\ 3,  3\Rightarrow  n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow  n = 15\\10.\ 3, 7 \Rightarrow  n = 21\\

11.\ 3,  11\Rightarrow  n = 33\\12.\ 3, 13 \Rightarrow  n = 39\\13.\ 5, 5 \Rightarrow  n = 25\\14.\ 5, 7 \Rightarrow  n = 35\\15.\ 5, 11 \Rightarrow  n = 55\\16.\ 5, 13 \Rightarrow  n = 65\\17.\ 7, 7 \Rightarrow  n = 49\\18.\ 7, 11 \Rightarrow  n = 77

So,as shown above <em>18 values for n are possible.</em>

3 0
3 years ago
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
melomori [17]

Answer:

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  

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

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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